case study based questions class 9 physics work and energy

Case Study Questions Class 9 Science Work and Energy

Case study questions class 9 science chapter 11 work and energy.

CBSE Class 9 Case Study Questions Science Work and Energy. Important Case Study Questions for Class 9 Exam. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Work and Energy.

CBSE Case Study Questions Class 9 Science – Work and Energy

(i) Work done is

(b) Vector quantity

(a) positive

(b) Newton meter(N-m)

(2) A moving object can do work. An object moving faster can do more work than an identical object moving relatively slow. A moving bullet, blowing wind, a rotating wheel, a speeding stone can do work. How does a bullet pierce the target? How does the wind move the blades of a windmill? Objects in motion possess energy. We call this energy kinetic energy.

The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the same as that of work, that is, joule (J).

(b) Kinetic energy and work

to that point against gravity.Let the work done on the object against gravity beW. That is,

= 5 ×9.8 ×10

(a) Electric energy converted into heat energy

(ii) Total energy consumed divided by total time taken is called as

Time used, t = 5 h

Class 9 Science Case Study Questions Chapter 11 Work and Energy

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Case study Questions in Class 9 Science Chapter 11 are very important to solve for your exam. Class 9 Science Chapter 11 Class 9 Science Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 11 Work and Energy

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In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Work and Energy Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 11 Work and Energy

Case Study/Passage-Based Questions

Case Study 1: The figure shows a watch glass embedded in clay. A tiny spherical ball is placed at edge B at a height h above the center A

case study based questions class 9 physics work and energy

The kinetic energy of the ball, when it reaches point A is (a) zero (b) maximum (c) minimum (d) can’t say

Answer: (b) maximum

The ball comes to rest because of (a) frictional force (b) gravitational force (c) both (a) and (b) (d) none of these

Answer: (c) both (a) and (b)

The energy possessed by the ball at point C is (a) potential energy (b) kinetic energy (c) both potential and kinetic energy (d) heat energy.

Answer: (a) potential energy

Case Study 2: The principle of conservation of energy states that the energy in a system can neither be created nor be destroyed. It can only be transformed from one form to another, but the total energy of the system remains constant. Conservation of electrical energy to various forms or vice versa along with devices is illustrated in the figure given below.

Water stored in a dam possesses (a) no energy (b) electrical energy (c) kinetic energy (d) potential energy.

Answer: (d) potential energy.

A battery lights a bulb. Describe the energy changes involved in the process. (a) Chemical energy →Light energy → Electrical energy (b) Electrical energy → Chemical energy → Electrical energy (c) Chemical energy → Electrical energy → Light energy (d) None of these.

Answer: (c) Chemical energy → Electrical energy → Light energy

Name a machine that transforms muscular energy into useful mechanical work. (a) A microphone (b) Bicycle (c) Electric torch (d) An electric bell

Answer: (b) Bicycle

A body is falling from a height of h. After it has fallen a height h/2 , it will possess (a) only potential energy (b) only kinetic energy (c) half potential and half kinetic energy (d) more kinetic and less potential energy.

Answer: (c) half potential and half kinetic energy

Case Study 3: An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.4 m s –1 . For this purpose, a motor with the required horsepower is used

The power of motor in hp is (a) 2.33 (b) 2.43 (c) 2.53 (d) 2.63

Answer: (d) 2.63

Case Study 4: Work and energy are fundamental concepts in physics that help us understand the physical world and the processes happening around us. Work is done when a force is applied to an object, and the object moves in the direction of the applied force. It is calculated as the product of force and displacement. The unit of work is joule (J). Energy, on the other hand, is the ability to do work. It exists in different forms, such as kinetic energy, potential energy, and various other forms like thermal energy, electrical energy, and chemical energy. The law of conservation of energy states that energy cannot be created or destroyed, but it can be transformed from one form to another. Understanding the concepts of work and energy helps us analyze the efficiency of machines, calculate the amount of work done, and comprehend various physical phenomena.

When is work considered to be done on an object? a) When a force is applied to the object b) When the object moves in the direction of the applied force c) When the object remains stationary d) When the object changes its shape Answer: b) When the object moves in the direction of the applied force

How is work calculated? a) Force multiplied by velocity b) Force multiplied by acceleration c) Force multiplied by displacement d) Force divided by time Answer: c) Force multiplied by displacement

What is the unit of work? a) Newton (N) b) Meter (m) c) Joule (J) d) Watt (W) Answer: c) Joule (J)

What is energy? a) The ability to do work b) The force applied to an object c) The distance traveled by an object d) The mass of an object Answer: a) The ability to do work

According to the law of conservation of energy, what happens to energy? a) It can be created b) It can be destroyed c) It can be transformed from one form to another d) It remains constant Answer: c) It can be transformed from one form to another

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 11 Work and Energy with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Science Work and Energy Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

Class 9 geography case study questions chapter 5 natural vegetation and wildlife, class 9 science case study questions chapter 3 atoms and molecules.

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Class 9th Science - Work and Energy Case Study Questions and Answers 2022 - 2023

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QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Science Subject - Work and Energy, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Work and energy case study questions with answer key.

9th Standard CBSE

Final Semester - June 2015

(ii) What is the value of total energy of the bob at position A ?

(iii) What is the value of kinetic energy of the bob at mean position 'O' ?

(iv) What is the value of kinetic energy and potential energy of the bob at the position 'P' whose height above 'O' is 2 cm ?

(v) What is kinetic energy? (a) Energy acquired due to motion (b) Energy acquired due to rest (c) Sum of Potential and mechanical energy (d) It is the energy stored inside a body.

*****************************************

Work and energy case study questions with answer key answer keys.

(i) (a) 0.05 J The work done in raising the bob through a height of 5 cm (against the gravitational attraction) gets stored in the bob in the form of its potential energy. PE = mgh = 0.1 x 10 x 5 x 10-2 = 0.05 J (ii) (b) 0.05 J At position A, PE = 0.05 J, KE = 0 So, Total energy = 0.05 J (iii) (c) 0.05 J At mean position, potential energy is zero, hence KE at O = 0.05 J (iv) (d) P.E. = 0.02 J and K.E. = 0.03 J PE at P = mgh = 0.1 x 10 x 2 x 10-2 = 0.02 J K.E = Total energy – PE = 0.05 – 0.02 = 0.03 J (v) (a) Energy acquired due to motion

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Class 9 Science Case Study Questions

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If you are wondering how to solve class 9 science case study questions, then myCBSEguide is the best platform to choose. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added Class 9 Science case study questions.

You can find a wide range of solved case studies on myCBSEguide, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

The rationale behind Science

Science is crucial for Class 9 students’ cognitive, emotional, and psychomotor development. It encourages curiosity, inventiveness, objectivity, and aesthetic sense.

In the upper primary stage, students should be given a variety of opportunities to engage with scientific processes such as observing, recording observations, drawing, tabulating, plotting graphs, and so on, whereas in the secondary stage, abstraction and quantitative reasoning should take a more prominent role in science teaching and learning. As a result, the concept of atoms and molecules as matter’s building units, as well as Newton’s law of gravitation, emerges.

Science is important because it allows Class 9 Science students to understand the world around us. It helps to find out how things work and to find solutions to problems at the Class 9 Science level. Science is also a source of enjoyment for many people. It can be a hobby, a career, or a source of intellectual stimulation.

Case study questions in Class 9 Science

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Examples of Class 9 science class case study questions

Class 9 science case study questions have been prepared by myCBSEguide’s qualified teachers. Class 9 case study questions are meant to evaluate students’ knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. We hope you find Class 9 science case study questions beneficial and that they assist you in your exam preparation.

The following are a few examples of Class 9 science case study questions.

Class 9 science case study question 1

due to its high compressibility
large volumes of a gas can be compressed into a small cylinder
transported easily
all of these
shape, volume
volume, shape
shape, size
size, shape
the presence of dissolved carbon dioxide in water
the presence of dissolved oxygen in the water
the presence of dissolved Nitrogen in the water
liquid particles move freely
liquid have greater space between each other
both (a) and (b)
none of these
Only gases behave like fluids
Gases and solids behave like fluids
Gases and liquids behave like fluids
Only liquids are fluids

Answer Key:

(d) all of these
(a) shape, volume
(b) the presence of dissolved oxygen in the water
(c) both (a) and (b)
(c) Gases and liquids behave like fluids

Class 9 science case study question 2

12/32 times
18 g of O 2
18 g of CO 2
18 g of CH 4
1 g of CO 2
1 g of CH 4 CH 4
2 moles of H2O
20 moles of water
6.022 × 1023 molecules of water
1.2044 × 1025 molecules of water
(I) and (IV)
(II) and (III)
(II) and (IV)
Sulphate molecule
Ozone molecule
Phosphorus molecule
Methane molecule
(c) 8/3 times
(d) 18g of CH 4
(c) 1g of H 2
(d) (II) and (IV)
(c) phosphorus molecule

Class 9 science case study question 3

collenchyma
chlorenchyma
It performs photosynthesis
It helps the aquatic plant to float
It provides mechanical support
Sclerenchyma
Collenchyma
Epithelial tissue
Parenchyma tissues have intercellular spaces.
Collenchymatous tissues are irregularly thickened at corners.
Apical and intercalary meristems are permanent tissues.
Meristematic tissues, in its early stage, lack vacuoles, muscles
(I) and (II)
(III) and (I)
Transpiration
Provides mechanical support
Provides strength to the plant parts
None of these
(a) Collenchyma
(b) help aquatic plant to float
(b) Sclerenchyma
(d) Only (III)
(c) provide strength to plant parts

Cracking Class 9 Science Case Study Questions

There is no one definitive answer to Class 9 Science case study questions. Every case study is unique and will necessitate a unique strategy. There are, nevertheless, certain general guidelines to follow while answering case study questions.

To begin, double-check that you understand the Class 9 science case study questions. Make sure you understand what is being asked by reading it carefully. If you’re unclear, seek clarification from your teacher or tutor.
It’s critical to read the Class 9 Science case study material thoroughly once you’ve grasped the question. This will provide you with a thorough understanding of the problem as well as the various potential solutions.
Brainstorming potential solutions with classmates or other students might also be beneficial. This might provide you with multiple viewpoints on the situation and assist you in determining the best solution.
Finally, make sure your answer is presented simply and concisely. Make sure you clarify your rationale and back up your claim with evidence.

A look at the Class 9 Science Syllabus

The CBSE class 9 science syllabus provides a strong foundation for students who want to pursue a career in science. The topics are chosen in such a way that they build on the concepts learned in the previous classes and provide a strong foundation for further studies in science. The table below lists the topics covered in the Class 9 Science syllabus of the Central Board of Secondary Education (CBSE). As can be seen, the Class 9 science syllabus is divided into three sections: Physics, Chemistry and Biology. Each section contains a number of topics that Class 9 science students must study during the course.

CBSE Class 9 Science (Code No. 086)


I	Matter- Its Nature and Behaviour	25
II	Organization in the Living World	22
III	Motion, Force and Work	27
IV	Food; Food Production	06
		80
		20
		100

Theme: Materials Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Particle nature and their basic units: Atoms and molecules, Law of constant proportions, Atomic and molecular masses. Mole concept: Relationship of mole to mass of the particles and numbers. Structure of atoms: Electrons, protons and neutrons, valency, the chemical formula of common compounds. Isotopes and Isobars.

Theme: The World of the Living Unit II: Organization in the Living World Cell – Basic Unit of life: Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, People and Ideas Unit III: Motion, Force and Work Motion: Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, derivation of equations of motion by graphical method; elementary idea of uniform circular motion. Force and Newton’s laws: Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Elementary idea of conservation of Momentum. Gravitation: Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, energy and power: Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy. Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

PRESCRIBED BOOKS:

Science-Textbook for class IX-NCERT Publication
Assessment of Practical Skills in Science-Class IX – CBSE Publication
Laboratory Manual-Science-Class IX, NCERT Publication
Exemplar Problems Class IX – NCERT Publication

myCBSEguide: A true helper

There are numerous advantages to using myCBSEguide to achieve the highest results in Class 9 Science.

myCBSEguide offers high-quality study materials that cover all of the topics in the Class 9 Science curriculum.
myCBSEguide provides practice questions and mock examinations to assist students in the best possible preparation for their exams.
On our myCBSEguide app, you’ll find a variety of solved Class 9 Science case study questions covering a variety of topics and concepts. These case studies are intended to help you understand how certain principles are applied in real-world settings
myCBSEguide is that the study material and practice problems are developed by a team of specialists who are always accessible to assist students with any questions they may have. As a result, students may be confident that they will receive the finest possible assistance and support when studying for their exams.

So, if you’re seeking the most effective strategy to study for your Class 9 Science examinations, myCBSEguide is the place to go!

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NCERT Solutions
NCERT Class 9
NCERT 9 Science
Chapter 11: Work and Energy

NCERT Solutions for Class 9 Science Chapter 11: Work and Energy

Ncert solutions class 9 science chapter 11 – cbse free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

NCERT Solutions for Class 9 Science Chapter 11 Work and Energy help you lay a good foundation for your CBSE exam preparation. Students who refer to NCERT Solutions regularly are benefited from the comprehensive methodology and the detailed step-by-step procedure, which will fetch them good marks in their board examinations.

Download Exclusively Curated Chapter Notes for Class 9 Science Chapter – 11 Work and Energy

Download most important questions for class 9 science chapter – 11 work and energy.

NCERT Solutions for Class 9 Science Chapter 11: Work and Energy help students to practise and gain more confidence in their preparations. All the NCERT Class 9 solutions are thoroughly prepared by our team of subject experts, and it includes objectives, diagram-based explanations, and short and long-type questions, which can be useful for both CBSE exams and other competitive exams.

Chapter 1 Matter in Our Surroundings
Chapter 2 Is Matter Around Us Pure?
Chapter 3 Atoms and Molecules
Chapter 4 Structure of the Atom
Chapter 5 The Fundamental Unit of Life
Chapter 6 Tissues
Chapter 7 Diversity in Living Organisms
Chapter 8 Motion
Chapter 9 Force and Laws of Motion
Chapter 10 Gravitation
Chapter 12 Sound
Chapter 13 Why Do We Fall ill?
Chapter 14 Natural Resources
Chapter 15 Improvement in Food Resources

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ncert solutions for class 9 march 29 science chapter 11 work and energy 01

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Access Answers of Science NCERT Class 9 Chapter 11: Work and Energy (All intext and exercise questions solved)

Exercise-11.1 Page: 148

1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?

When a force F acts on an object to move it in its direction through a distance S, work is done.

The work on the body is done by force.

Work done = Force × Displacement

F = 7 N S = 8 m

So, work done,

Exercise-11.2 Page: 149

1. When do we say that work is done?

Work is completed whenever the given conditions are satisfied:

(i) A force acts on the body.

(ii) There’s a displacement of the body by applying force in or opposite to the direction of the force.

2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.

When a force F displaces a body through a distance S within the direction of the applied force, then the work done W on the body is given by the expression:

3. Define 1 J of work.

1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long.

How much work is done in ploughing the length of the field?

Work done by the bullocks is given by the expression:

Applied force, F = 140 N

Displacement, d = 15 m

W = 140 x 15 = 2100 J

Therefore, 2100 J of work is done in ploughing the length of the field.

Exercise-11.3 Page: 152

1. What is the kinetic energy of an object?

The energy possessed by a body by virtue of its motion is termed mechanical energy or kinetic energy. Every moving object possesses mechanical energy. A body uses mechanical energy to try to work. The kinetic energy of the hammer is employed in driving a nail into a log of wood, the mechanical energy of air is employed to run wind mills, etc.

2. Write an expression for the kinetic energy of an object.

If a body of mass m is moving with a speed v, then its K.E. E k is given by the expression,

E k = 1/2 m v 2

Its SI unit is Joule (J).

3. The kinetic energy of an object of mass, m moving with a velocity of 5 ms -1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

K.E. of the object = 25J

Velocity of the object (v) = 5 m/s

K.E. = (½) mv 2

25 = (½) m (5) 2

50 = 25 x m

Now, when velocity is doubled

K.E. = (½) x 2 x (10) 2

K.E. = 10 2

K.E. = 100 J

When velocity is increased three times, then

K.E. = (½) x 2 x (15) 2

K.E. = (15) 2

K.E. = 225 J

Exercise-11.4 Page: 156

1. What is power?

Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work W in time t, then power is given by:

NCERT Solutions for Class 9 Science - Chapter 11 Image 2

It is expressed in watt (W).

2. Define 1 watt of power.

A body is claimed to possess power of one watt if it works at the speed of 1 joule in 1 s.

One W = 1 J/1 S

3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?

Power = Work/Time

Time = 10 s

Work done = Energy consumed by the lamp = 1000 J

Power = 1000/10 = 100 Js -1 =100 W

Hence, the power of the lamp is 100 W

4. Define average power.

Average power is defined as the ratio of total work done by the body to the total time taken by the body.

NCERT Solutions for Class 9 Science - Chapter 11 Image 3

Exercises – 11.5 Page: 158

1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

(a) Suma is swimming in a pond.

(b)A donkey is carrying a load on its back.

(d) A green plant is carrying out photosynthesis.

(e) An engine is pulling a train.

(f) Food grains are getting dried in the sun.

(g) A sailboat is moving due to wind energy.

Work is finished whenever the given 2 conditions are satisfied:

(a) While swimming, Suma applies a force to push the water backwards. Therefore, Suma swims in the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, the work is done by Seema while swimming.

(b) While carrying a load, the donkey has to apply a force in the upward direction. But, displacement of the load is in the forward direction. Since displacement is perpendicular to force, the work done is zero.

(c) A windmill works against gravity to elevate water. The windmill lift water by applying a force in an upward direction, and thus the water is moving in the same upward direction itself. Hence, work is done by the windmill to lift water from the well.

(d) No force is required when a green plant is carrying out photosynthesis. The plant does not exert any force to move. Since there is no displacement or force. Hence, no work is done.

(e) When an engine is pulling a train, it is applying a force in the forward direction. So, it is moving in the forward direction. Since displacement and force are in the same direction. Hence, work is done by the engine.

(f) There is no force involved in the process of drying food grains in the sun and the grains do not move. Since there is no force or displacement. Hence, no work is done.

(g) When a sailboat is moving due to wind energy, it is applying force in the forward direction. So, it is moving in the forward direction. Since displacement and force are in the same direction. Hence, work is done.

2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

Work done by the force of gravity on an object depends solely on vertical displacement. Vertical displacement is given by the distinction in the initial and final positions/heights of the object which is zero.

Work done by gravity is given by the expression,

W = m × g × h

h= Vertical displacement = 0

W = m g × 0 = 0 J

Hence, the work done by the gravity on the given object is zero joule.

3. A battery lights a bulb. Describe the energy changes involved in the process.

When a bulb is connected to a battery, then the energy of the battery is transferred into voltage. Once the bulb receives this voltage, then it converts it into light and heat energy. Hence, the transformation of energy in the given situation can be shown as:

Chemical Energy → Electrical Energy → Light Energy + Heat Energy.

4. Certain force acting on a 20 kg mass changes its velocity from 5 m s -1 to 2 m s -1 . Calculate the work done by the force.

Initial velocity u = 5 m/s

Mass of the body = 20kg

Final velocity v = 2 m/s

The initial kinetic energy

E i = (1/2) mu 2 = (1/2) × 20 × (5) 2

Final kinetic energy

E f = (1/2) mv 2 = (1/2) × 20 × (2) 2

Work done = Change in kinetic energy

Work done = E f – E i

Work done = 40 J – 250 J

Work done = -210 J

5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.

Work done by gravity depends solely on the vertical displacement of the body. It doesn’t rely on the trail of the body. Therefore, work done by gravity is given by the expression,

Vertical displacement, h = 0

∴ W= m × g × zero = 0

Therefore the work done on the object by gravity is zero.

6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

No, the method doesn’t violate the law of conservation of energy. This is because when the body falls from a height, its potential energy changes into kinetic energy progressively. A decrease in the potential energy is equal to an increase in the kinetic energy of the body. Throughout the method, the total mechanical energy of the body remains conserved. Therefore, the law of conservation of energy isn’t desecrated.

7. What are the various energy transformations that occur when you are riding a bicycle?

During riding a bicycle, the muscular energy of the rider regenerates into heat and mechanical energy. Kinetic energy provides a rate to the bicycle, and warmth energy heats our body.

Muscular energy → mechanical energy + heat

8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

When we push a huge rock, there’s no transfer of muscular energy to the stationary rock. Also, there’s no loss of energy since muscular energy is transferred into heat energy, which causes our body to become hot.

9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

1 unit of energy = 1kWh

Energy (E) = 250 units

1 unit = 1 kWh

1 kWh = 3.6 x 10 6 J

Therefore, 250 units of energy = 250 × 3.6 × 10 6

= 9 × 10 8 J.

10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

Given Mass (m) = 40 kg

Acceleration due to gravity (g)= 10m/s²

Height (h)= 5m

Potential energy= m × g× h

P.E= 40 × 10 × 5 = 2000J

Potential energy = 2000J ( 2000 joules)

At a height of 5 metres, the object has a potential energy of 2000 J.

When this object is allowed to fall and it is halfway down, its height above the ground will be half of 5 m= 5/2= 2.5m.

P.E at Halfway down= m× g×h

P.E= 40× 10 × 2.5= 1000J

Potential Energy halfway down= 1000 joules.

According to the law of conservation of energy:

Total potential energy= potential energy halfway down+ kinetic energy halfway down

2000 = 1000 + K.E halfway down

K.E at halfway down= 2000- 1000= 1000 J

Kinetic energy at halfway down= 1000 joules .

11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Work is completed whenever the given two conditions are satisfied:

If the direction of force is perpendicular to displacement, then the work done is zero. When a satellite moves around the Earth, then the force of gravity on the satellite is perpendicular to its displacement. Therefore, the work done on the satellite by the Earth is zero.

12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher

Yes, there can be displacement of an object in the absence of any force acting on it. If a single force acts on an object, the object accelerates. If an object accelerates, a force is acting on it.

Assume an object is moving with constant velocity. The net force acting on it is zero. But, there is a displacement along with the motion of the object. Therefore, there can be a displacement without a force.

13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.

Work is completed whenever the given 2 conditions are satisfied.

When an individual holds a bundle of hay over his head, there is no displacement in the hay bundle. Although the force of gravity is acting on the bundle, the person isn’t applying any force on it. Therefore, in the absence of force, work done by the person on the bundle is zero.

14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Power of the heater = 1500 W = 1.5 kW

Time taken = 10 hours

Energy consumed by an electric heater can be obtained with the help of the expression,

Power = Energy consumed / Time taken

Energy consumed = Power x Time taken

Energy consumed = 1.5 x 10

Energy consumed = 15 kWh

Therefore, the energy consumed by the heater in 10 hours is 15 kWh.

15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

Consider the case of an oscillation pendulum.

NCERT Solutions for Class 9 Science - Chapter 11 Image 4

When an apparatus moves from its mean position P to either of its extreme positions A or B, it rises through a height h on top of the mean level P. At this time, the K.E. of the bob changes fully into P.E. The K.E. becomes zero, and also, the bob possesses P.E. solely. Because it moves towards purpose P, its P.E. decreases increasingly. Consequently, the K.E. will increase. Because the bob reaches purpose P, its P.E. becomes zero, and also, the bob possesses K.E. solely. This method is perennial as long as the apparatus oscillates.

The bob doesn’t oscillate forever. It involves rest as a result of air resistance resisting its motion. The apparatus loses its K.E. to beat this friction and stops once in a while. The law of conservation of energy isn’t desecrated because the energy lost by the apparatus to beat friction is gained by its surroundings. Hence, the overall energy of the apparatus and also the encompassing system stay preserved.

16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?

The kinetic energy of an object of mass m, moving with a velocity, v, is given by the expression,

K.E = 1/2 mv 2

In order to bring it to rest, its velocity has to be reduced to zero, and in order to accomplish that, the kinetic energy has to be drained off and sent somewhere else.

An external force has to absorb energy from the object, i.e. do negative work on it, equal to its kinetic energy, or

– 1/2 mv 2 .

17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h.

Given data:

The mass of the body = 1500kg

Velocity v = 60km/hr

NCERT Solutions for Class 9 Science - Chapter 11 Image 7

The work required to stop the moving car = change in kinetic energy

18. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

NCERT Solutions for Class 9 Science - Chapter 11 Image 9

In this case, the direction of force functioning on the block is perpendicular to the displacement. Therefore, work done by force on the block will be zero.

In this case, the direction of force functioning on the block is in the direction of displacement. Therefore, work done by force on the block will be positive.

In this case, the direction of force functioning on the block is contrary to the direction of displacement. Therefore, work done by force on the block will be negative.

19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Acceleration in an object could be zero even when many forces work on it. This happens when all the forces get rid of one another, i.e., the online force working on the object is zero. For a uniformly moving object, the online force working on the it is zero. Hence, the acceleration of the thing is zero. Hence, Soni is correct.

20. Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.

Power rating of the device (P) = 500 W = 0.50 kW

Time for which the device runs (T) = 10 h

Energy consumed by an electric device can be obtained by the expression

Power = Energy consumed/Time taken

∴ Energy consumed = Power × Time

Energy consumed = 0.50 × 10

Energy consumed = 5 kWh

Thus, the energy consumed by four equal rating devices in 10 h will be

⇒ 4 × 5 kWh

21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

When an object falls freely towards the ground, its potential energy decreases, and kinetic energy increases; as the object touches the ground, all its potential energy becomes kinetic energy. Since the object hits the ground, all its kinetic energy becomes heat energy and sound energy. It can also deform the ground depending upon the ground’s nature and the amount of kinetic energy possessed by the object.

Work and Energy is one of the important topics in the Class 9 Science curriculum, and the expected weightage is 27. Every student should practise these NCERT Solutions as there are more solved numericals which are repetitively asked in the exams. Apart from the solved examples, these solutions also include key notes and important terminologies from the exam point of view.

Topic covered in NCERT Solutions

Work – 5 Questions (3 long, 2 short)
Power -5 Questions (1 long, 4 short)
Energy 1 Question (1 short)
Unit of Energy 1 Question (1 MCQ)
Forms of Energy 1 Question (1 MCQ)
Kinetic Energy and its expression 1 Question (1 short)
Potential Energy and its expression 1 Question (1 short)
Conservation of Energy 1 Question (1 short)

Class 9 Chapter 11: Work and Energy is an important topic that provides a foundation for all your future studies. Work and Energy are closely related terms which are quite often used in our daily lives. Here, in this topic, students learn more in detail about work, power, force, energy, and how they are interrelated to each other. NCERT Solutions for Class 9 Science Chapter 11: Work and Energy include more solved problems and other daily basis examples, which help students to learn the topic effectively.

Key Features of NCERT Solutions for Class 9 Science Chapter 11: Work and Energy

Provides detailed explanations for all complex topics.
Provides completely solved solutions to all the questions in the Class 9 Science NCERT textbooks.
These solutions include important questions at the end of every chapter
The language used in NCERT Solutions for Class 9 Science is easy and simple to understand for the students.
All these solutions are prepared by the subject experts after extensive research on every topic in order to provide appropriate and genuine information to the students.

Disclaimer:

Dropped Topics – 11.3.1 Commercial Unit of Energy.

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NCERT Solutions for Class 9 Science Chapter 10 - Work and Energy

NCERT Solutions
Chapter 11 Work And Energy

Work and Energy Class 9 Questions and Answers NCERT Solutions FREE PDF Download

Work and Energy Class 9 NCERT Solutions introduces fundamental concepts like work, energy, and power. It covers various forms of energy, the work-energy theorem, and the law of energy conservation. This chapter helps students grasp these essential principles through practical examples and mathematical formulations. Work and Energy Class 9 Questions and Answers provide detailed answers to all the chapter questions, helping students understand concepts thoroughly and prepare effectively for exams. Download the FREE PDF to access comprehensive explanations, solved examples, and practice problems to enhance your learning experience.

NCERT Science Class 9 Solutions are prepared by experts and updated according to the new CBSE Class 9 syllabus for the academic year 2024-25.

Quick Insights of Work and Energy Class 9 NCERT Solutions

Work and Energy Class 9 Questions and Answers will give you insights into the General Introduction: Grasp the fundamental concepts of force and its effects on motion and help you understand how objects move and interact in the physical world.

Class 9 science chapter 10 will give you crisp learnings about Newton's laws of motion, which is essential for understanding how forces influence the motion of objects. Explore practical applications and numerical problems related to these laws.

The understanding related to topics like the concept of inertia and the relationship between force, mass, and acceleration.

Class 9 Work and Energy question-answer solutions can help students analyse their level of preparation and understanding of concepts.

Access NCERT Answers for Class 9 Science Chapter 10 – Work and Energy

Intext exercise 1.

1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through displacement. What is the work done in this case?

Ans: In the above question it is given that:

Force is $F=7N$.

Displacement is $S=8m$

Work done is given by the formula:

$\text{Work done = Force }\!\!\times\!\!\text{ Displacement}$

$W=F\times S=7\times 8=56Nm$

Hence, the work done in this case is $56J$.

Intext Exercise 2

1. When do we say that work is done?

Ans: Work is said to be done if the following two conditions are satisfied:

The force must act on the object

The object should be displaced.

2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.

Ans: Consider a constant force $F$ acting on an object. If the object is displaced by a distance $s$ in the direction of the force (Fig. 11.1) and \[W\] is the work done. Hence work done is equal to the product of the force and displacement.

$\Rightarrow W=F\times S$

3. Define 1 J of work.

Ans: $1J$ of work is the amount of work done on an object when a force of $1N$ displaces it by $1m$ along the line of action of the force.

4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?

Applied force is $F=140N$.

Displacement is $s=15m$.

We know that

$\Rightarrow W=F\times S=140\times 15=2100J$

Therefore, work done in ploughing the length of the field is $2100J$.

Intext Exercise 3

1. What is the kinetic energy of an object?

Ans: The energy possessed by an object due to its motion is called kinetic energy.

2. Write an expression for the kinetic energy of an object.

Ans: Consider a body of mass $'m'$ to be moving with a velocity $'v'$. Hence its kinetic energy ${{E}_{k}}$ is given by:

${{E}_{k}}=\frac{1}{2}m{{v}^{2}}$

The SI unit of kinetic energy is the Joule $\left( J \right)$.

3. The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25J.What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Velocity of the object is $5m/s$.

Kinetic energy of the object is ${{E}_{k}}=25J$.

Mass of the object is $m$.

We know that,

Case 1: When the velocity of an object is doubled,

Kinetic Energy is given by:

${{E}_{k}}\propto {{v}^{2}}$

Hence, when $v=10m/s$

${{E}_{k}}=25\times 4=100J$

The kinetic energy becomes 4 times its initial value.

Case 2: If velocity is increased three times, then its kinetic energy will be nine times its original value, as ${{E}_{k}}\propto {{v}^{2}}$.

Hence, kinetic energy will be:

${{E}_{k}}=25\times 9=225J$

Thus, kinetic energy becomes nine times its initial value.

Intext Exercise 4

1. What is power?

Ans: Power is defined as the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by:$\text{Pow}er=\frac{Work\text{ }done}{Time}=\frac{Energy}{Time}$

$P=\frac{W}{T}$

The S.I. unit power is watt $\left( W \right)$.

2. Define 1 watt of power.

Ans: The power of an agent, which does work at the rate of 1 joule per second is defined as $1$ watt. It is also said that power is $1W$ when the rate of consumption of energy is $1J/s$.

$1W=\frac{1J}{1s}$

3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?

Energy consumed is $E=1000J$

Time $T=10s$

$\text{Pow}er=\frac{Work\text{ }done}{Time}=\frac{Energy}{Time}$

$\Rightarrow Power=\frac{1000}{10}=100W$

Hence, the power generated is$100W$.

4. Define average power.

Ans: Average power is defined as the power obtained by dividing the total amount of work done in the total time taken to do this work.$\text{Average power}=\frac{\text{Total work done}}{\text{total time taken}}$

NCERT Exercise

1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

Suma is swimming in a pond.

A donkey is carrying a load on its back.

A wind mill is lifting water from a well.

A green plant is carrying out photosynthesis.

An engine is pulling a train.

Food grains are getting dried in the sun.

A sailboat is moving due to wind energy.

Ans: According to the definition, work is done whenever the following two conditions are satisfied:

A force acts on the body.

There is a displacement of the body by the application of force in or opposite to the direction of force.

In this case of swimming, Suma applies a force to push the water backwards. This allows Suma to swim in the forward direction by the forward response of water. Hence, the force causes a displacement. Thus, while swimming work is done by Suma.

In this case the donkey applies a force in the upward direction while carrying a load. But the displacement of the load is in the forward direction. As the displacement is perpendicular to force, the work done is zero.

In this case work is done by the windmill in lifting water from the well as a windmill works against the gravitational force to lift water.

In this case there is no displacement of the leaves of the plant. Hence, the work done is zero.

In this case an engine applies force to pull the train. This makes the train move in the direction of the applied force. Thus, there is a displacement in the train in the same direction. Hence the work is done by the engine on the train.

In this case during the process of food grains getting dried in the sun, no work is done as food grains do not displace in the presence of solar energy.

In this case wind energy applies a force on the sailboat to push it in forward direction. Therefore, the displacement of the boat is along the direction of the force. Hence, the work is done by wind on the boat.

Ans: In the above question the object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line.

Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement is given by the difference in the initial and final positions (height) of the object, which is zero.

Work done by gravity is expressed as, \[W\text{ }=\text{ }mgh\]

Where,

m is the mass of the object,

g is the acceleration due to gravity

h is the vertical displacement, which is zero.

\[\Rightarrow W\text{ }=\text{ }mg\left( 0 \right)=0J\]

Thus, the work done by gravity on the given object is zero joules.

3. A battery lights a bulb. Describe the energy changes involved in the process.

Ans: Once a bulb is connected to a battery its chemical energy is converted into electrical energy. When the bulb receives this electrical energy, it is converted into light and heat energy. Therefore, the transformation of energy in the given situation can be shown as:\[\text{Chemical Energy}\to \text{Electrical Energy}\to \text{Light Energy}\to \text{Heat energy}\]

4. Certain force acting on a 20 kg mass changes its velocity from $5m/s$ to $2m/s$. Calculate the work done by the force.

Ans: Change in kinetic energy is defined as the work done.

It is given that a 20 kg mass changes its velocity from $5m/s$ to $2m/s$.

Kinetic energy is given by the expression,

$\left( {{E}_{k}} \right)v=\frac{1}{2}m{{v}^{2}}$

${{E}_{k}}=$ Kinetic energy of the object moving with a velocity, v

$m=$ Mass of the object

Kinetic energy when the object was moving with a velocity $5m/s$,${{\left( {{E}_{k}} \right)}_{5}}=\frac{1}{2}\times 20\times {{\left( 5 \right)}^{2}}=250J$

Kinetic energy when the object was moving with a velocity $2m/s$,${{\left( {{E}_{k}} \right)}_{2}}=\frac{1}{2}\times 20\times {{\left( 2 \right)}^{2}}=40J$

Therefore, work done by force $=40-250=-210J$

Here, the negative sign shows that the force is acting in the opposite direction of the motion of the object.

Ans: Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Hence, work done by gravity is given by:

\[W\text{ }=\text{ }mgh\]

Vertical displacement, $h=0$

Thus, the work done by gravity on the object is zero joules.

6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Ans: When the body drops from a height, its potential energy changes into kinetic energy progressively. A decrease in the potential energy is equivalent to an increase in the kinetic energy of the body.

During this process, the total mechanical energy of the body is conserved. Therefore, the law of conservation of energy is not violated.

7. What are the various energy transformations that occur when you are riding a bicycle?

Ans: When we ride a bicycle, our muscular energy is transferred into the bicycle's heat energy and kinetic energy. Heat energy heats the body and kinetic energy provides a velocity to the bicycle. The transformation can be shown as:\[Muscular\text{ Energy}\to Kinetic\text{ Energy}\to \text{Heat energy}\]

The total energy is conserved in the entire transformation of energies.

8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

Ans: When we push a huge rock there is no transfer of energy because there is no transfer of muscular energy to the stationary rock. Instead muscular energy is transferred into heat energy, which causes our body to become hot.

9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

Ans: We know that \[1\] unit of energy is equal to \[1\] kilowatt hour (kWh).\[1\text{ }unit\text{ }=\text{ }1\text{ }kWh\]

$1kWh=3.6\times {{10}^{6}}J$

Hence, $250$ units of energy $=250\times 3.6\times {{10}^{6}}=9\times {{10}^{8}}J$.

10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

\[h\text{ }=\text{ }Vertical\text{ }displacement\text{ }=\text{ }5\text{ }m\]

\[m\text{ }=\text{ }Mass\text{ }of\text{ }the\text{ }object\text{ }=\text{ }40\text{ }kg\]\[g\text{ }=\text{ }Acceleration\text{ }due\text{ }to\text{ }gravity\text{ }=\text{ }9.8\text{ }m/{{s}^{2}}\]

Gravitational potential energy is given by the expression,

\[\Rightarrow W\text{ }=\text{ }40\times 5\times 9.8\text{ }=\text{ }1960\text{ }J\]

At half-way down, the potential energy of the object will reduce to half i.e., $\frac{1960}{2}=980J$.

At this point, the object has potential energy equal to kinetic energy. This is due to the law of conservation of energy. Hence, half-way down, the kinetic energy of the object will be \[980J\].

11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Ans: When the direction of force is perpendicular to displacement, the work done is zero. If a satellite moves around the Earth the direction of force of gravity on the satellite is perpendicular to its displacement. Therefore, the work done on the satellite by the Earth is zero.

12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

Ans: Yes, there be displacement of an object in the absence of any force acting on it by moving with uniform velocity. Suppose an object is moving with constant velocity, then net force acting on it is zero. Hence, there can be a displacement without a force.

13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.

Ans: If a person holds a bundle of hay over his head, gravitational force is acting on the hay downwards. But, there is no displacement in the bundle of hay in the direction of force. Hence, no work is done.

14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Ans: Energy consumed by an electric heater is given by the expression,

\[P\text{ }=\text{ }1500\text{ }W\text{ }=\text{ }1.5\text{ }kW\]

\[T\text{ }=\text{ }10\text{ }hrs\]

\[Work\text{ }done\text{ }=\text{ }Energy\text{ }consumed\text{ }by\text{ }the\text{ }heater\]

Hence, \[energy\text{ }consumed\text{ }=\text{ }Power\text{ }\times \text{ }Time=1.5\times 10=15kWh\]= 1.5 × 10 = 15 kWh.

Ans: According to the “law of conservation of energy”: Energy can be neither created nor destroyed. It can only be converted from one form to another. Consider the case of an oscillating pendulum.

When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height h above the mean level P. At the extreme point, bob comes to rest and the kinetic energy of the bob is transformed into potential energy.

As it moves towards point P, its potential energy decreases progressively. Simultaneously, the kinetic energy increases.

As the bob reaches point P, its potential energy is converted to kinetic energy. At this point bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates. The bob does not oscillate forever. It comes to rest due to some friction in the air.

The pendulum loses energy overcoming this friction. After all the energy is lost, it comes to rest. The law of conservation of energy holds here as the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system is conserved.

16. An object of mass m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?

Ans: When a body is in motion it possesses kinetic energy. Kinetic energy of an object of mass, m moving with a velocity, v is given by the expression,

Therefore, $\frac{1}{2}m{{v}^{2}}$ amount of work is required to be done on the object to bring the object to rest.

17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Mass of car, \[m\text{ }=\text{ }1500\text{ }kg\]

Velocity of car, \[v\text{ }=\text{ }60\text{ }km/h\text{ }=60\times \frac{5}{18}m/s\]

Kinetic energy is given by:

$\Rightarrow {{E}_{k}}=\frac{1}{2}\left( 1500 \right){{\left( 60\times \frac{5}{18} \right)}^{2}}=20.8\times {{10}^{4}}J$

Hence, $20.8\times {{10}^{4}}J$ needs to be done to stop the car.

Different positions of force and displacement w.r.t. each other

Ans: Following cases are explained below:

Force and displacement are perpendicular

Here the direction of force acting on the block is perpendicular to the displacement.

Hence, work done by force on the block will be zero.

Here the direction of force acting on the block is in the direction of displacement. So, work done by force on the block will be positive.

Force and displacement are anti-parallel

Here the direction of force acting on the block is opposite to the direction of displacement. Hence, work done by force on the block will be negative.

19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Ans: Acceleration of an object will be zero when net force acting on an object will be zero. Net force can be zero even when there are multiple forces acting on the body. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

20. Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.

\[P\text{ }=\text{ }500\text{ }W\text{ }=\text{ }0.50\text{ }kW\]

Energy consumed by an electric device can be obtained with the help of the expression given below:

\[Work\text{ }done\text{ }=\text{ }Energy\text{ }consumed\text{ }by\text{ }the\text{ }device\]\[Energy\text{ }consumed\text{ }by\text{ }each\text{ }device\text{ }=\text{ }Power\text{ }\times \text{ }Time\]\[\text{Work done}=\text{ }0.50\text{ }\times \text{ }10\text{ }=\text{ }5\text{ }kWh\]

Hence, the energy consumed by four equal rating devices in \[10\text{ }hrs\] will be \[4\times 5kWh=20kWh\]

21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Ans: Once the object hits the hard ground, the kinetic energy gets converted into heat energy and sound energy. Further, entire energy is lost to the environment. This energy can also deform the ground with respect to the nature of the ground and the amount of kinetic energy possessed by the object.

Class 9 Chemistry Chapter 10 Quick Overview of Topics

Class 9 Chemistry Chapter 10 NCERT Solutions - Quick Overview of Detailed Structure of Topics and Subtopics Covered.

Topic	Subtopics
Work	The scientific conception of Work
Work	Work done by a constant force
Energy	Forms of energy
	Kinetic energy
	Potential energy
	Potential energy of an object at a height
Law of Conservation of Energy	Explanation and examples
Rate of Doing Work	Definition and examples

Dropped Topics: 11.3.1 Commercial Unit of Energy.

Class 9 NCERT Solutions chapter 10 Important Topics

Class 9 NCERT solutions help the students to go through the Important Highlights easily. Here find the Important topics of chapter 10 - Work and Energy to crack your exams.

Scientific Conception of Work : Work is defined as the product of force and displacement in the direction of the force. It is mathematically expressed as $(W = F \times d \times \cos(\theta))$ , where (W) is work, (F) is force, (d) is displacement, and $(\theta)$ is the angle between the force and the displacement direction. Work is done only when the force causes displacement.

Forms of Energy

Kinetic Energy : Kinetic energy is the energy an object possesses due to its motion. It is given by the formula $(KE = \frac{1}{2}mv^2)$ , where (m) is the mass of the object and (v) is its velocity.

Potential Energy : Potential energy is the stored energy of an object due to its position or state. One common form is gravitational potential energy, calculated using $(PE = mg)$ , where (m) is mass, (g) is the acceleration due to gravity, and (h) is height above the ground.

Law of Conservation of Energy

Explanation and Examples : The law of conservation of energy states that energy cannot be created or destroyed; it can only be transformed from one form to another. The total energy in an isolated system remains constant.

Benefits of NCERT Solutions Class 9 Science Chapter 10

The Vedantu’s Class 9 NCERT Solutions For Science Chapter 10 Work and Energy answers provided herein PDFs offer various benefits, including:

Detailed explanations and step-by-step solutions for all topics in chapter 10.

Work and energy class 9 questions and answers pdf curated by experienced educators to ensure accuracy and clarity.

The fundamental concepts of force and its effects on motion will help you understand how objects move and interact in the physical world.

Clear and concise explanations using precise chemical terminology.

In-depth analysis of key concepts and their applications in real-life scenarios.

Learn about Newton's laws of motion, which is essential for understanding how forces influence the motion of objects.

Solutions to a variety of problems to strengthen analytical and problem-solving abilities.

Step-by-step solutions for numerical problems and reaction mechanisms.

Important Study Materials for Class 9 Chemistry Chapter 10 Work and Energy

Students can access extra study materials on Work and Energy. These resources are available for download and offer additional support for your studies.

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The NCERT Solutions for Class 9 Chapter 10, Science - Word and Energy, provided by Vedantu, is a valuable tool for Class 9 students. It helps introduce Science concepts in an accessible manner. The provided solutions and explanations simplify complex ideas, making it easier for Class 9 Students to understand the material. By using Vedantu's resources, Students can develop a deeper understanding of NCERT concepts. These solutions are a helpful aid for class 9 students, empowering them to excel in their studies and develop a genuine appreciation for Work and Energy.

NCERT Solutions for Class 9 Science - Chapter-Wise Links


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NCERT Study Resources for Class 9 Science

For complete preparation of Science for CBSE Class 9 exams, check out the following links for different study materials available at Vedantu.

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FAQs on NCERT Solutions for Class 9 Science Chapter 10 - Work and Energy

1. How Many Exercises Does this Chapter Include?

Work and energy is a vast chapter that includes about 5 exercises in total. These are mentioned below.

Exercise 11.1 - 1 Question on work and power.

Exercise 11.2 - 4 questions on energy and units of energy.

Exercise 11.3 - 3 questions on different forms of energy.

Exercise 11.4 - 4 questions on kinetic energy and its expression.

Exercise 11.5 - 21 questions on potential energy and its expression and conservation of energy.

2. How is work related to energy in physics?

Work and energy are closely related concepts in physics. Work is done when a force acts on an object to move it over a distance. The work done on an object results in a change in its energy. According to the work-energy theorem, the work done by the net force on an object is equal to the change in its kinetic energy. Mathematically, this is expressed as $(W = \Delta KE)$ . This means that when work is done on an object, it can increase or decrease its energy, demonstrating the transfer and transformation of energy through work.

3. What is the kinetic energy of an object in class 9 science chapter 10?

The Kinetic Energy of a particular object is the energy that it has because of its motion. It is known as the work required to accelerate the respective body of a given mass from rest to its actual velocity. Thus, any object in motion always has kinetic energy. For instance, a person walking or running, a thrown baseball, a crumb falling from a tabletop are all examples of Kinetic Energy. To know more about it, visit Vedantu website or Vedantu Mobile app. The solutions provided by Vedantu are free of cost.

4. What is work according to Class 9 Science Chapter 10 Question Answer NCERT?

Work done is as the product of two components- force (F) acting in the direction of the displacement and the magnitude of displacement (s) of the object. Thus, the two conditions required for work to be done are the action of force on an object and the displacement of the object. The formula for work done is Work Done = Force x Displacement, W = Fs. It is a scalar quantity (because it has only magnitude and no direction) and it is measured in Joules (J). class 9 science chapter 10 question answer is a important chapter from exams point of you to Score well.

5. What is energy according to Work and Energy Class 9 NCERT Solutions?

A body's Energy is defined, in Physics, as the capacity of that body to perform work. Energy has only magnitude and no direction i.e., it is a scalar quantity and is measured in Joules (J). The Sun constitutes the largest natural source of energy for all living creatures. Some other energy sources are nuclei of atoms, tides, and the interior of the Earth. There are several types of energy, such as Potential Energy, Kinetic Energy, Mechanical Energy, Light Energy, Chemical Energy, Electrical Energy, and Heat Energy.

6. What is the formula of Energy according to Chapter 11 of NCERT Solutions for Class 9 Science?

A body's Energy, in Physics, is defined as the capacity of that body to perform work. Energy’s SI unit is Joules. It is scalar in nature as it has only magnitude and no direction. The formula for Potential Energy, Ep, of an object is equivalent to the work done over an object of mass 'm' to raise it by a height 'h' when 'g' is the acceleration due to gravity. Therefore, Ep = mgh. Work and energy class 9 exercise answers provided by Vedantu will help you to crack the exams.

7. What is Work and Energy in Physics of Class 9?

In physics, work is defined as the sum of two factors: the force acting in the direction of displacement (F) and the amount of the object's displacement (s). The capacity of a body to produce work is defined as its energy in physics. Both Work and Energy are scalar quantities that are measured in Joules (J). Class 9 work and energy question answer gives you better practice to get expertise in the chapter.

8. What is the difference between kinetic energy and potential energy?

Kinetic energy is the energy an object possesses due to its motion. It depends on the mass of the object and the square of its velocity, given by the formula $(KE = \frac{1}{2}mv^2)$ . On the other hand, potential energy is stored in an object because of its position or configuration. A common type is gravitational potential energy, which depends on the object's mass, the height above the ground, and the acceleration due to gravity, calculated as $(PE = mgh)$ . While kinetic energy is associated with motion, potential energy is associated with the position or state of an object. Students can also refer to class 9 work and energy question answers provided by NCERT Solution for better understanding.

NCERT Solutions for Class 9 Science

Ncert solutions for class 9.

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Case Study Questions of Chapter 11 Work and Energy PDF Download

Case study Questions on Class 9 Science Chapter 11 are very important to solve for your exam. Class 9 Science Chapter 11 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 11 Work and Energy

Work and Energy Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 11 Work and Energy

Case Study/Passage Based Questions

Figure shows a watch glass embedded in clay. A tiny spherical ball is placed at the edge B at a height h above the centre A

The kinetic energy of the ball, when it reaches at point A is (a) zero (b) maximum (c) minimum (d) can’t say

Answer: (b) maximum

The ball comes to rest because of (a) frictional force (b) gravitational force (c) both (a) and (b) (d) none of these

Answer: (c) both (a) and (b)

The energy possessed by ball at point C is (a) potential energy (b) kinetic energy (c) both potential and kinetic energy (d) heat energy.

Answer: (a) potential energy

The principle of conservation of energy states that the energy in a system can neither be created nor be destroyed. It can only be transformed from one form to another, but total energy of the system remains constant. Conservation of electrical energy to various forms or vice versa along with devices is illustrated in the figure given below.

Water stored in a dam possesses (a) no energy (b) electrical energy (c) kinetic energy (d) potential energy.

Answer: (d) potential energy.

Answer: (c) Chemical energy → Electrical energy → Light energy

Name a machine that transforms muscular energy into useful mechanical work. (a) Amicrophone (b) Bicycle (c) Electric torch (d) An electric bell

Answer: (b) Bicycle

Answer: (c) half potential and half kinetic energy

An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.4 m s –1 . For this purpose a motor of required horse power is used

The power of motor in hp is (a) 2.33 (b) 2.43 (c) 2.53 (d) 2.63

Answer: (d) 2.63

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 11 Work and Energy with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 9 Science Work and Energy Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible

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Extra Questions for Class 9 Science Chapter 11 Work and Energy

Extra questions for Class 9 Science Chapter 11 Work and Energy with answers is given below. Our subject expert prepared these solutions as per the latest NCERT textbook. These questions will be helpful to revise the all topics and concepts. CBSE Class 9 extra questions are the most simple and conceptual questions that are prepared by subject experts for the students to study well for the final exams. By solving these extra questions, students can be very efficient in their exam preparations.

Work and Energy Class 9 Science Extra Questions and Answers

Very short answer questions.

1: What is the work done against the gravity when a body is moved horizontally along a frictionless surface? Answer: Zero

2: When displacement is in a direction opposite to the direction of force applied, what is the type of work done? Answer: Negative work. 3: A 40 kg girl is running along a circular path of radius 1 m with a uniform speed. How much work is done by the girl in completing one circle? Answer: Zero

4: Seema tried to push a heavy rock of 100 kg for 200 s but could not move it. Find the work done by Seema at the end of 200 s. Answer: Work done = 0 Since displacement, s = 0

5: Identify the kind of energy possessed by a running athlete. Answer: Kinetic energy.

Short Answer Type Questions

1: An electrical heater is rated 1200 W. How much energy does it use in 10 hours?

Answer: Electrical energy = Power × time taken = 1.2 × 10 = 12 kWh

2: If an electric appliance is rated 1000 W and is used for 2 hours. Calculate the work done in 2 hours.

Answer: Work done = Energy consumed Energy = Power × Time taken = 1000 W × 2 hour = 2000 W-hr or 2 kW-hour or 2 kWh

3: A man of mass 62 kg sums up a stair case of 65 steps in 12 s. If height of each step is 20 cm, find his power.

Answer: m = 62 kg, g = 10 m/s 2 h = 65 × 20/100 = 13 m

P.E. = mgh P.E. = 62 × 10 × 13 = 8060 J

Power (𝑃) = (𝑃.𝐸.)/𝑡 = 8060/12 = 671.67 𝑊

4: How is work done by a force measured? A porter lifts a luggage of 20 kg from the ground and puts it on his head 1.7 m above the ground. Find the work done by the porter on the luggage. (g = 10m/s 2 )

Answer: Work done is product of force and displacement W=F × s m=20 kg g=10m/s 2 h =1.7𝑚 The work done by the porter = mgh = 20 × 10 × 1.7 = 340 J

5: The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 1

6: When a force retards the motion of a body, what is the nature of work done by the force? State reason. List two examples of such a situation.

Answer: The work done by the force is negative because the displacement is opposite to the direction of force applied. Example: (i) Work done by the force of friction; (ii) Work done by applying brakes.

7: When is the work done by a force said to be negative ? Give one situation in which one of the forces acting on the object is doing positive work and the other is doing negative work.

Answer: Negative work: When the force is acting opposite to the direction of the displacement, the work done by the force is said to be negative. When we lift an object, two forces act on the (i) Muscular force: Doing positive work in the direction of the displacement. (ii) Gravitational force: Doing negative work opposite to the direction of the displacement.

8: (a) Under what conditions work is said to be done? (b) A porter lifts a luggage of 1.5 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.

Answer: (a) (i) Force should be applied. (ii) Body should move in the line of action of force. (iii) Angle between force and displacement should not be 90°.

(b) Mass of luggage, m = 15 kg and displacement, s = 1.5 m. Work done, W = F×s = mg × s = 15 × 10 × 1.5 = 225 J

9: Four persons jointly lift a 250 kg box to a height of 1 m and hold it. (i) Calculate the work done by the persons in lifting the box. (ii) How much work is done for just holding the box ? (iii) Why do they get tired while holding it ? (g = 10ms 2 )

Answer: (i) F = 250 × 10 = 2500𝑁 s =1 m W = F ×s = 2500 ×1| =2500𝐽

(ii) Zero, as there is no displacement. (iii) To hold the box, men are applying a force which is opposite and equal to the gravitational force acting on the box. While applying the force muscular effort is involved, and so they feel tired.

10: A boy is pulling a cart by supplying a constant force of 8 N on a straight path of 20 m. On a round about of 10 m diameter he forgets the path and takes 1½ turns and then continues on the straight path for another 20 m. Find the net work done by the boy on the cart.

Answer: F = 8N Work done, W = F × s

W 1 = 8 × 20 = 160 J

D =10 m So radius, D/2 = 10/2 = 5m

Circumference of a circle = 2πr = 2 × 22/7 × 5 = 31.43

Distance in 1⁄2 circle = πr = 22/7 × 5 = 15.71

Total distance for 1 ½ circle = 31.43 + 15.71 = 47.14 m

W 2 = F × s = 8 × 47.14 = 376 J

W 3 = 20 × 8 = 160 J

Total work done = 160 + 376 + 160 = 696 J

Long Answer Type Questions

1: Calculate the electricity bill amount for a month of 30 days, if the following devices are used as specified: (i) 2 bulbs of 40 W for 6 hours. (ii) 2 tubelights of 50 W for 8 hours. (iii) A TV of 120 W for 6 hours.

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 2

Given the cost of electricity is ₹2.50 per unit.

Answer: Given the cost of electricity is ₹2.50 per unit. (i) 2 bulbs of 40 watts for 6 hrs. Energy consumed by Bulbs E 1 = 2 × 40 × 6 = 480 W = 0.48 kWh

(ii) Energy consumed by 2 tubelights E 2 = 50 × 8 × 2 = 0.800 kWh

(iii) Energy consumed by TV E 3 = 120 × 6 = 0.720 kWh Total Energy = 0.48 + 0.80 + 0.72 = 2.00 units rate = 2.50 per unit

Cost per day = 2 × 2.50 = 5.00 Cost 30 days = 5.00 × 30 = 150

2: (i) What is meant by mechanical energy? State its two forms. State the law of conservation of energy. Give an example in which we observe a continuous change of one form of energy into another and vice-versa. (ii) Calculate the amount of work required to stop a car of 1000 kg moving with a speed of 72 km/h. Answer: (i) Sum of kinetic energy and potential energy of an object is the total mechanical energy. Its two forms are kinetic energy and potential energy. Energy can neither be created nor be destroyed but can be transformed from one form to another. One example is simple pendulum.

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 3

3. A boy pushes a book by applying a force of 40N. Find the work done by this force as the book is displaced through 25 cm along the path.

Answer: Here, force acting on the book, F = 40N distance through which book is displaced, s = 25 cm = 0·25 m Work done by the force, i.e., W = F × s = (40 N) (0·25 m) = 10J

4. A ball of mass 1 kg thrown upwards, reaches a maximum height of 4 m. Calculate the work done by the force of gravity during the vertical displacement. (g = 10 m/s 2 ).

Answer: Here, force of gravity on the ball, F = mg = (1 kg) (10 m/s 2 ) = 10N vertical displacement of the ball, s = 4m Since the force and the displacement of the ball are in opposite directions, work done by the force of gravity, i.e., W= F × s = 10 × 4 = 40J Obviously, work done against the force of gravity = 40J

5. An engine pulls a train 1 km over a level track. Calculate the work done by the train given that the frictional resistance is 5 × 10 5 N.

Answer: Here, frictional resistance, F = 5 × 10 5 N distance through which the train moves, s = 1 km = 1000 m Work done by the frictional force, i.e., W = Fs = (5 × 10 5 ) × 1000 = 5 × 10 8 J (F and s are in opposite directions) Obviously, work done by the train is 5 × 10 8 J

6. A man weighing 70 kg carries a weight of 10 kg on the top of a tower 100 m high. Calculate the work done by the man. (g = 10 m/s 2 ).

Answer: Here, force exerted by the man, F = (70 + 10) = 80 kg wt = 80 × 10 = 800 N vertical displacement, s = 100 m Work done by the man, i.e., W = F × s = (800N) (100m) = 80000 J

7. How fast should a man of mass 60 kg run so that his kinetic energy is 750 J ?

Answer: Here, mass of the man, m = 60 kg kinetic energy of the man, E k = 750J If v is the velocity of the man, then

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 4

8. Find the mass of the body which has 5J of kinetic energy while moving at a speed of 2 m/s. Answer: Here, kinetic energy of the body, E k = 5J speed of the body, v = 2 m/s If m is the mass of the body, then

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 5

9. A player kicks a ball of mass 250 g at the centre of a field. The ball leaves his foot with a speed of 10 m/s, Find the work done by the player on the ball.

Answer: The ball, which is initially at rest, gains kinetic energy due to work done on it by the player.

Thus, the work done by the player on the ball, W = kinetic energy (E k ) of the ball as it leaves his foot, i.e., W = E k = mv 2

Here, m = 250 g = 0·25 kg, v = 10 m/s

W = (0·25) ×(10) 2 = 12·5 J

10. A body of mass 5 kg, initially at rest, is subjected to a force of 20N. What is the kinetic energy acquired by the body at the end of 10s?

Answer: Here, mass of the body, m = 5 kg initial velocity of the body, u = 0 force acting on the body, F = 20 N time for which the force acts, t = 10 s If a is the acceleration produced in the body,

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 6

Let v be the velocity of the body after 10 s. Clearly, v = u We know that V = u + at When u = 0 v = 0 + at ⇒ v = 4 × 10 ⇒ v = 40 m/s

Kinetic energy acquired by the body, E k = mv 2 = 5 × (40) 2 = 4000J

11. A bullet of mass 20 g moving with a velocity of 500 m/s, strikes a tree and goes out from the other side with a velocity of 400 m/s. Calculate the work done by the bullet in joule in passing through the tree.

Answer: Here, mass of the bullet, m = 20 g = 0·02 kg initial velocity of the bullet, u = 500 m/s final velocity of the bullet, v = 400 m/s If W is the work done by the bullet in passing through the tree, then according to work-energy theorem W = mu 2 – mv 2 = m(u 2 – v 2 ) = (0·02) [(500) 2 – (400) 2 ] = 900J

12. A body of mass 4 kg is taken from a height of 5 m to a height 10 m. Find the increase in potential energy.

Answer: Here, mass of the body, m = 4 kg increase in height of the body, h = (10 – 5) = 5m Increase in potential energy, E p = mgh = 4 × 10 × 5 = 200J

Initial potential energy of the body, E pi = mgh = 4 × 10 × 5 = 200J

Final potential energy of the body, E pf = mgh f = 4 × 10 × 10 = 400J

Increase in potential energy, E p = E pf – E pi = 400J – 200J = 200J

13. A 5 kg ball is thrown upwards with a speed of 10 m/s. (a) Find the potential energy when it reaches the highest point. (b) Calculate the maximum height attained by it.

Answer: (a) Here, mass of the ball, m = 5 kg, speed of the ball, v = 10 m/s

Kinetic energy of the ball, E k = mv 2 = 5 × (10) 2 = 250J

When the ball reaches the highest point, Its kinetic energy becomes zero as the entire kinetic energy is converted into its potential energy (E p ) i.e., E p = 250J ….(i)

(b) If h is the maximum height attained by the ball, E p = mgh …. (i) From eqn. (i) and (ii),

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 7

14. A 5 kg ball is dropped from a height of 10m. (a) Find the initial potential energy of the ball. (b) Find the kinetic energy just before it reaches the ground and (c) Calculate the velocity before it reaches the ground.

Answer: Here, mass of the ball, m = 5 kg Height of the ball, h = 10m

(a) Initial potential energy of the ball, E p = mgh = 5 × 10 × 10 = 500J

(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted into its kinetic energy (E k ), i.e., E k = 500J

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 8

15. A body is thrown up with a kinetic energy of 10 J. If it attains a maximum height of 5 m, find the mass of the body.

Answer: Here, kinetic energy of the body, E k = 10J maximum height attained by the body, h = 5m When the body attains maximum height, its entire kinetic energy is converted into its potential energy (E p )

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 9

16. A rocket of mass 3 × 10 6 kg takes off from a launching pad and acquires a vertical velocity of 1 km/s and an altitude of 25 km. Calculate its (a) potential energy (b) kinetic energy.

Answer: Here, mass of the rocket, m = 3 × 10 6 kg velocity acquired by the rocket, v = 1 km/s = 1000 m/s height attained by the rocket, h = 25 km = 25000 m

(a) Potential energy of the rocket, E p = mgh = (3 × 10 6 ) × (10 2 ) × 25000 = 7.5 × 10 11 J

(b) Kinetic energy of the rocket, E k = mv 2 = (3 × 10 6 ) × (1000) 2 = 1.5 × 10 12 J

17. A boy of mass 40 kg runs up a flight of 50 steps, each of 10 cm high, in 5 s. Find the power developed by the boy.

Answer: Here, mass of the boy, m = 40 kg total height gained, h = 50 × 10 cm = 500 cm = 5m time taken to climb, t = 5s Work done by the boy, W = mgh = 40 × 10 × 5 = 2000J

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 10

18. What should be the power of an engine required to lift 90 metric tonnes of coal per hour from a mine whose depth is 200 m

Answer: Here, mass of the coal to be lifted, m = 90 metric tonnes = 90 × 1000 kg = 9 × 10 4 kg

height through which the coal is to be lifted, h = 200m time during which the coal is to be lifted, t = 1h = 60 × 60 = 3600 s

work done to lift the coal, i.e., W = mgh = 9 × 10 4 × 10 × 200m = 18 × 10 7 J

Power of the engine required i.e.,

19. How much time does it take to perform 500J of work at a rate of 10W ?

Answer: Here, work to be performed, W = 500J rate at which work is to be performed, i.e., power, P = 10W

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 12

20. Calculate the units of energy consumed by 100 W electric bulb in 5 hours.

Answer: Here, power of the electric bulb, P = 100 W = 0·1 kW time for which bulb is used, t = 5h As P = W/t, ⇒ W = Pt

Energy consumed by the bulb, W = Pt = 0·1 × 5 = 0·5 kWh = 0·5 units

21. A lift is designed to carry a load of 4000 kg through 10 floors of a building, averaging 6 m per floor, in 10 s. Calculate the power of the lift.

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 13

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Important Questions for CBSE Class 9 Science Work, Power and Energy

December 11, 2019 by Sastry CBSE

Important Questions for CBSE Class 9 Science Chapter 11 Work, Power and Energy

Important questions.

1-MARK QUESTIONS Question.1 Does work done depend upon the velocity of the body. [SAII-2014] Answer. No.

Question.2 State the law of conservation of energy. [SAII-2014] Answer. It states that energy can neither be created nor destroyed. It can only change its form.

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Question.3 In a tug-of-war one team gives way to the other. What work is being done and by whom ? [SAII-2014] Answer. The winning team does work. The work is equal to the product of the resultant force and the displacement undergone by the losing team.

Question.4 What will cause greater change in kinetic energy of a body? Changing its mass or changing its velocity ? Answer. Changing its velocity.

Question.5 List two essential conditions for work to be done. [SAII-2010] Answer. (i) A force must act and (ii) There should be displacement in the body.

Question.6 When is 1 joule of work said to be done ? Answer. When a force of 1 newton acting on a body displaces it in its own direction.

Question.7 What is the SI unit of work done and power ? Answer. Joule and Watt.

Question.8 What is power? What is its SI unit ? Answer. It is defined as the rate of doing work. Its unit is watt.

Question.9 Find the energy in kWh consumed in 10 hours by a machine of power 500 W. [SAII-2011] Answer. W = P x t = 500 x 10 = 5000 Wh – 5 kWh.

Question.10. When is work said to be done against the force of gravity ? Answer. When a body lifted the work is done against the force of gravity.

Question.11 Write an expression for the work done in lifting a body of mass ‘m’ through a vertical height ‘h’. [SAll-2012] Answer. Work done W = mgh, where g is acceleration due to gravity.

Question.12 When a book is lifted from a table, against which force work is done ? Answer. Work is done against the force of gravity.

Question.13 Will work be done by a man who pushes a wall ? Answer. No.

Question.14 What is the work done when the force acting on the body and the displacement produced in the body are at right angles to each other ? Answer. Zero.

Question.15 Is it possible that some force is acting on a body but still the work done is zero ? Answer. Yes, when force acts at an angle of 90° with the displacement.

Question.16 What is the work done on a body moving in a circular path ? Answer. Zero, because force and displacement are perpendicular to each other.

Question.17 Does every change in energy of the body involve work ? Answer. Yes.

important-question-for-cbse-class-9-science-work-power-and-energy-1

Question.19 A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case ? Answer. Given, displacement = 8 m, Force = 7N Now, Work done = Force x Displacement = 7 x 8 = 56 J

Question.20 When do we say that work is done ? ~ Answer. Work is said to be done when a force causes displacement of an object in the direction of applied force.

Question.21 Write an expression for the work done when a force is acting on an object in the direction of its displacement. Answer. Work done = Force x Displacement

Question.22 A pair of bullocks exert a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field ? Answer. Work done = Force x Displacement = 140 x 15 = 2100 J

Question.23 What is the kinetic energy of an object? Answer. The energy possessed by a body by virtue of its motion is called kinetic energy.

Question.24 Write an expression for the kinetic energy of an object. Answer. The expression is KE = 1/2 mv 2 , where ‘m’ is the mass and V is the velocity of the body.

Question.25 Define 1 watt of power. Answer. When a work of 1 joule is done in 1s, the power is said to be one watt.

Question.26 A lamp consumes 1000 J of electrical energy in 10 s. What is its power ? Answer. Given, W = 1000 J, t = 10 s, R = ? Using p = W/t = 1000/10 = 100 W

Question.27 Define average power. Answer. When a machine or person does different amounts of work or uses energy in different intervals of time, the ratio between the total work or energy consumed to the total time is average power.

Question.28 Define energy. Answer. Energy is the ability of a body to do work. It is also defined as the capacity to do work.

Question.29 A body performs no work. Does it imply that the body possesses no energy ? Answer. When a body does not perform any work, it never implies that the body has no energy. The body may have energy but still does not perform any work, e.g., a book placed on a table has potential energy but is not performing any work.

Question.30 What is the SI unit of energy? Answer. The SI unit of energy is joule.

Question.31 Does a body at rest possess any kinetic energy ? Answer. No.

Question.32 What will happen to the kinetic energy of a body if its mass is doubled ? Answer. Its kinetic energy will be doubled.

Question.33 What will happen to the kinetic energy of a body if its velocity is halved ? [SAII-2012] Answer. The kinetic energy of the body will become one-fourth.

Question.34 By how much will the speed of a body, of fixed mass, increase if its kinetic energy becomes four times its initial kinetic energy ? Answer. The speed is doubled.

Question.35 Can a body possess energy even if it is not in motion ? Answer. Yes, it can possess potential energy.

Question.36 Define potential energy. Answer. It is defined as the energy possessed by a body by virtue of its position or change in shape.

Question.37 Name the energy possessed by a stretched rubber band lying on the table. Answer. Potential energy.

Question.38 Give the SI unit of potential energy. Answer. The SI unit of potential energy is.joule.

Question.39 What do you mean by trans- formation of energy ? Answer. It is the change of energy from one form of energy into another form of energy.

Question.40 Can energy be destroyed? Can energy be created ? Answer. No,

Question.41 A cell converts one form of energy into another. Name the two forms. Answer. It converts chemical energy into electrical energy.

Question.42 Name one unit of power bigger than watt. Answer. A unit bigger than watt is kilowatt.

Question.43 When an arrow is shot from its bow, it has kinetic energy. From where does it get the kinetic energy ? [SAll-2010] Answer. A stretched bow possesses potential energy on account of a change in its shape. To shoot an arrow; the bow is released. The potential energy of the bow is converted into the kinetic energy of the arrow.

Question.44 Name at least three commonly used units of energy. Answer. (i) Joule (ii) Erg (iii) Kilowatt hour.

Question.45 Name the practical unit of power in engineering. Answer. Horsepower.

Question.46 Name at least six forms of energy. Answer. (i) Chemical energy (ii) Heat energy (iii) Light energy (iv) Electrical energy (v) Sound energy (vi)Solar energy

Question.47 How many watt are there in 1 horse – power ? Answer. 746 watt.

Question.48 What is horsepower ? Answer. It is a unit of power.

Question.49 A light and a heavy body have equal kinetic energy. Which one is moving fast ? [SAII-2011] Answer. The lighter body is moving fast.

2 MARKS QUESTIONS Question.1 State the relation between kW h and joule. Define 1 watt. [SAII-2014] Answer. 1 kW h – 1000 W h = 1000 Js -1 x 60 x 60 s = 3.6 x 10 6 J 1 watt is the power of an agent which can do one joule of work in one second.

Question.2 Is it possible that a body be in accelerated motion under a force acting on the body, yet no work is being done by the force? Explain your answer giving a suitable example. [SAII-2012] Answer. Yes, it is possible, when the force is perpendicular to the direction of motion. The moon revolving round the earth under the centripetal force of attraction of the earth but earth does not do any work on the motion of The moon.

Question.3 Define work. How is work measured ? When is work done by a force negative? [SAII-2013] Answer. Work is said to be done if force acting on an object displaces it through a certain distance. It is measured as the product of force and displacement. Work done is negative if force and displacement are in the opposite direction.

Question.4 What is the work done by the force of gravity in the following cases ? (a) Satellite moving around the earth in a circular orbit of radius 35000 km. (b) A stone of mass 250 g is thrown up through a height of 2.5 m. [SAII-2013] Answer. (a) Zero, as the displacement in one complete revolution is zero. (b) Given m = 250 g = 0.25 kg, h = 2.5 m, g = 10 ms -2 , W = ? Now, W = FS = mg x h = 0.25 x 10 x 2.5 = 6.25 J

Question.5 A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer. Answer. The work done is zero. This is because the gravitational force and displacement are perpendicular to each other.

Question.6 The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why? [SAII-2010] Answer. It does not violate the law of conservation of energy. Whatever, is the decrease in PE due to loss of height, same is the increase in the KE due to increase in velocity of the body.

Question.7 What are the various energy transformations that occur when you are riding a bicycle? Answer. The chemical energy of the food changes into heat and then to muscular energy. On paddling, the muscular energy changes into mechanical energy.

Question.8 Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going? Answer. Energy transfer does not take place as no displacement takes place in the direction of applied force; the energy spent is used to overcome inertia of rest of the rock.

Question.9 An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object ? [SAll-2011 ] Answer. Since the body returns to a point which is on the same horizontal line through the point of projection, no displacement has taken place against the force of gravity; therefore, no work is done by the force due to gravity.

Question.10 A battery lights a bulb. Describe the energy changes involved in the process. Answer. Within the electric cell of the battery the chemical energy changes into electrical energy. The electric energy on flowing through the filament of the bulb, first changes into heat energy and then into the light energy.

Question.11 What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer. Answer. The work done by the force of gravity on the satellite is zero because the force of gravity acts at right angles to the direction of motion of the satellite. Therefore, no displacement is caused in the direction of applied force. The force of gravity only changes the direction of motion of the i satellite.

Question.12 Can there be displacement of an object in the absence of any force acting on it? Think; discuss this question with your friends and teacher. [SAII-2012] Answer. The answer is. both Yes and No. Yes, because when an object moves in deep space from one point to another point in a straight line, the displacement takes place, without the application of force. No, because force cannot be zero for displacement on the surface of earth. Some force is i essential.

Question.13 A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not ? justify your answer. Answer. The person does not do work because no displacement takes place in the direction of applied force as the force acts in the vertically upward direction.

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Question.15 Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her ? Why ? Answer. Yes, we do agree when the number of forces act on a body, such that they constitute balanced forces, then net force acting on the body is zero. In such a situation no acceleration acts on the object.

Question.16 A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy ? [SAI1-2011] Answer. The KE on reaching the ground changes into heat energy, sound energy etc. and, therefore, gets dissipated in air.

Question.17 What kinds of energy transformations take place at a thermal power station ? Answer. At a thermal power station, the chemical energy of coal is changed into heat energy which is further changed into electrical energy with the help of an electric generator.

Question.18 Name the transformation of energy involved in the following cases : (a) When a body is thrown upwards. (b) When a body falls from the top of a hill. (c) When coal burns. (d) When a gas bums. (e) When water falls from a height. Answer. (a) Kinetic energy into potential energy. (fa) Potential energy into kinetic energy. (c) Chemical energy into heat energy. (cf) Chemical energy into heat energy. (e) Potential energy into kinetic energy.

Question.19 What are the factors on which the work done depends ? [SAII-2010] Answer. The work done by a force depends upon: (i) The magnitude of the force. (ii) The magnitude of the displacement and (iii) The angle between force and displacement.

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Question.21 What is the work done by a coolie walking on a horizontal platform with a load on his head ? Answer. In order to balance the load on his head, the coolie applies a force on it in the upward direction, equal to its weight. His displacement is along the horizontal direction. Thus, the angle between force F and displacement is 90°. Therefore, work done W = FS cos θ = FS cos 90° =0.

Question.22 We wind our watch once a day, what happens to the energy ? Answer. When we wind our watch, we wind the spring inside the watch. As a result, energy is stored in the spring in the form of elastic potential energy. This elastic potential energy is used to make the watch work the whole day. .

Question.23 What is the amount of work done by a force when a body moves in a circular path ? [SAll-2012] Answer. Work done is given by the expression W = FS cos θ. When a body moves in a circular path, then the displacement (S) is zero. Therefore, work done is W = F x 0 = 0.

3 MARKS QUESTIONS Question.1 Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’ (i) Suma is swimming in a pbnd. (ii) A donkey is carrying a load on its back. (iii) A wind-mill is filling water from a well. (iv) A green plant is carrying out photosynthesis. (v) An engine is pulling a train. (vi)Food grains are getting dried in the sun. (vii) A sailboat is moving due to wind energy. Answer. (i) Work is done because the displacement of swimmer takes place in the direction of applied force. (ii) If the donkey is not moving, no work is done as the displacement of load does not take place in the direction of applied force. (iii) Work is done, as the displacement takes place in the direction of force. (iv) No work is done, because no displacement takes place. (v) Work is done, because displacement takes place in the direction of applied force. (vi)No work is done, because displacement does not take place. (vii)Work is done because displacement takes place in the direction of the force.

Question.2 Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest ? What happens to its energy eventually ? Is it a violation of the law of conservation of energy ? Answer. When the pendulum bob is pulled (say towards left), the energy supplied is stored in it is the form of PE on account of its higher position. When the pendulum is released so that it stars moving towards right, then its PE changes into KE, such that in mean position, it has maximum KE, and zero PE. As the pendulum moves towards extreme right, its -KE changes into PE such that at the extreme position, of has maximum PE and zero KE. When it moves from this extreme position to mean position, its PE again changes to KE. This illustrates the law of conservation of energy. Eventually, the bob comes to rest, because during each oscillation a part of the energy possessed by it transferred to air and- m overcoming friction at the point of suspension. Thus, the energy of the pendulum is dissipated in air. The law of conservation of energy is not violated because the energy merely changes its form and is not destroyed.

important-question-for-cbse-class-9-science-work-power-and-energy-4

Question.4 Distinguish between work, energy and power. State the SI units for each of these quantities. Answer. Work: It is defined as the product of force applied and the distance moved by the body on the application of the force. In SI it is measured in joule. Energy : It is defined as the capacity of a body to do work. In SI it is measured in joule. Power: It is defined as the rate of doing work. It measures how fast or slow the work is done. In SI it is measured in watt.

important-question-for-cbse-class-9-science-work-power-and-energy-6

Question.4 (a)Derive an expression for kinetic energy of a body having mass m and moving with a velocity v. (b)When velocity of a body is increased 5 times, what is the change in its kinetic energy ? (c)Two masses m and 2m are dropped from heights h and 2h. On reaching the ground, which will have greater kinetic energy and why ? [SAII-2013] Answer. (a)For derivation see above questions. (b)Kinetic energy is given by the expression KE = 1/2 mv 2 , therefore, if velocity is made 5 times KE will increase by 25 times. (c)More the potential energy more will be the kinetic energy of the body when it falls. Hence, the body with mass 2m will have greater kinetic energy as it has more potential energy.

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APPLICATION BASED QUESTIONS

Question.4 When an arrow is shot from its bow, it has kinetic energy. From where does it get the kinetic energy ? Answer. A stretched bow possesses potential energy on account of a change in its shape. To shoot an arrow; the bow is released. The potential energy of the bow is converted into the kinetic energy of the arrow.

Question.5 A spring which has been kept compressed by tying its ends together is allowed to be dissolved in an acid. What happens to the potential energy of the spring ? Answer. The PE of the spring gets converted into KE of acid molecules whose temperature rises.

Higher Order Thinking Skills (HOTS) Questions

Question.1 justify giving proper reasoning whether the work done in the following cases is positive or negative : (a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) Work done by gravitational force in the above case. (c) Work done by friction on a body sliding down an inclined plane. (d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity. (e) Work done by resistive force of air on a vibrating pendulum in bringing it to rest. Answer. (a) Work done is positive as the bucket moves in the direction of force applied by the man. (b) Work done by the gravitational force is negative, as the bucket moves upwards i.e., opposite to the gravitational force. (c) Work done is negative, as frictional force acts opposite to the direction of motion of the body. (d) Work done is positive because applied force acts along the same direction as the direction of motion of the body. (e) Work done is negative because the resistive force of air always acts opposite of the direction of motion of the vibrating*pendulum.

Question.2 What is the work done by a coolie walking on a horizontal platform with a load on his head? [SAII-2010] Answer. In order to balance the load on his head, the coolie applies a force on it in the upward direction, equal to its weight. His displacement is along the horizontal direction. Thus, the angle between force F and displacement is 90°. Therefore, work done W = FS cos θ = FS cos 90° = 0.

Question.3 The work done in lifting a box on to a platform does not depend upon how fast it is lifted up. Explain your answer giving proper reasoning. [SAll-2013] Answer. The work done (W) in lifting a box through a distance (S) against the gravitational force (F) is given by W = FS. Hence, it is obvious that it is independent of the rate at which the box is lifted.

Question.4 Is it possible that a body be in accelerated motion under a force acting on the body, yet no work is being done by the force ? Explain your answer giving a suitable example. Answer. Yes, it is possible, when the force is perpendicular to the direction of motion. The moon revolving round the earth under the centripetal force of attraction of the earth, but earth does no work on the motion.

Question.5 A body moves along a circular path. How much work is done in doing so? Explain. Answer. In case of a body moving along a circular path, the force (centripetal) is always along the radius while displacement is tangential. Hence, work done W = FS cos 90° = 0 as angle between F and S is 90°.

Question.6 A man rowing a boat upstream is at rest with respect to the shore. Is he doing work ? [SAII-2012] Answer. The man is doing work relative to the stream because he is applying force to produce relative motion between the boat and the stream. But he does zero work relative to the shore as the displacement relative to the shore is zero.

Question.7 What type of energy is stored in the spring of a watch? [SAII-2013] Answer. When we wind a watch, the configuration of its spring is changed. The energy stored in the spring is obviously potential in nature (elastic potential energy to be more accurate).

important-question-for-cbse-class-9-science-work-power-and-energy-15

Question.10 A spring which is kept compressed by tying its ends together is allowed to be dissolved in an acid. What happens to the potential energy of the spring ? Answer. The potential energy of the spring gets converted into heat energy (kinetic energy of acid molecules). Due to this heat, the temperature of the acid rises.

important-question-for-cbse-class-9-science-work-power-and-energy-17

Question.2 (a) How much work is done when a force of 1 N moves a body through a distance of 1 m in its direction? (b) Is it possible that a force is acting on a body but still the work done is zero? Explain giving one example. [SA II – 2011] Answer. (a) 1 J of work is done. (b) Yes, it is possible when force acts at right angles to the direction of motion of the body. Example Gravitational force of earth acts on a satellite at right angles to its direction of motion.

Question.3 (a) What is meant by potential energy? Is potential energy vector or scalar quantity? (b) Give one example of a body having potential energy. [SA II – 2011] Answer. (a) The energy possessed by a body by virtue of its position or configuration. It is a scalar quantity. (b) Stretched string of a bow.

Question.4 When is the work done by a force said to be negative? Give one situation in which one of the forces acting on the object is doing positive work and the other is doing negative work. [SA II – 2012] Answer. We know that work done W = FS cos 0, where 0 is the angle between F and S. Clearly, W will be -ve, if 0 is between 90° and 180° because then cos 0 will be -ve. Consider the case of a body falling under gravity. The body experiences an upward frictional force and downward force due to gravity. Since the body is moving downwards, the work done by force to gravity will be +ve but that is against the upward thrust will be -ve.

important-question-for-cbse-class-9-science-work-power-and-energy-18

Important Topics/Areas/Questions which are frequently asked in the examination

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Question.6 Define : (a) power (b) work done (c) kinetic energy. Give SI unit of each. [SAII-2014] Answer. (a) The rate of doing work is called power. Its SI unit is watt. (b) Work is the product of force and displacement. Its SI unit is joule. (c) It is the energy possessed by a body by virtue of its motion. Its SI unit is joule.

important-question-for-cbse-class-9-science-work-power-and-energy-24

NUMERICAL PROBLEMS

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Question.12 A man weighing 70 kg carries a weight of 10 kg to the top 6f a tower 100 m high. Calculate the work done. Solution. We know that work done is given by W = FS = mgh i.e., change in potential energy, therefore, we have W = mgh = (70 + 10) x 9.8 x100 = 78400 J

important-question-for-cbse-class-9-science-work-power-and-energy-38

Question.16 Which would have greater effect on kinetic energy of an object – doubling the mass, or doubling the velocity? Solution. We .know that KE∝ m, KE ∝ v2 Therefore, by doubling the mass, the kinetic energy doubles, while by doubling the velocity, the kinetic energy increases four times. Therefore, doubling the velocity will have a greater effect on the kinetic energy of an object.

important-question-for-cbse-class-9-science-work-power-and-energy-41

Question.18 A crane pulls up a car weighing 500 kg to a vertical height of 4 m. Calculate the work done by the crane. Solution. In order to lift the car, the crane has to do work against the force of gravity. Therefore, the force required = Mg = 500 x 9.8 = 4900 N Now, displacement undergone by the car, S = 4 m Hence, work done = FS = 4900 x 4 = 19600 J

Question.19 A force of 10 N displaces a body by a distance of 2 m at an angle 60° to its own direction. Find the amount of work done. [SAII-2012] Solution. By definition, Work = Force x Displacement in the direction of force = FD cos θ Given, F = 10 N;S = 2m;θ = 60°. Therefore, W = 10 x 2 x cos 60° = 10 x 2 x 1/2 =10J

important-question-for-cbse-class-9-science-work-power-and-energy-42

Question.24 If you apply 1 J of energy to lift a book of 0.5 kg, how high will it rise? [Take g = 10 ms -2 ] Solution. We know that PE = mgh 1 = 0.5 x 10 x h Therefore, h = 0.2 m = 20 cm

important-question-for-cbse-class-9-science-work-power-and-energy-46

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Important Questions for CBSE Class 9 Science Chapter 11 – Work and Energy

Home » CBSE » Important Questions for CBSE Class 9 Science Chapter 11 – Work and Energy

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Work and Energy Class 9 Important Questions for Class 9 Science

Science is an important subject we study in school. We need Science in every aspect of our life. From technology to medicines, it has improved human lives. In this chapter, students will study work and energy.

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In Physics, energy is the ability to do work. Now the question is, what is work? Work is the measure of energy transferred to or from an object by an external force that causes displacement of the object. That means work is done only when there is a displacement of the object. The general sense of “work” is different from how physics defines it. Students must practise questions from this chapter to score better in exams.

Extramarks is a leading company that provides all the important study materials. Our experts have made the Important Questions Class 9 Science Chapter 11 to help students. They have collected the questions from sources such as the textbook exercises, CBSE sample papers , CBSE past years’ question papers , NCERT Exemplars, and important reference books. They have also solved the questions so that students can follow the answers. Experienced professionals have further checked the answers to ensure the best quality of the content.

Extramarks is a leading company that provides all the important study materials related to CBSE and NCERT. You can download the study materials after registering on our official website. You can find CBSE syllabus , CBSE extra questions , CBSE revision notes , CBSE sample papers, CBSE past years’ question papers, CBSE sample papers, CBSE past years’ question papers, NCERT books, NCERT exemplars, NCERT solutions, NCERT important questions, vital formulas, and many more.

Get Access to CBSE Class 9 Science Important Questions 2022-23 with Solutions

Also, get access to CBSE Class 9 Science Important Questions for other chapters too:



1	Chapter 1
2	Chapter 2
3	Chapter 3
4	Chapter 4
5	Chapter 5
6	Chapter 6
7	Chapter 7
8	Chapter 8
9	Chapter 9
10	Chapter 10
11	Chapter 11	Work and Energy
12	Chapter 12
13	Chapter 13
14	Chapter 14
15	Chapter 15

Work and Energy Class 9 Important Questions and Answers

The experts at Extramarks have made this question series to help students prepare. They have collated the questions from the textbook exercise, CBSE sample papers, CBSE past years’ question papers, NCERT Exemplars, and important reference books. They have also solved the questions so that students can follow the answers, while experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Important Questions Class 9 Science Chapter 11 will help students score better in exams. The important questions are-

Question 1. Explain work. Give the SI unit of work. Enumerate, write briefly about positive, negative, and zero work. Answer 1: When the object is displaced from its position in the direction of force due to the application of force to the object, then the work is said to be done. The Newton meter, also known as the joule, is the SI unit of work.There are two scientific concepts of work. First of all, there should be a force acting on the object, and second, the object must be displaced.

Work is a scalar quantity with only one magnitude.Work has no direction.

Let us suppose that F is the force and s is the displacement.

When the force and displacement are in the same direction, which means the angle between the direction of force and the displacement is 0 degrees, then the work done is positive. This basically means that the work is done by force.

As an example, consider a boy pulling an object towards himself.

W = + F × s

When the force and the displacement are in the opposite direction, which means the angle between the direction of force and the displacement is 180 degrees, then the work done is negative. This means that the work is done against the force.

In this case, W = – F × s

When the force and the displacement are in the perpendicular direction, which means the angle between the direction of force and the displacement is 90 degrees, then the work done is zero. For example, if a coolie is carrying the load on his head, the force of gravity is acting vertically downward and the displacement is along the horizontal direction, this shows that the force and the displacement are perpendicular to each other. So in this case, it could be said that the work done by the gravitational force is zero. This implies that W = 0.

Question 2. Explain energy. Give an example and mention its SI unit.

Answer 2: Energy is defined as the ability to do work. For performing any mechanical work, energy is always required. An object that is capable of doing work is said to possess energy. When the object does the work, it loses energy and the body on which the work is done gains energy. So it could be said that the energy of an object is measured in terms of how much capacity it has to do work.

The SI unit of energy is the joule. It is the same as that of the work. In order to do 1 joule of work, you need 1 joule of energy. A kilojoule (kJ) is the largest unit of energy.

1 kilo joule = 103 J

It is always important to remember that the work done against the force is stored as energy.

For example,

In cricket matches, when the fast-moving ball hits a stationary wicket, the wicket is thrown away.
When a hammer is raised and falls on a nail that is placed on a piece of wood, it drives the nail into the wood.

Question 3. Answer the following questions.

Enumerate the forms of energy?
What is kinetic energy?
Give the formula to calculate kinetic energy.
What is the work energy theorem?
Calculate the kinetic energy of a bullet of mass 8 g which is fired with a velocity of 80 m/s.
Let us suppose that the weight of a body has mass 5 kg, it has a velocity of 10 m/s and is moving in a straight line with the acceleration of 20 m/s 2 . Calculate the kinetic energy after 10 seconds.
If the mass of the van is 2000 kg, then calculate the work done to increase its velocity from 10 m/s to 20 m/s.
The forms of energy are:
Mechanical energy
Heat energy
Chemical energy
Electrical energy
Light energy
Kinetic energy is defined as the energy which is possessed by an object due to its motion.

According to the equation of motion,

V2 – u2 = 2as

S = v2 – u2/2a

Here, a is the uniform acceleration

u is the initial velocity

v is the final velocity.

So on substitution of the values, we get,

W = ma.v2 -u2/2a = ½ m(v2 – u2)

If initial velocity = 0 then

Since work done is equal to the kinetic energy, so

According to the work energy theorem,

Total work is equal to the change in kinetic energy.

Mass of the bullet = 8g

To convert it to kg, we get

8g = 8/1000 kg

Velocity = v = 80 m/s

To calculate the KE and applying the formula, we get

KE of bullet = ½ mv2 = ½ × 8/1000 × 80 × 80

= ½ × 8/1000 × 1600 = 25.6 J

Mass of the body = m = 5 kg

Initial velocity = u = 10 m/s

Acceleration = a = 20 m/s2

Velocity = v = ?

We will use two formulas here,

First, we will calculate the v using the formula of equation of motion,

V = (u + at) m/s

We know the formula for kinetic energy,

KE = ½ m × (u + at)2

From both the above equation we will get,

KE = ½ × 5 × (10 + 20 × 10 )2

KE = ½ × 5 × 210 × 210 = 110250 J

As per given in the question,

Mass = m = 2000 Kg

v1 = 10 m/s

v2 = 20 m/s2

To calculate the initial KE of the van,

KE1 = ½ mv2

KE = ½ × 2000 kg × (10 m/s)2

KE = 100000 J = 100 kJ

In order to calculate the final kinetic energy of the van, we will use the equation as,

KE2 = ½ mv2

KE2 = ½ × 2000 × (20 m/s)2 = 400000 J = 400 Kj

The work done will be calculated as,

Work done = change in KE = 400 kJ – 100kJ = 300kJ

So, it could be said that the KE of the van increases by 300 kJ when it speeds up from 10 m/s to 20 m/s.

Question 4. Answer the following questions in short.

What is the SI unit of kinetic energy?
Give examples of kinetic energy.
What is potential energy?
What is the SI unit of potential energy?
Give examples of potential energy.
What is the formula for potential energy?
State the law of conservation of energy.
Mention some important results which can be derived from the formula of the kinetic energy.
The SI unit of kinetic energy is Joules.
Some of the examples of kinetic energy are: a target pierced by a bullet fired from a gun, a hammer which drives a nail into the wood, a running horse and a flowing river. All these have kinetic energy.
The energy which is acquired or possessed by the body due to the change in position or shape is known as potential energy.
The SI unit of potential energy is Joules.
Some of examples of potential energy are like water stored in a dam and a stone lying on the roof of a building has potential energy due to its height. A wound spring of a watch has potential energy due to the change of its shape.
Potential energy or PE = mgh. The work done is equal to the gain in energy of the body.
According to the law of conservation of energy,

The energy can neither be created nor destroyed. It can only be transformed from one form to the other. It is very important to consider that the total energy before and after transformation remains constant always.

Some of the important results which could be derived from the formula of kinetic energy are:
Kinetic energy of KE is doubled when the mass of an object is doubled.
The KE gets halved when the mass of the object is halved.
KE becomes the four times when the speed of an object is doubled.
KE becomes one-fourth when the speed of the object is halved.
The heavy objects move at high speed so they have more KE.

Question 5. An object is given as shown in the figure. The force of 7 N acts on the object, the displacement is 8 m . and it is in direction of the force. Let us suppose that the force is acting on the object through the displacement. In this case and the condition depicted in the figure, calculate the work done.

As shown in the figure above,

Work done = W = Fs

Here the work done is positive because the force and the displacement are in the same direction.

W = 7 × 8 = 56 J

Question 6. When do we say that the work is done?

Answer 6: The work is said to be done when the force acting on the body causes some displacement. So the work is done by the force on the body.

Question 7. Present the equation of work done when the force is acting on the object in the direction of the displacement.

Answer 7: Work done will be positive because the force and displacement are in the same direction.

Here F is the force and s is the displacement.

Question 8. What is 1 Joule of work?

Answer 8: This is the amount of work done on the object when a force of 1 Newton displaces it by 1 metre along the line of action of the force. This is called 1 Joule of work.

Question 9. A force of 140 N is exerted by a pair of bullocks on a plough. The field is ploughed 15 m long. Calculate how much work is done in ploughing the length of the field.

Answer 9: As given in the question,

Force = F = 140 N

Displacement = s = 15 m

Work done = W = + Fs

F and s are in the same direction.

W = Fs = 140 × 15 = 2100 J.

Question 10. Suppose that the kinetic energy of an object of mass m moving with a speed of 5 m/s is 25 J. When the velocity is increased three times what will be the kinetic energy?

Mass of the body = m

Initial velocity = vi

KEi = ½ mvi2

The values given in the question are,

KEi = 25 J, vi = 5 m/s

25 J = ½ × m × 5 × 5

m = 25 × 2/ 5 × 5

final velocity is becoming three times of initial velocity

vf = 3vi = 3 × 5 = 15 m/s

KE = ½ × m ×vf2 = ½ × 2 × 15 × 15 = 225 J

Question 11. Explain power. What is 1 watt of power? Calculate the power if the lamp consumes 1000 J of electrical energy in 10 s.

Answer 11: The power is defined as the rate of doing work or the rate at which the energy is transferred or used or transformed to other forms.

Suppose the work done is W in time t, then

Power = work/time = W/t

The unit of power is Watt or W. This is in honour of James Watt.

Kilowatt is used to express a larger rate of energy transfer.

1 Watt = 1 J/s

1kW = 1000 W = 1000 J/s

1 MW = 106 W

1 horsepower or HP = 746 W

To solve the problem in the given question,

Energy = 1000 J

Work done = W = 1000 J

Time = t = 10 s

Power of lamp = W/t = 1000/10 = 100-watt

Question 12. What is the average power?

Answer 12: The ratio of total work done to the total time taken is known as the average power.

Average power = total energy consumed/total time taken

Question 13. Justify the given question with the reason. What is the work done by the force of gravity on a satellite moving around the earth?

Answer 13: Work done in this case will be zero. This is because the angle between the force and the displacement in case of circular motion so the work done is zero.

Question 14. Multiple choice questions:

Various factors are given below, find out the one on which the work done on an object does not depend.
Displacement
Force applied
The angle between force and displacement
Initial velocity of the object

Answer: (d) Initial velocity of the object

Explanation:

Work =W = F. d cosϴ

In the above equation, W is the work done

F is the force which is applied on the object. The displacement is d and ϴ is the angle between the force and the displacement. So it is clear that the work done on an object does not depend upon the initial velocity of the object.

What about the total energy of the body when it falls freely towards the earth?
Remain constant
It first increases but then it decreases.

Answer: c) Remains constant

As we know, the total energy of the system is always conserved. So when the object falls freely towards the earth, the energy remains constant. This means that the sum of the potential energy and the kinetic energy of the body would be the same at all the points.

Which type of energy is possessed by the water stored in a dam?
Kinetic energy
Potential energy

Answer: (d) Potential energy.

Water stored in the dam has potential energy because potential energy is stored energy, or, in other words,it could be said that it is the energy of position.

In the case of negative work, what is the angle between the force and the displacement?

Answer 4: (d) 180 degree

A girl is carrying a school bag which has a mass of 3 kg on her back. The girl moves 200 m on a levelled road. If g = 10 m/s 2 then what is the work done against the gravitational force?

Answer 5: (d) Zero

An electric bulb of 60 W burns for 5 hours a day. The electricity cost in 30 days of the month is at Rs 3 per unit is

Answer 6: (b) 27

Cost of electricity = P × t × cost per kW

= 0.06 kW × (5 × 30) × 3 = 27

A tank has water which has the mass m to a height h. Then the potential energy associated is
this depends on how much radius the tank has

Answer: (b) mg h/2

This is the potential energy calculated based on the height will be up to the centre of gravity of tank.

What is the total energy at any point on the path when the mass m falls from a height h.
It depends on the height

Answer: C – mgh

This is according to the law of conservation.

Question 15. An example of some of the activities is given below. Just find out whether the work has been done or not.

Aryan is swimming in a pond.
A donkey is carrying the load on the back.
A windmill is lifting water from the well.
A green plant is carrying out photosynthesis.
An engine is pulling a train
Food grains are getting dried in the sun
A sailboat is moving due to wind energy.
In this example, work is done because Aryan gets displaced due to the reaction force of water while swimming in the pond.
The donkey is moving in the horizontal direction, and the force is being applied to the load in the vertical direction, so in this case the work done is zero.
In this example, the windmill is doing the work of lifting water, and the moving blades of the windmill displace the water from the well in the direction of upwards.
In this case, no work is being done because there is no displacement and no force is acting .
The engine is exerting force, and the train is being displaced, so the work is being completed in this case.
As seen, force and displacement are zero, so no work is done.
The sailboat is moving due to the applied wind force, so the work is completed in this case.

Question 16. Describe the energy changes involved when the battery lights the bulb.

Answer 16: In this case, chemical energy is converted to electrical energy. So the final result is the combination of heat energy and light energy.

Question 17. Calculate the work done by the force when the force is acting on a mass of 20 kg and there is a change in velocity from 5 m/s to 2 m/s.

Answer 17: In the question, the following values are given,

Mass = m = 20 kg

Work = Change in Kinetic energy = ½ m (v2 – u2) = ½ × 20 × (22 – 52) = -210 J

This shows that the work is done in the opposite direction to the displacement of mass.

Question 18. Is there any violation of the law of conservation of energy when the potential energy of a freely falling object decreases progressively?

Answer 18: No, in this case, the law of conservation of energy is not violated because a freely falling body gains kinetic energy in an amount that is equal to the loss of potential energy. So overall, the total energy of the object remains conserved.

Question 19. When riding a bicycle, what type of energy transformation occurs?

Answer 19: While riding the bicycle, there is a conversion of the muscular energy of the rider to the kinetic energy of the bicycle.

Question 20. While pushing a heavy rock, you might fail to move it, so in this case, is there any transfer of energy that takes place? And the energy you spent, where did it go?

Answer 20: Yes, the energy that is used up is responsible for causing muscular contractions, and it is dissipated in the form of heat or sweating as well.

Question 21. In a house, 250 units of energy are consumed in a month. So how much energy is this in Joules?

Answer 21: As per the question,

Units consumed = 250 = 250 kWh

1 unit = 1 kWh = 3.6 × 106 J

250 kWh = 250 × 3.6 × 106 J = 9 × 108 J

Question 22. A 40 kg object is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half – way down.

Answer 22: m = 40 kg, h = 5 m, g = 9.8 m/s2

Potential energy = mgh = 40 × 5 × 9.8 m/s2 = 1960 J

When the height becomes half, potential energy gets halved = 1920/2 = 980 J

Loss in potential energy is equal to the gain in kinetic energy equal to 980 J.

Question 23. When there is no force acting on the object, can there be any displacement of the object? Clarify the answer.

Answer 23: If the object moves with constant velocity, then according to Newton’s first law of motion, the net force acting on it is zero. This states that an object will remain in a state of rest or of uniform motion in a straight line unless a net external unbalanced force acts on it. Here the force acting is zero, but the displacement is non – zero.

Question 24. A woman is holding a bundle of hay over her head for 30 minutes, and she gets tired. Has she done any work or not? Justify your answer.

Answer 24. The displacement of the person is zero, so this means that the person has not done any work.

Question 25. How much energy does an electric heater use in 10 hours when the heater is rated at 1500 watts? Answer 25: Power = P = 1500 W

On conversion P = 1.5 kW, t = 10 hour

Energy = Pt = 1.5 kW × 10 h = 15 kWh

Question 26. An object of mass m is moving with a constant velocity v. calculate the work done on the object in bringing it to rest.

Answer 26: Work = ½ × mass × (velocity)2

Question 27. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Answer 27: Mass = m = 1500 kg

Velocity = v = 60 km/h = 60 × 5/18 m/s

Work done = kinetic energy = ½ mv2

Work done = ½ × 1500 × ( 60 × 5/18 ) × (60 × 5/18) = 208333.3 J = 208.3 kJ

Question 28. What is the difference between a kilowatt and a kilowatt hour? Suppose a waterfall has a height of 20 m and 20,00 tonnes of water falls from it in a minute. If this energy is utilised then how much is the equivalent power? Calculate.

Answer 28: Kilowatt is the unit of power, while kilowatt hour is the unit of electrical energy consumed.

m = 2000 × 103 kg

t = 1 min = 60 s

Power of falling water = mgh/t

= 2000 × 103 × 10 × 20/ 60

= 6.67 × 106 W or 6.67 MW

Question 29. How can power be related to the speed at which a body could be lifted? A man is working at a power of 100 W. How many kilograms will he be able to lift at a constant speed of 1 m/s 2 . The value of g is 10 m/s 2 .

We know the formula that,

Power = force × velocity

P = 100 Watt, v = 1 m/s2

F = P/v =100N

So, m = F/g = 100N/10 ms2

The answer is 10 kg can be lifted by the man.

Question 30. Compare the power for (a) and (b). These are moving upward against the force of gravity.

A butterfly has a mass of 1 gram and is flying upward at a rate of 0.5 m/s.
A 250 g of a squirrel climbing up a tree at a rate of 0.5 m/s
P = Fv = mgv

m = 1 g = 10-3 kg, v = 0.5 m/s

Power of the butterfly will be calculated as

P = 10-3 × 10 × 0.5 = 5 × 10-3 = 0.005 Watt

Mass = m = 250 g = 250 × 10-3 kg

v = 0.5 m/s

Power of squirrel = 250 × 10-3 × 10 × 0.5

P = 1250 × 10-3 = 1.25 watt

The answer could be written as the power of squirrels is more than that of the butterfly.

Question 31. When the momentum of the object is zero, can the object have any mechanical energy?

Answer 31: Yes, it is possible. When a body is thrown up, potential energy in the form of mechanical energy is non-zero at the highest point of a projectile, even if kinetic energy is zero due to the zero velocity.

Question 32. Is it possible for the object to have momentum even if its mechanical energy is zero? Explain

Answer 32. No, because when energy is zero, both potential energy and kinetic energy are zero. So this makes both the momentum and the velocity zero.

Question 33. The motor pump has a power of 2kW. The pump can raise how much water per minute to a height of 10 m?

Answer 33: As per the question,

m is raised to the height h

PE gained = mgh

Power = work done / time = mgh/t

= 2 × 103/10 × 10 = 20 kg/s

Mass of water pumped per minute

= 20 × 60 = 1200 kg.

Question 34. For the work to be done on an object list two conditions which need to be satisfied.

Answer 34: There are two conditions here, a force should act on the object and there should be a displacement in the object.

Question 35. A few situations are given below. Analyse them and comment on where work is said to be done. Mention the answer with an explanation.

A man is pushing the rock, which is hard and huge, but it is not moving.
A bullock is pulling a cart up to 1 km on the road.
A girl is pulling a trolley for about 2 m distance.
A person standing with a heavy bag on his head.
Since the displacement is zero, the work done is also zero.
In this situation, the work done is positive.
In this case, since the displacement is zero, the work done is also zero.

Question 36. Define what is 1kWh?

Answer 36: If a 1 kW device is operated for 1 hour, the energy consumed by the device is 1 kWh.

Question 37. Suppose an object is lifted by the crane in the time t to a height of h. What is the relation between the power of a crane to the speed at which it is lifting the object?

Answer 37: power = work done / time taken

Mass = m, height = h

Work done = mgh

P = mgh / t = mg (h/t)

Is the speed at which the body is lifted, v = h/t

Since it is distance / time

Force = F = mg

Power = (mg) (h/t)

Power = mg v = Fv

Question 38. An electric iron of 1600 W is used for 45 minutes per day. Find the amount of electric energy consumed in the month of March.

Answer 38: Power = P = 1600 W = 1.6 kW

t = 45 = 45/60 h

The time of consumption in March will be calculated as follows: March has 31 days

T = (45/60 × 31) h

Energy consumed = PT = 1.6 × 45/60 × 31 = 37.2 units

Question 39. How much work is done by the moon when the moon is experiencing a gravitational force because of the earth and it is revolving around the earth in a circular orbit.

Answer 39: In this case, the work done by the moon will be zero because, towards the centre of the orbit, centripetal force is acting and the direction of the displacement is tangential to the orbit.

Question 40. What are the conditions under which the work done is zero.

Answer 40. The work done by body is zero, if

There is no displacement occurring despite the application of force.
W = Fs cos ϴ and cos 90 degree = 0, so displacement in the body is perpendicular to the applied force.

Question 41. How much work is done by the person when a 2 m high person is holding a trunk on his head that has a weight of 25 kg and is standing at roadway bus terminals.

Answer 41: There are various conditions mentioned in the question, but there has been no displacement, so the work done in this case will be zero.

Question 42. State whether the given statement is true or false.

A student is studying hard for the exam and is doing more work than the baby who is crawling from one corner of the room to the other corner of the room.
Energy is converted from one form to the other but only in a few cases.
Work done is a vector quantity.
Energy is a scalar quantity while momentum is a vector quantity.
A moving truck has more kinetic energy than a bicycle which is moving with the same speed.
Horse power is used to measure the power of heavy machines.
False. Here the work done by the student is zero because there is no displacement. Baby is crawling and is displaced from one corner to the other.

Question 43. Fill in the blanks:

Work done can be__________, _________ or___________.
_________work is done by the satellite when it is revolving around the earth.
60 W power means the device uses 60 Joule of energy is _________hour.
When non-zero force and displacement are observed with zero work done, the angle between the force and displacement is __________.
Engine require fuel like for example petrol and diesel. The combustion of fuel provides ____________ to them.
Living beings and machines as well need _________ to work.
Living beings get energy from _________
Machines get energy from __________ and _________
Work done can be positive, negative, or zero.
Zero work is done by the satellite when it is revolving around the earth.
60 W power means the device uses 60 Joule of energy is 1/3600 hour.
When non-zero force and displacement are observed with zero work done, the angle between the force and displacement is 90 degrees
Engine require fuel like for example petrol and diesel. The combustion of fuel provides energy to them.
Living beings and machines as well need energy to work.
Living beings get energy from food
Machines get energy from electricity and fuel

Question 44. Answer the following questions:

1.A raised hammer possesses which type of energy?

What happens to the potential energy in each case when the slinky is compressed or stretched.

3.A moving car possesses which type of energy?

What happens to the rubber band when it is stretched, mention about the work done.
Which type of kinetic energy is possessed by the running athlete.
energy is possessed by the different objects due to their motion. Give examples in this case.
Which is the term used to sum of kinetic energy and potential energy.
1 calorie equals how many joules?
I kilojoule equals how many joules?
What is the CGS unit of energy and how it is related to the SI unit.
Two persons A and B are asked to complete a work in 1 hour and 2 hours respectively. Who is more powerful and why?

1. The raised hammer possesses potential energy.

The potential energy of the slinky will increase in each case.
The type of energy by a moving car is kinetic energy.
Suppose a rubber band is stretched, work done on the rubber band is stored as its potential energy since it is not used to cause a change in speed or velocity of any object.
The energy possessed by a running athlete is the kinetic energy.
Few of the examples are falling coconut, speeding car, flying aircraft, flowing water, etc.
The sum of the kinetic energy and the potential energy is the mechanical energy.
1 calorie is equal to the 4.186 Joule.
1 kilojoule is equal to the 103 Joule.
The CGS unit of energy is erg. The value of 1 Joule equals 107 erg.
Person A is more powerful because he does work in less time as compared to B.

Question 45. At what position the potential and kinetic energies are going to be maximum in an oscillating pendulum?

Answer 45: In this case of an oscillating pendulum, the potential energy will be maximum at the extreme position and the kinetic energy will be maximum at the mean position.

Question 46. The mass of a body is 1 kg, at what speed it will have the kinetic energy of 1 Joule?

Answer 46: Kinetic energy = 1 Joule

Mass = m = 1 kg

1 = ½ × 1v2

This implies that

Question 46. Mention the change in the kinetic energy when the speed of the body is halved?

From the above equation it is clear that the KE is directly proportional to the v2

So as mentioned if v is v/2, then v2 will become v2/4

Question 47. A horse and a dog are running at the same speed. The horse has a mass of 210 kg and the dog has a mass of 25 kg. Which out of these two have more kinetic energy and how?

Answer 47: As we know, kinetic energy and mass are directly proportional to each other.

Mass of the horse is greater than the mass of the dog.

So it could be said that horses possess greater kinetic energy.

Question 48: The examples are given below. Mention the type of energy transformation.

Arrow is released from the bow.
Solar cell use.
Electric motor
Electric generator
Steam engine
Electric bulb
Potential energy of the bow string will convert to the kinetic energy of the arrow.
Light energy will get converted to electrical energy.
Electrical energy will get converted to mechanical energy.
Mechanical energy will get converted to electrical energy.
Heat energy will get converted to kinetic energy.
Electrical energy will get converted to light energy.
Chemical energy will get converted to electrical energy.

Question 49. A force of 10 N acts on a body which has a mass of 2 kg for a period of 3 seconds. Find out the kinetic energy acquired by the body in 3 seconds.

Answer 49:

Kinetic energy = work done on the body

Mass = m = 2 kg

Time = t = 2 seconds

u = 0 and F = 10 N

acceleration = a = F/m = 10/2 = 5 m/s2

displacement = s = ut + ½ at2

= 0 + ½ × 5 × 3 × 3 = 22.5 m

Work done = Fs = 10 N × 22.5 m = 225 J

Question 50. What is the force which is applied by the brakes of the truck which has a mass of 1800 kg and is moving with a speed of 54 km/h. When the brakes are applied it stops with a uniform acceleration at a distance of 200 m. Calculate the force applied by the brakes of the truck and the work done before stopping.

Answer 50: mass = m = 1800 kg

u = 54 km/h = 54 × 5/18 = 15 m/s

retardation, a = v2 – u2/2s

a = -9/16 m/s2

F = ma = 1800 × -9/16

= -1012.5 N

Since the force is acting in the opposite direction to motion so the negative sign is used.

Work done = Fs = 1012.5 × 200 = 202500 J

Question 51. The mass of the scooter and the bike are in the ratio of 2 : 3, they are moving with the same speed at 108 km/h. What is the ratio of their kinetic energy?

Kinetic energy is directly proportional to the mass of the body

Suppose mass of scooter = ms = 2m

Mass of the bike = mb = 3m

Kinetic energy of the scooter/ kinetic energy of the bike = ms/mb = 2m/3m = 2:3

Question 52. A heavy ball falls on the thick bed of wet sand from different heights-50 cm, 1 m, and 2 m, respectively. The depression created in the sand is measured. Identify which depression is the deepest and which one is the shallowest. Give a reason for the answer.

When the ball is dropped from a height of 2 m, the impression made by the wall is the deepest, as it transfers the most energy when it hits the wet sand.On the other hand, when the ball is dropped from a height of 50 cm, the impression made by the ball is the shallowest because the energy transferred by the ball to the wet sand is the least.

Question 53. Suppose there is a trolley and a block of a mass m is placed in front of it. When the mass is kept on the trolley it moves, hits the block and then it gets displaced. From where does the block get the energy? If the mass on the pan increases, will there be more work done, and what type of energy is possessed by the trolley?

Answer 53: First of all, the block is getting the energy because work is done on it by moving the trolley. When more work is done on the block as the trolley moves faster, then there will be more work done. The energy possessed by the trolley is kinetic energy.

Question 54. Explain what you understand of energy conversion by the phenomena which are given below.

How do green plants prepare food?
How the air moves from one place to the other.
How are fossil fuels formed?
The green plants prepare food by converting light energy from the sun into the chemical energy of food in the form of sugars.
Thermal energy from the sun gets converted into the kinetic energy of air, so convection currents are set up in the air due to the uneven heating of the earth by the sun.
Fossil fuels are formed when solar energy is stored as chemical energy in plants. These are buried in the earth over millions of years, and eventually they get converted into fossils.

Benefits of Solving Class 9 Work and Energy Extra Questions

Students must answer as many questions as possible. The habit of practicing develops their confidence and interest in the subject matter. Students often need more than textbook exercises to be better prepared for exams. The experts at Extramarks understand the needs of students, and they have created this question series. Work and Energy Class 9 Questions will help students in several ways. The important questions are-

The exercises in the textbook have limited questions, and students must seek help from other sources. They need help searching for answers from different sources. They collected different types of questions from different sources. They have taken help from the textbook exercises, CBSE sample papers, CBSE past years’ question papers, and important reference books. Thus, the students will find the questions in a single pdf, and they can practice the question. Thus, the Class 9 Science Chapter 11 Important Questions provided all the vital questions.
The experts have also written answers to the questions. The students can follow the answers if needed. The answers will also help them improve their writing skills and clear up any confusion. Experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Class 9 Science Chapter 11 Important Questions will help the students strengthen their ideas about the subject matter. The students can also learn better answer-writing techniques if they follow the answers closely.
Practice is very important, and it helps students in several ways. Many students tend to be afraid of the subject because they face problems understanding it. But Science is an interesting and scoring subject. So, students must practise questions to score better on tests. Also, they need to understand the subject and develop an interest in it. Solving questions will help to clarify their doubts, and they will like the subject too. So, Science Class 9 Chapter 11 Important Questions will help them understand the chapter.

Extramarks is a leading company that provides all the important study materials for students. You can download these after registering on our official website. We provide CBSE syllabus, CBSE sample papers, CBSE past years’ questions papers, CBSE revision notes, CBSE extra questions, NCERT important questions, NCERT solutions, NCERT Exemplars, NCERT books, vital formulas and many more. Like the Important Questions Class 9 Science Chapter 11, you will also find important questions for other chapters. The links to the study materials are given below-

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Q.1 The types of waves, which are produced when a slinky is plucked as shown below, are

transverse and longitudinal waves, respectively

longitudinal and transverse waves, respectively

transverse wave only

longitudinal wave only

Marks: 1 Ans

Transverse and longitudinal waves, respectively

Echo is the repetition of sound caused by the reflection of sound waves. It is found that we can hear two sounds separately only if there is a time gap of 1/10 th of a second or more between them. By knowing the minimum time interval for an echo and the speed of sound in air, the minimum distance from a sound reflecting surface can be evaluated. The minimum distance to hear an echo can be calculated as:

Hence the distance traveled by sound in 1/10 th of a second is 34.4 m. This distance is traveled by the sound from the source of sound to the sound reflecting surface and then back to the source of sound.

So, to hear an echo the distance between the source of sound to the sound reflecting surface, = (34.4/2) m = 17.2 m

Q.3 Tortoises use ______ for navigation and location of food in dark.

Audible sound

Supersonic sound

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Cbse class 9 science important questions, chapter 1 - matter in our surroundings.

Chapter 2 - Is Matter Around Us Pure

Chapter 3 - atoms and molecules, chapter 4 - structure of atom, chapter 5 - the fundamental unit of life, chapter 6 - tissues, chapter 7 - diversity in living organisms, chapter 8 - motion, chapter 9 - force and laws of motion, chapter 10 - gravitation, chapter 12 - sound, chapter 13 - why do we fall ill, chapter 14 - natural resources, chapter 15 - improvement in food resources, faqs (frequently asked questions), 1. what are the main concepts of class 9 science chapter 11.

Chapter 11 is about work and energy. We have a general concept of work. But physics defines work differently. Work is the measurement of energy transferred onto or from an object that causes displacement in physicsFor the transfer of energy, force is needed. Students have studied force in previous chapters and will study it further. On the other hand, energy is the ability to do work. Thus, work and energy are interrelated. Students can solve the Chapter 11 Class 9 Science Important Questions to better understand the subject matter.

2. How can the Important Questions Class 9 Science Chapter 11 help students?

The experts at Extramarks have collected all the important questions from different sources. They have collated the questions from the textbook exercise, CBSE sample papers, CBSE past years’ question papers, and important reference books. They have also solved the questions, and experienced professionals have further checked the content to ensure the best quality. Thus, students solve different types of questions that are important for the exams, and they can follow the answers if they cannot solve any questions. So, the Important Questions Class 9 Science Chapter 11 will improve their confidence and help them score better in exams.

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NCERT Solutions for Class 9 Science Chapter 11 Work and Energy

Study Reference for Class 9 Chapter 11 Work and Energy

Chapter 11 Work and Energy NCERT Solutions for Class 9 Science

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Work and Energy Class 9 Extra Questions with Answers

We have provided Work and Energy Class 9 Science Extra Questions with Answers to help students understand the concept very well.

Work, Power And Energy Class 9 Extra Questions Very Short Answer Type

Question 1. Define the following terms. (a) Work was done (b) Energy (c) Mechanical energy (d) Kinetic energy (e) Potential energy (f) Power (g) Commercial unit of energy. Answer: (a) Work done: Work done by a force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. (b) Energy: Energy of a body is defined as the capacity or ability of the body to do work. (c) Mechanical energy: Mechanical energy includes kinetic energy and potential energy. (d) Kinetic energy: The energy possessed by a body by virtue of its motion. (e) Potential energy: The energy possessed by a body due to its position or configuration. (f) Power: Power is defined as the rate of doing work or the rate of transfer of energy. (g) Commercial unit of energy: The energy used in households, industries, and commercial establishment are usually expressed in kilowatt-hour. 1 kWh 1 unit = 3.6 x 10 6 J

Question 2. Write down the type of energy stored in (a) spring of a watch (b) flowing water (c) rolling stone (d) raised hammer (e) running athlete Answer: (a) potential energy (b) kinetic energy (c) kinetic energy (d) potential energy (e) kinetic energy.

Question 3. What will be the kinetic energy of a body when its mass is made four-time and the velocity is doubled? Answer: Initial kinetic energy, $E_{K_{i}}=\frac{1}{2} m v^{2}$ Final kinetic energy, $E_{K_{f}}=\frac{1}{2}(4 m) \times(2 v)^{2}$ = 16 x $\frac {1}{2}$mυ 2 $E_{K_{f}}=16 E_{K_{i}}$

Question 4. If we lift a body of 7 kg vertically upwards to a height of 10 m, calculate the work done in lifting the body. Answer: Given, m = 7 kg s = 10m Workdone, W = F x s E = mg x s W = 7 x 10 x 10 J w = 7000 J

Question 5. State the transformation of energy that takes place when

Green plants prepare their food.
Head of a nail hammered hard and it becomes hot.
Solar energy of sun into chemical energy.
The kinetic energy of the hammer into heat energy.

Question 6. How much work is done by a man who tries to push the wall of a house but fails to do so? Answer: W = Fs = 0 As there is no displacement.

Question 7. Establish a relationship between SI unit and commercial unit of energy. Answer: SI unit of energy is joule and the commercial unit of energy is the joule. 1kWh = 1000 W x 3600 s = 3.6 x 10 6 J

Question 8. Write down the energy transformation taking place (a) In electric bulb (b) In torch (c) In the thermal power station (d) In solar cell (e) Electric heater Answer: (a) Electricity into light energy (b) The chemical energy of the cell into light and heat energy (c) The chemical energy of fuel into electricity (d) Solar energy into electricity (e) Electricity into heat energy.

Question 9. A body of mass m is moving in a circular path of radius r. How much work is done on the body? Answer: Zero. This is because the centripetal force acting on the body is perpendicular to the displacement of the body.

Question 10. A horse of mass 200 kg and a dog of mass 20 kg are running at the same speed. Which of the two possesses more kinetic energy? How? Answer: The kinetic energy of the horse is more as kinetic energy is directly proportional to mass.

Question 11. What is the condition for work done to be positive? Answer: For positive work, the angle between force and displacement should be acute.

Question 12. Write down the relation between kinetic energy and momentum of a body. Answer: $E_{K}=\frac{p^{2}}{2 m}$ E k = kinetic energy of a body p = momentum of the body m = mass of the body.

Question 13. A cyclist comes to a skidding stop at 50 m. During this process, the force on the cycle due to the road is 1000 N and is directed opposite to the motion. How much work does the road do on the cycle? Answer: Given, Displacement, s = 50 m Force, F = – 1000 N Workdone, W = F x s W = -1000 x 50J W = 50000J

Question 14. A boy pushes a book by applying a force of 40 N. Find the work done by this force as the book is displaced through 25 cm along the path. Answer: Given, Force (F) = 40 N Displacement (s) = 25 cm = x 10 -2 m Workdone, W = F x s = 40 x 25 x 10 -2 = 10J ∴ W = 10J

Work, Power And Energy Class 9 Extra Questions Short Answer Type 1

Question 1. State law of conservation of energy and law of conservation of mechanical energy. Answer: Law of conservation of energy: Energy can neither be created nor be destroyed, it can only be transformed from one form to another. Conservation of mechanical energy: If there is no energy, then the mechanical energy of a system is always constant.

Question 2. Define (a) 1 joule (b) 1 watt. Answer: (a) 1 joule is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force. (b) 1 watt is the power of an agent, which does work at the rate of 1 joule per second.

Question 3. Write down SI unit of the following quantities. (a) work (b) kinetic energy (c) potential energy (d) power Answer: (a) joule (J) (b) joule (J) (c) joule (J) (d) watt (W).

Question 4. What is the sequence of energy change that takes place in the production of electricity from adam? Answer: The potential energy of stored water is converted into the rotational kinetic energy of turbine blades. The rotational kinetic energy of turbine blades is finally converted into electric energy by the generator.

Question 6. Why a man does not do work when he moves on a level road while carrying a box on his head? Answer: When a man carries a load on his head, the angle between displacement (s) and force (F) is 900. Therefore work done is zero.

Work, Power And Energy Class 9 Extra Questions Numericals

Question 7. If an electric iron of 1200 W is used for 30 minutes every day, find electric energy consumed in the month of April. Answer: Given, Power, P = 1200 W time, t = 30 minutes Power, p = $\frac {W}{t}$ = $\frac {E}{t}$ E = P x t Energy consumed, E = 1200 x 30 x 60 = 2.16 x 10 6 J = 2.16MJ

Question 8. What is work done by a force of gravity in the following cases? (a) Satellite moving around the Earth in a circular orbit of radius 35000 km. (b) A stone of mass 250 g is thrown up through a height of 2.5 m. Answer: (a) Zero (b) Given, mass (m) = 250 g = 0.25 kg height (h) = 2.5 m Workdone, W = Fs = mgh = 0.25 x 10 x 2.5 = 6.25 J W = 625 J

Work, Power And Energy Class 9 Extra Questions Short Answer Type 2

Question 3. A girl having a mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 ms1 by applying a force. The trolley comes to rest after traversing a distance of 16 m. (a) How much work is done on the trolley? (b) How much work is done by the girl? Answer: Given, mass of girl, m = 35 kg mass of trolley, m = 5 kg. initial velocity of trolley, u = 4 m/s

(a) using work done = change in kinetic energy W = $E_{K_{f}}-E_{K_{i}}$ = o – $\frac {1}{2}$ x 40 x (4) 2 = – 320 ∴ W = 320 J_J (b) Work done by the girl = 0.

Question 4. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What forces does it exert in moving the car at a speed of 20 ms -1 ? Answer: 1 kW = iooo w = 10 3 J/S Given, mass (m) = 150 kg power (P) = 500 W velocity (u) = 20 m/s Using P = Fυ or, 500 = F x 20 = F =$\frac {500}{20}$ = 25 F= 25 N

Question 6. A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 1 minute. A second crane does the same job in 2 minutes. What is the power applied by each crane? Answer: Given, mass of the car to be lifted, m = 2000 kg height through which the car is to be lifted, h = 30 m Time taken by first crane, t 1 = 1 minute = 60 s time taken by second crane, t 1 = 2 minutes = 120 s Amount of work done by each crane, W = mgh = 2000 x 10 x 30 J W= 6 x 10 6 J Power of first crane, ‘ P 1 = $\frac{W_{1}}{t_{1}}=\frac{6 \times 10^{5} \mathrm{J}}{60 \mathrm{s}}$ = 10kW Power of second crane P 2 = $\frac{W_{2}}{t_{2}}=\frac{6 \times 10^{5} \mathrm{J}}{120 \mathrm{s}}$ = 10kW

Question 7. Calculate the electricity bill amount for a month of June, if 6 bulbs of 100 W for5 hours, 4 tube lights of 60 W for 5 hours, a TV of 50 W for 6 hours are used per day. The cost per unit is ₹ 5. Answer: Total energy used in a day = (6 x 100 x 5 + 4 x 60 x 5 + 1 x 50 x 6) Wh = (3000+ 1200 +300)Wh = 4500 Wh Total energy = 4.5 kWh = 4.5 unit Total energy used in 30 days = 4.5 x 30 = 135 units Bill amount = 135 units x ₹ 5 unit = ₹ 675.

Work, Power And Energy Class 9 Extra Questions Long Answer Type

Question 2. Give a reason for the following: (a) A bullet is released on firing the pistol. (b) An arrow moves forward when released from the stretched bow. (c) Winding the spring of a toy car makes it to run on the ground. (d) Falling water from a dam generates electricity. (e) Winding the spring of our watch, the hands of the watch movement. Answer: (a) The chemical energy of gun powder is converted into kinetic energy of the bullet. (b) The elastic potential energy in a stretched bow is converted into kinetic energy of the arrow. (c) The potential energy of a spring is converted into kinetic energy of the toy. (d) The kinetic energy of water is converted into electric energy. (e) The potential energy of spring due to its windings is converted into mechanical energy of the watch.

$E_{T_{A}}=E_{K}+E_{P}$ or, $E_{T_{A}}=0+m g h$ ∴ $E_{T_{A}}=m g h$ Total energy at point B, $E_{T_{B}}=E_{T}+E_{p}$ For finding out velocity at point B apply υ 2 – u 2 = 2as $v_{\mathrm{B}}^{2}=2 g h$ = 2gh Hence, $E_{T_{B}}=\frac{1}{2} m \mathrm{V}_{\mathrm{B}}^{2}+m g$ $E_{T_{B}}=\frac{1}{2} m(2 g h)=m g h$ Here, $E_{T_{A}}=E_{T_{B}}$ Hence if there is no energy loss, total energy is conserved.

Work, Power And Energy Class 9 Extra Questions HOTS

Question 1. A running man has half the kinetic energy that a body of half of his mass has. The man speeds up by 1 m/s and then has the same kinetic energy as the boy. What are the original speeds of the man and the boy? Answer: Let us take mass of boy = m mass of man = M velocity of boy = u velocity of man = υ Here, m = $\frac {M}{2}$ Initially E K of man = $\frac {1}{2}$ E K of boy $\frac {1}{2}$ Mυ 2 = $\frac {1}{2}$ mu 2 x $\frac {1}{2}$ and $\frac {1}{2}$ Mυ 2 = $\frac {1}{2}$ $\frac{1}{2}\left(\frac{M}{2}\right) u^{2} \times \frac{1}{2}$ ∴ υ 2 = $\frac{u^{2}}{4}$ ………….(1)

Finally E K of man = E K of boy $\frac {1}{2}$M(υ +1) 2 = $\frac {1}{2}$ mu 2 $\frac {1}{2}$M(υ +1) 2 = $\frac{1}{2}\left(\frac{M}{2}\right) u^{2}$ (υ +1) 2 = $\frac{u^{2}}{2}$ ∴ υ +1 = $\frac{u}{\sqrt{2}}$ …………(2) υ = ${\sqrt{2}}$ + 1 = 2.41 m/s and u = 4.82 m/s

Question 2. Avinash can run with a speed of 8 m/s against the frictional force of 10 N, and Kapil can move with a speed of 3 m/s against the frictional force of 25 N. Who is more powerful and why? Answer: Power can be expressed as, P = Fu Power of Avinash, P = Fu = 10 x 8 = 80 W Power of Kapil, P = Fu = 25 x 3 = 75 W Avinash is more powerful than Kapil.

Question 3. The weight of a person on a planet A is about half that on the earth. He can jump up to 0.4 m height on the surface of the earth. How high can he can jump on planet A? Answer: For the case, of jump, the energy imparted by the person is converted into potential energy, Hence, (m A g A )h A = (m e g e )h e ………(1) Given, m A g A = $\frac{m_{e} g_{e}}{2}$ ………….(2) Using (1) and (2), $\frac{h_{\mathrm{A}}}{2}=h_{e}$ ∴ h A = 2h e = 2 x 0.4 = 0.8m

Question 4. A ball is dropped from a height of 10 m. If the energy of the bal] is reduced by 40% after striking the ground, how much high can the ball bounce back? (g 10 m/s 2 ) Answer: Given, height, h = 10 If the energy of the ball is reduced by 40%, the remaining energy of the ball is 60% of initial. Hence ball will rebound to 60% of the initial height h = $\frac {60}{100}$ x 10 m ∴ h = 6m

Question 5. Four men lift a 250 kg box to a height of 1 m and hold it. Without raising or lowering it (a) How much work is done by men in lifting the box? (b) How much work they do in just holding it? Answer: Given, mass of block, m = 250 kg height, h = 1 m (a) work done in lifting, W = Fs = mgh = 250 x 10 x 1 W = 2500 J (b) work done in holding, W = 0

Question 6. What is power? How do you differentiate kilowatt from kilowatt-hour? The Jog falls in Karnataka state are nearly 20 m high. 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized. (g = 10 ms -2 ) Answer: 1. Power is rate of doing work. 2. Kilowatt is the unit of power and kilowatt-hour is unit of energy. 3. Given, height, h = 20 m Mass per unit time, m/t = 2000 tonnes per minutes = $\frac{2000 \times 10^{3}}{60}$ kg/s Power P = $\frac{W}{t}=\frac{m g h}{t}=\frac{2000 \times 10^{3}}{60}$ = 10 x 20 ∴ P = 6.67 x 10 6 W = 6.67 MW

Question 7. What happens to the kinetic energy when: 1. the mass of the body is doubled at constant velocity? 2. the velocity of the body is doubled at constant mass? 3. the mass of the body is doubled but velocity is reduced to one fourth? Answer: 1. The kinetic energy of body is given by, E k = $\frac {1}{2}$ mυ 6 , E k ∝ m. If the mass of the body is doubled its kinetic energy is also doubled.

2. Kinetic energy, E k ∝ u 2 If velocity of the body is doubled, its kinetic energy becomes four times.

3. Initial kinetic energy, $E_{K_{i}}$= $\frac {1}{2}$ mu 2 Final kinetic energy, $E_{K_{f}}$ = $\frac {1}{2}$(2m) $\left(\frac{v}{4}\right)^{2}=\frac{m v^{2}}{16}$ Kinetic Energy becomes one eighth.

Question 8. Why is the water at the bottom of a waterfall warmer than the water on the top? Answer: When waterfalls, its potential energy is converted into kinetic energy of molecules, and the kinetic energy of molecules is converted into heat energy.

Work, Power And Energy Class 9 Extra Questions Value Based (VBQs)

Question 1. Aman is a student of class IX. He saw an old man trying to keep his box on the roof of a bus but was unable to do so. Aman picked up his box and placed the box on the roof of the bus. The old man thanked Aman. Answer the following questions based on the given paragraph:

Is the work done by Aman while placing the box on the roof of the bus positive or negative?
Is the work done by gravity on the box positive or negative?
What values are shown by Aman?
Aman is a kind and helpful person.

Question 2. In the winter season, John gifted an electric heater to his grandfather. The electric heater uses electricity to increase room temperature. Answer the following questions based on the above paragraph:

Write down the energy transformation in the electric heater.
What values are shown by John?
Electric heater converts electricity into heat energy.
John is an intelligent and caring person.

Question 3. The government of a state decided to construct dams on the river for power generation. Nowadays the demand of electricity is continuously increasing and therefore more generation is required. Answer the following questions based on the above paragraph: (a) Write down the type of energy conversion is taking place in dams. (b) Write down the values shown by the state government. Answer: (a) The potential energy of stored water is converted into electricity. (b) The state government is working efficiently and showing good governance.

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Test: Work and Energy- Case Based Type Questions- 1 - Class 9 MCQ

10 questions mcq test - test: work and energy- case based type questions- 1, direction: study the given figure and answer the following question. (i) (ii) q. what will be the work done in the above situations state whether it will be positive, negative or zero..

A. In fig (i), Work down is zero. In fig (ii), work done is positive.
B. In fig (i), Work down is zero. In fig (ii), work done is negative.
C. In fig (i), Work down is positive . In fig (ii), work done is negative.
D. In fig (i), Work down is negative. In fig (ii), work done is positive.

Example: When an object moves on horizontal surface, force and displacement acts in same direction. So, work done is positive.

NEGATIVE WORK: The work done is said to be negative work when force and displacement are in opposite direction.

Example: When an object is thrown upwards,the force of gravity is in downward direction whereas displacement acts in upward direction.

ZERO WORK: The work done is said to be zero when force and displacement are perpendicular to each other or when either force or displacement is zero.

Example: When we hold an object and walk, the force acts in downward direction whereas displacement acts in forward direction.

Direction: Study the given figure and answer the following question. (i) (ii) Q. Write an expression for work in terms of force and displacement.

P.E.= m × g × h

d = s × t

F = m × a

W = F × s

when a force F displaces a body through a distance S in the direction of applied force, then the work done W on the body is given by the expression

Workdone = Force × displacement or W = F × S or W = Fs

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Direction: Study the given figure and answer the following question. (i) (ii) Q. In case of negative work, the angle between the force and displacement is

Direction: Study the given figure and answer the following question.

Q. Give reason for the same in each case.

A. In figure (i), the direction of force, (F) and displacement are parallel to each other. There is displacement in the direction of force so the work done is zero. In fig (ii), the direction of force F and displacement are in the same direction. Hence, work done by the force is positive.
B. In figure (i), the direction of force, (F) and displacement are perpendicular to each other. There is displacement in the direction of force so the work done is zero. In fig (ii), the direction of force F and displacement are in the same direction. Hence, work done by the force is negative.
C. In figure (i), the direction of force, (F) and displacement are parallel to each other. There is no displacement in the direction of force so the work done is positive. In fig (ii), the direction of force F and displacement are in the same direction. Hence, work done by the force is negative.
D. In figure (i), the direction of force, (F) and displacement are perpendicular to each other. There is no displacement in the direction of force so the work done is negative. In fig (ii), the direction of force F and displacement are in the opposite direction. Hence, work done by the force is positive.

Q. Define one joule of work.

A. One joule of work is said to be done on an object when a force of one hundred Newton displaces it by one meter along the line of action of the force.
B. One joule of work is said to be done on an object when a force of one Newton displaces it by one hundred meters along the line of action of the force.
C. One joule of work is said to be done on an object when a force of one Newton displaces it by one meter along the line of action of the force.
D. One joule of work is said to be done on an object when a force of ten Newton displaces it by one meter along the line of action of the force.

Direction: 75 kg missile is dropped downwards from an air plane, and has a speed of 60 m/s at an altitude of 850 m above the ground. Determine:

(a) The K.E. possessed by the missile at 850 m.

(b) The P.E. possessed by the missile at 850 m.

(d) The K.E. and velocity with which it strikes the ground.

Q. The K.E. possessed by the missile at 850 m.

B. 135000 J
C. 153000 J

Mass of the missile, m = 75 kg and Speed, v = 60 m/s

Therefore, K.E. = 1/ 2 mv 2 = 1 /2 × 75 × 60 × 60 = 13500 J.

Q. The total mechanical energy possessed by the missile.

B. 7725000 J
C. 772500 J
D. 775200 J

= 135000 + 637500 = 7.7 × 10 5 J = 772500 J.

Q. The P.E. possessed by the missile at 850 m.

B. 637050 J
D. 637500 J

Mass of the missile, m = 75 kg

Acceleration due to gravity, g = 10 m/s 2 and Height, h = 850 m

Therefore, P.E. = m × g × h = 75 × 10 × 850 = 637500 J.

Q. The energy possessed by an object due to motion is ………

A. chemical energy
B. kinetic energy
C. potential energy

Q. The K.E. and velocity with which it strikes the ground.

A. 135000 J
B. 637500 J
D. 635700 J

Therefore, v 2 = 2 K.E. / m

Velocity with which missile strikes the ground = 143.2 m/s.

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Case Study and Passage Based Questions for Class 9 Science Chapter 9 Force and Laws of Motion

Last modified on: 2 years ago
Reading Time: 6 Minutes

Case Study/Passage Based Questions:

Question 1:

Read the following and answer any four questions from (i) to (v) given below :

In the figure below the card is flicked with a push. It was observed that the card moves ahead while coin falls in glass.

(i) Give reason for the above observation. (a) The coin possesses inertia of rest, it resists the change and hence falls in the glass. (b) The coin possesses inertia of motion; it resists the change and hence falls in the glass. (c) The coin possesses inertia of rest, it accepts the change and hence falls in the glass. (d) The coin possesses inertia of rest, it accepts the change and hence falls in the glass.

(ii) Name the law involved in this case. (a) Newton’s second law of motion. (b) Newton’s first law of motion. (c) Newton’s third law of motion. (d) Law of conservation of energy

(iii) If the above coin is replaced by a heavy five-rupee coin, what will be your observation. Give reason. (a) Heavy coin will possess more inertia so it will not fall in tumbler. (b) Heavy coin will possess less inertia so it will fall in tumbler. (c) Heavy coin will possess more inertia so it will fall in tumbler. (d) Heavy coin will possess less inertia so it will not fall in tumbler.

(iv) Name the law which provides the definition of force. (a) Law of conservation of mass (b) Newton’s third law. (c) Newton’s first law (d) Newton’s second law.

(v) State Newton’s first law of motion. (a) Energy can neither be created nor be destroyed, it can be converted from one form to another, total amount of energy always remains constant. (b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force. (c) For every action in nature there is an equal and opposite reaction. (d) The acceleration in an object is directly related to the net force and inversely related to its mass.

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Last modified on:2 years agoReading Time:4MinutesCase Study Questions for Class 9 Science Chapter 11 Work and Energy In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage based as well. In that, a paragraph will be given, and then…

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Last modified on:2 years agoReading Time:4MinutesCase Study Questions for Class 9 Science Chapter 13 Why Do We Fall Ill In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage based as well. In that, a paragraph will be given,…

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Case Study Questions On Work And Energy
case study questions class 9 physics cbse science
case study questions for 9th maths
Physics, class 9, work and energy, pls help, see the attachment below
Work and Energy Class 9 Extra Questions with Answers
CBSE Class 9 Science Notes Chapter 11

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Work and Energy : 25
Motion Class 9: Top 5 Case Study Based Questions
कार्य की परिभाषा
case study based questions for class 9 maths chapter 7(triangles)
case study questions class 9 physics cbse science
MOST IMPORTANT QUESTIONS OF WORK ENERGY

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Case Study Questions Class 9 Science
CBSE Case Study Questions Class 9 Science - Work and Energy. (1) Work done by force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. Work has only magnitude and no direction. Work done is negative when the force acts opposite to the direction of displacement.
Class 9 Science Case Study Questions Chapter 11 Work and Energy
Work and Energy Case Study Questions With Answers. Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 11 Work and Energy. Case Study/Passage-Based Questions. Case Study 1: The figure shows a watch glass embedded in clay. A tiny spherical ball is placed at edge B at a height h above the center A.
Case Study and Passage Based Questions for Class 9 Science Chapter 11
Case Study Questions for Class 9 Science Chapter 11 Work and Energy. In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage based as well. In that, a paragraph will be given, and then questions based on it will be asked.
Class 9th Science
Answers. <. Work and Energy Case Study Questions With Answer Key Answer Keys. Case Study. (i) (a) 0.05 J. The work done in raising the bob through a height of 5 cm (against the gravitational attraction) gets stored in the bob in the form of its potential energy. PE = mgh. = 0.1 x 10 x 5 x 10-2. = 0.05 J.
Top 5 case based questions from work & Energy
Top 5 case based questions from work & Energy | CBSE Class 9 Physics ️ ⏮️ Subscribe to Vedantu 9 Channel to get Daily LIVE Classes Notifications👉 https://w...
Case Study Questions On Work And Energy
Case Study Questions On Work And Energy | Class 9 Physics | Class 9 Science Chapter 11.Topic covered in this video is-(1) - Case Study Questions On Work And ...
CBSE Class 9 Science Chapter 11
Ans: The chemical energy of the battery is converted into heat and light energy of the bulb in the given case. 12. Calculate the work done by the force that changes the velocity of a moving body from 5 ms-1 to 2 ms-1. The body has a mass of 20 kg. =$\frac {1} {2}\times 20\times (5^2-2^2)$=10× (25−4)=10×21=210 J. 13.
Class 9 Science Case Study Questions
If you are wondering how to solve class 9 science case study questions, ... As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it. ... Work, energy and power: Work done ...
NCERT Solutions for Class 9 Science Chapter 11: Work and Energy
NCERT Solutions Class 9 Science Chapter 11 - CBSE Free PDF Download *According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10. NCERT Solutions for Class 9 Science Chapter 11 Work and Energy help you lay a good foundation for your CBSE exam preparation. Students who refer to NCERT Solutions regularly are benefited from the comprehensive methodology and the ...
Category: Case Study Questions for Class 9 Science
Case Study and Passage Based Questions for Class 9 Science Chapter 11 Work and Energy. February 4, 2022 Physics Gurukul Leave a Comment.
NCERT Solutions for Class 9 Science Chapter 10
Table of Content. 1. Work and Energy Class 9 Questions and Answers NCERT Solutions FREE PDF Download. 2. Quick Insights of Work and Energy Class 9 NCERT Solutions. 3. Access NCERT Answers for Class 9 Science Chapter 10 - Work and Energy. 4. Class 9 Chemistry Chapter 10 Quick Overview of Topics.
CBSE Class 9 Physics Work and Energy Competency Based Questions
Work and Energy - Competency Based Questions. Select the number of questions for the test: 5. 10. Strengthen your understanding of Work and Energy in CBSE Class 9 Physics through competency based questions. Acquire in-depth knowledge and improve problem-solving abilities with comprehensive solutions.
NCERT Solutions for Class 9 Science Chapter 11 Work and Energy
Answer: Battery converts chemical energy into electrical energy. This electrical energy is further converted into light and heat energy. Question 4: Certain force acting on a 20 kg mass changes its velocity from 5 m s -1 to 2 m s -1. Calculate the work done by the force. Answer: Mass of the body = 20 kg.
Case Study Questions of Chapter 11 Work and Energy PDF Download
Case study Questions on Class 9 Science Chapter 11 are very important to solve for your exam. Class 9 Science Chapter 11 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 11 Work and Energy
Extra Questions for Class 9 Science Chapter 11 Work and Energy
Calculate the work done by him on the luggage. Answer: (a) (i) Force should be applied. (ii) Body should move in the line of action of force. (iii) Angle between force and displacement should not be 90°. (b) Mass of luggage, m = 15 kg and displacement, s = 1.5 m. Work done, W = F×s = mg × s = 15 × 10 × 1.5 = 225 J.
Assertion and Reason Questions for Class 9 Science Chapter 11 Work and
Here we are providing assertion reason questions for class 9 science. This article covers assertion reason questions based on Class 9 Science Chapter 11 Work and Energy. To check the answer, click on 'answer' given below each question. After clicking it will expand. Assertion and Reason Questions for Class 9 Science Chapter 11 Work and Energy … Continue reading Assertion and Reason ...
Important Questions for CBSE Class 9 Science Work, Power and Energy
Answer. The expression is KE = 1/2 mv 2, where 'm' is the mass and V is the velocity of the body. Question.25 Define 1 watt of power. Answer. When a work of 1 joule is done in 1s, the power is said to be one watt. Question.26 A lamp consumes 1000 J of electrical energy in 10 s.
CBSE Class 9 Science Chapter 11 Work and Energy Important Questions and
Answer 16: In this case, chemical energy is converted to electrical energy. So the final result is the combination of heat energy and light energy. Question 17. Calculate the work done by the force when the force is acting on a mass of 20 kg and there is a change in velocity from 5 m/s to 2 m/s.
NCERT Solutions for Class 9 Science Chapter 11 Work and Energy
Write an expression for the kinetic energy of an object. Answer. If a body of mass mis moving with a velocity v, then its kinetic energy E k is given by the expression, E k = 1/2 mv2. Its SI unit is Joule (J). 3. The kinetic energy of an object of mass, m moving with a velocity of 5 m s−1 is 25 J.
Work and Energy Class 9 Extra Questions with Answers
Work, Power And Energy Class 9 Extra Questions Long Answer Type. Question 1. State the conditions for positive, negative, and zero work. Give at least one example of each. Answer: 1. Zero work: If the angle between force and displacement is 90°, then work done is said to be zero work.
Test: Work and Energy- Case Based Type Questions- 1
Detailed Solution for Test: Work and Energy- Case Based Type Questions- 1 - Question 2. when a force F displaces a body through a distance S in the direction of applied force, then the work done W on the body is given by the expression. Workdone = Force × displacement or W = F × S or W = Fs. View Solution.
Problems Based on Work and Energy for Class 9 Science
Work is a scalar quantity and is typically measured in the unit of Joules (J). The formula for calculating work is: W = F x d. where W is the work done, F is the force applied, and d is the distance over which the force is applied. Work is a measure of the energy transferred to or from an object. When work is done on an object, energy is ...
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Case Study and Passage Based Questions for Class 9 Science Chapter 9
Last modified on:2 years agoReading Time:4MinutesCase Study Questions for Class 9 Science Chapter 11 Work and Energy In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage based as well. In that, a paragraph will be given, and then…