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.
As an example, consider a boy pulling an object towards himself.
W = + F × s
In this case, W = – F × s
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,
Question 3. Answer the following questions.
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
Total work is equal to the change in kinetic energy.
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
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
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.
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.
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:
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.
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.
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.
Answer 4: (d) 180 degree
Answer 5: (d) Zero
Answer 6: (b) 27
Cost of electricity = P × t × cost per kW
= 0.06 kW × (5 × 30) × 3 = 27
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.
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.
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.
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
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.
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
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.
Question 43. Fill in the blanks:
Question 44. Answer the following questions:
1.A raised hammer possesses which type of energy?
3.A moving car possesses which type of energy?
1. The raised hammer possesses potential energy.
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.
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.
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-
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-
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
Cbse class 9 science important questions, chapter 1 - matter in our surroundings.
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.
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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Study Reference for Class 9 Chapter 11 Work and Energy |
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We have provided Work and Energy Class 9 Science Extra Questions with Answers to help students understand the concept very well.
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
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
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
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.
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.
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.
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:
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:
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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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..
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.
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.
Q. Give reason for the same in each case.
Q. Define one joule of work.
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.
(c) The total mechanical energy possessed by the missile.
(d) The K.E. and velocity with which it strikes the ground.
Q. The K.E. possessed by the missile at 850 m.
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.
= 135000 + 637500 = 7.7 × 10 5 J = 772500 J.
Q. The P.E. possessed by the missile at 850 m.
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 ………
Q. The K.E. and velocity with which it strikes the ground.
Therefore, v 2 = 2 K.E. / m
Velocity with which missile strikes the ground = 143.2 m/s.
Work and energy- case based type questions- 1 mcqs with answers, online tests for work and energy- case based type questions- 1.
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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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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.
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 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.
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 | CBSE Class 9 Physics ️ ⏮️ Subscribe to Vedantu 9 Channel to get Daily LIVE Classes Notifications👉 https://w...
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 ...
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.
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 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 ...
Case Study and Passage Based Questions for Class 9 Science Chapter 11 Work and Energy. February 4, 2022 Physics Gurukul Leave a Comment.
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.
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.
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 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
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.
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 ...
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.
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.
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, 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.
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.
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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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…