case study questions work and energy class 9

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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.

At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks.

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 done is positive when the force is in the direction of displacement.The unit of work is newton-metre (N m)or joule (J).

(i) Work done is

(a) Scalar quantity

(b) Vector quantity

(c) Tensor quantity

(d) None of these

(ii) When force acts against the direction of displacement then work done will be

(a) positive

(b) negative

(c) both a and b can possible

(iii) SI unit of work is

(a) Joule(J)

(b) Newton meter(N-m)

(c) both a and b

(iv)You are lifting stone from floor. Work is done by theforce exerted by you on the stone. Theobject moves upwards. The force youexerted is in the direction ofdisplacement. However, there is theforce of gravity acting on the object. Which one of these forces is doingpositive work?

 Which one is doing negative work?

(v) Define 1J of work.

Answer key-1

(iv) Here work done by you is positive work as work is being done in the direction of displacement unlike in case of gravitational force which acts in downward direction against the direction of displacement which is in upward direction.

(v) When 1 Newton of force acts on body and body displaces from its position by 1 meter then the work done is said to be 1 joule (J).

(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.

Thus, the kinetic energy possessed by an object of mass, m and moving with a uniform velocity, v is

KE = ½ *mv 2

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).

(i) Energy possessed by body which is in motion is called

(a) Potential energy

(b) Kinetic energy

(c) Nuclear energy

(ii) Which of the following has same unit?

(a) Potential energy and Force

(b) Kinetic energy and work

(c) Both a and b

(iii) Kinetic energy depends

(a) Inversely on velocity of body

(b) Directly on square of velocity of body

(c) Directly on  velocity of body

(iv) Define kinetic energy of body. Give its SI unit

(v) Is kinetic energy scalar or vector? Justify your answer

Answer key-2

(iv) Energy possessed by object due to its motion is called as kinetic energy. Its SI unit is N-m or Joule(J).

(v) kinetic energy is scalar quantity as it is a work done and work done is scalar quantity hence kinetic energy is also scalar quantity and doesn’t have any direction.

(3) Lift an object through a certain height. The object can now do work. It begins to fall when released. This implies that it has acquired some energy. If raised to a greater height it can do more work and hence possesses more energy. From where did it get the energy? In the above situations, the energy gets stored due to the work done on the object. The energy transferred to an object is storedas potential energy if it is not used to cause a change in the velocity or speed of the object.An object increases its energy when raisedthrough a height. This is because work isdone on it against gravity while it is being raised. The energy present in such an objectis the gravitational potential energy.The gravitational potential energy of anobject at a point above the ground is definedas the work done in raising it from the ground by height h

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

work done, W = force × displacement

Therefore potential energy (PE)= mg*h.

(i) Energy possessed by body due to its position is called

(ii) SI unit of potential energy is

(iii)You do work while winding the key of a toy car. The energy transferred to the spring inside is stored as

(iv)Find the energy possessed by an object of mass 5kg when it is at a height of 10 m above the ground. Given, g = 9.8 m/s 2 .

(v)Find the work done by Gravity on an object of mass 5 kg which moves from height 10m to ground when it is released from height of 10 m. Given, g = 9.8 m/s 2 .

Answer key-3

(iv) we have potential energy as

=5 ×9.8 ×10

(v) work done, W = force × displacement

= 5 ×9.8 ×10

(4) The form of energy can be changed from one form to another. What happens to the totalenergy of a system during or after the process?Whenever energy gets transformed, the totalenergy remains unchanged. This is the law ofconservation of energy. According to this law, energy can only be converted from one form to another it can neither be created nor destroyed. The total energy before and after the transformation remains the same.The lawof conservation of energy is valid inall situations and for all kinds of transformations. Thus during motion the sum of the potential energy and kinetic energy of the object would be the same at all points. That is, potential energy + kinetic energy = constant.Andcalled as mechanical energy.

(i) Which of the energy conversion occur in electric iron?

(a) Electric energy converted into heat energy

(b) Electric energy converted into light energy

(c) Heat energy converted into electrical energy

(ii) When ball drops from height which of the energy conversion takes place

(a) Gravitational potential energy converted into kinetic energy

(b) Kinetic energy converted into Gravitational potential  energy

(iii) When ball is thrown vertically upward which of the following quantity remains constant?

(c) Potential energy

(iv) State law of conservation of energy.

(v) In hydroelectric power plant which energy conversion happens?

Answer key-4

(iv) This is the law ofconservation of energy. According to this law,energy can only be converted from one form to another it can neither be created nor destroyed. The total energy before and after the transformation remains the same.

(v)  In hydroelectric power plant potential energy of water reservoir is converted into electric energy.

(5 ) A more powerful vehiclewould complete a journey in a shorter timethan a less powerful one. We talk of the powerof machines like motorbikes and motorcars.The speed with which these vehicles changeenergy or do work is a basis for theirclassification. Power measures the speed ofwork done, that is, how fast or slow work isdone. Power is defined as the rate of doingwork or the rate of transfer of energy. If anagent does a work W in time t, then power isgiven by

P= work/time

P= W/T. The unit of power is watt.

(i) The rate of doing work is defined as

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

(a) Average power

(b) Instantaneous power

(iii) Let A and B having same weight start climbing the rope and reach height of 10m. Let A takes 10sec while B takes 12sec then work done

(a) By both will be same

(b) By A is more than work done by B

(c) By B is more than work done by A

(iv) Define 1 Watt of power

(v) An electric bulb of 20W is used for 5h per day. Calculate the ‘units’ of energy consumed in one day by the bulb.

Answer key-5

(iv) A power is said to be 1 watt when 1 joule of work is done within 1 second of time.

(v) Power of electric bulb = 20 W

Time used, t = 5 h

Energy = power × time taken

= 0.02kW × 5 h

= 0.10 kW h

= 0.10‘units’.

The energy consumed by the bulb 0.10 units

Jaru mitaya

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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

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 the motor is (a) 1940 W (b) 1950 W (c) 1960 W (d) 1970 W

Answer: (c) 1960 W

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

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

Class 9th Science - Work and Energy Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

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.

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.

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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.

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• 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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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: 

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.

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.

(c) A wind-mill is lifting water from a well.

(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.

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

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.

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.

Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 11

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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)

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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Work and Energy

Class 9 - ncert science solutions, intext questions 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?

Force (f) = 7 N

Displacement (S) = 8 m

Work done = Force × Displacement

Substituting we get,

W = 7 × 8 = 56 Nm or 56 J

Hence, work done = 56 J

Intext Questions 2

When do we say that work is done?

Work is said to be done when force applied on an object shows the displacement in that object. It is equal to the product of force and displacement.

Work done = force x displacement

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

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.

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?

Force (F) = 140 N

Displacement (S) = 15 m

W = 140 x 15 = 2100 J

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

Intext Questions 3

What is the kinetic energy of an object?

The kinetic energy is the energy possessed by an object due to its motion. It increases when the speed increases.

Write an expression for the kinetic energy of an object.

Kinetic Energy (K E ) = 1 2 \dfrac{1}{2} 2 1 ​ mv 2 . Its SI unit is Joule (J).

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?

Kinetic Energy (K E ) = 25 J

Velocity (v) = 5 ms -1

Kinetic Energy (K E ) = 1 2 \dfrac{1}{2} 2 1 ​ mv 2 .

25 = 1 2 \dfrac{1}{2} 2 1 ​ x m x 5 2

25 x 2 = 25 x m

m = 50 25 \dfrac{50}{25} 25 50 ​

When velocity is doubled :

v' = 10 ms -1

K E = 1 2 \dfrac{1}{2} 2 1 ​ x 2 x 10 2

K.E. = 100 J

When velocity is increased three times, then

v'' = 15 ms -1

K E = 1 2 \dfrac{1}{2} 2 1 ​ x 2 x 15 2

K.E. = 225 J

Hence, Kinetic Energy becomes 100 J when velocity is doubled and it becomes 225 J when velocity is increased three times.

Intext Questions 5

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:

Power = Work done Time \dfrac{\text{Work done}}{\text{Time}} Time Work done ​

It is expressed in watt (W).

Define 1 watt of power.

1 watt is the power of an agent, which does work at the rate of 1 joule per second.

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

Time = 10 s

Work done = Energy consumed by the lamp = 1000 J

Power = 1000 10 \dfrac{1000}{10} 10 1000 ​ = 100 Js -1 or 100 W

Hence, the power of the lamp is 100 W

Define average power.

Average power is defined as the ratio of total energy consumed to the total time taken by the body.

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.

(c) A wind-mill is lifting water from a well.

(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 said to be done when force applied on an object shows the displacement in that object.

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

(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 .

(d) No force is required when a green plant is carrying out photosynthesis and there is no displacement of plant. Hence, no work is done .

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

(f) As there is no force applied when grains are dried and there is no displacement as well. 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. As, displacement and force are in the same direction. Hence, work is done .

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 gravity depends on the vertical displacement of the body.

So, work done by gravity is :

Vertical displacement, h = 0 [∵ initial and the final points of the path of the object lie on the same horizontal line.]

∴ W = m × g × 0 = 0

Hence, work done by the force of gravity on the object = 0.

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

A battery converts chemical energy into electrical energy. When the bulb receives this electrical energy, it converts it into light and heat energy. Hence, energy changes involved in the process are :

Chemical Energy ⟶ Electrical Energy ⟶ Light Energy + Heat Energy.

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

Initial velocity u = 5 ms -1

Mass of the body = 20 kg

Final velocity v = 2 ms -1

Initial kinetic energy

E i = 1 2 \dfrac{1}{2} 2 1 ​ mu 2

E i = 1 2 \dfrac{1}{2} 2 1 ​ x 20 x 5 2 = 10 × 25 = 250 J

Final kinetic energy

E f = 1 2 \dfrac{1}{2} 2 1 ​ mv 2 = 1 2 \dfrac{1}{2} 2 1 ​ x 20 x 2 2 = 10 × 4 = 40 J

As, Work done = Change in kinetic energy = E f – E i

Work done = 40 J - 250 J

Work done = -210 J

Hence, work done by the force = -210 J.

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.

Mass (m) = 10 kg

Work done by gravity depends on the vertical displacement of the body. It is independent of the horizontal path.

So, work done by gravity is = m g h

Vertical displacement, h = 0 [∵ the line joining A and B is horizontal. ]

Hence, work done on object by gravity is zero.

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

No. 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. Hence, throughout the process, the total mechanical energy of the body remains conserved.

Hence, the law of conservation of energy is not violated.

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

The rider's muscular energy is converted to heat energy and the bicycle's kinetic energy while riding a bicycle. Hence, energy transformations are :

Muscular energy ⟶ Kinetic energy + Heat energy

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 and fail to move it, our muscular energy is not transferred to the rock as kinetic energy as there is no displacement of the rock. However, as per the law of conservation of energy, our muscular energy is transformed into heat energy that heats up our body and makes us sweat.

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

Energy (E) = 250 units

1 kWh = 3.6 x 10 6 J

1 unit of energy = 1 kWh

So, 250 units of energy = 250 × 3.6 × 10 6 = 9 × 10 8 J.

Hence, 250 units of energy = 9 × 10 8 J.

Question 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.

Mass (m) = 40 kg

Acceleration due to gravity = 10 ms -2

Height (h) = 5 m

Potential energy = ?

Potential energy = m g h

P.E. = 40 × 10 × 5 = 2000 J

Height when halfway down : = 5 2 \dfrac{5}{2} 2 5 ​ = 2.5 m

Potential energy when halfway = ?

P.E. = 40 × 10 × 2.5 = 1000 J

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

Hence, Potential energy = 2000 J and kinetic energy at halfway down = 1000 joules .

Question 11

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

When a satellite revolves around the earth in a circular orbit the work done is zero as force of gravity acting on satellite is perpendicular to its displacement.

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

Yes, a uniformly moving object can experience displacement even when no force is acting on it. According to Newton's first law of motion, an object in motion will continue in its straight-line motion unless acted upon by an external force. Therefore, displacement of an object can occur in the absence of any force acting on it.

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.

Work is said to be done when force applied on an object shows the displacement in that object. In this case, as there is no displacement of the hay bundle, hence no work is done.

Question 14

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

Power of the heater = 1500 W

Converting W to kW

1000 W = 1 kW

So, 1500 W = 1500 1000 \dfrac{1500}{1000} 1000 1500 ​ = 1.5 kW

Time taken = 10 h

Power = Energy consumed Time taken \dfrac{\text{Energy consumed}}{\text{Time taken}} Time taken Energy consumed ​

Energy consumed = Power x Time taken

Energy consumed = 1.5 x 10 = 15 kWh

Converting kWh to J

So, 15 kWh = 3.6 x 10 6 x 15 = 5.4 x 10 7 J

Hence, the energy consumed = 5.4 x 10 7 J

Question 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?

The law of conservation of energy states that energy can only be converted from one form to another; it can neither be created or destroyed. The total energy before and after the transformation remains the same.

Refer the figure of an oscillating pendulum bob shown below:

The kinetic energy decreases and the potential energy becomes maximum at B.

After a moment the the to and fro movement starts again.

So, from B to A, again the potential energy changes into kinetic energy and this process repeats again and again.

So, when the bob is in its state of to and fro movement it has potential energy at the extreme position B or C and kinetic energy at resting position A.

It has both the kinetic energy and potential energy at an intermediate position. However, the sum of kinetic and potential energy remain same at every point of movement.

The bob will eventually come to rest due to the frictional resistance offered by air on the surface of bob and pendulum loses its kinetic energy to overcome this friction and finally comes to rest.

The law of conservation of energy is not violated because the kinetic energy lost by the pendulum to overcome the friction is gained by surroundings. Hence, total energy of the system will remain conserved.

Question 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,

Kinetic energy = 1 2 \dfrac{1}{2} 2 1 ​ mv 2

In order to bring it to rest, its velocity has to be reduced to zero.

An external force has to absorb energy from the object, i.e., do negative work on it, equal to its kinetic energy = − 1 2 -\dfrac{1}{2} − 2 1 ​ mv 2

Question 17

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 kmh -1 .

Mass (m) = 1500 kg

Initial velocity (v) = 60 kmh -1

Final velocity = 0

Converting kmh -1 to ms -1 : multiply by 5 18 \dfrac{5}{18} 18 5 ​

60 x 5 18 \dfrac{5}{18} 18 5 ​ = 50 3 \dfrac{50}{3} 3 50 ​ ms -1

Work required to stop the moving car = Kinetic energy of car (K.E.)

K.E. = 1 2 \dfrac{1}{2} 2 1 ​ mv 2

= 1 2 \dfrac{1}{2} 2 1 ​ x 1500 x ( 50 3 ) (\dfrac{50}{3}) ( 3 50 ​ ) 2

= 1 2 \dfrac{1}{2} 2 1 ​ x 1500 x 2500 9 \dfrac{2500}{9} 9 2500 ​

= 208333.3 J

Hence, work done = 208333.3 J.

Question 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.

The direction of force acting on the block is perpendicular to the displacement. Hence, work done by force on the block will be zero .

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

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.

Question 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 several forces are acting on it. This happens when all the forces cancel out each other i.e., the net force acting on the object is zero.

Consider the below object:

F = 0 [∵ equal and opposite forces cancel out each other]

∴ V final = V initial

Hence, if the object was initially moving, it will continue to move with uniform velocity without acceleration.

Therefore, I agree with Soni's statement.

Question 20

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

Power rating (P) = 500 W

Converting W into kW

500 W = 500 1000 \dfrac{500}{1000} 1000 500 ​ = 0.5 kW

Time (T) = 10 h

Energy consumed by each device = Power x Time taken

Energy consumed by each device = 0.5 x 10 = 5 kWh

Energy consumed by four devices = 4 x 5 = 20 kWh

So, 20 kWh = 3.6 x 10 6 x 20 = 7.2 x 10 7 J

Hence, the energy consumed = 7.2 x 10 7 J

Question 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. When the object hits the ground, all its kinetic energy gets converted into heat energy and sound energy.

NCERT Solutions for Class 9 Science Chapter 10 Work and Energy

• 9th June 2023

NCERT Solutions for Class 9 Science Chapter 10 Work and Energy provides detailed answers for all in-text and exercise Questions. These solutions contain an in-depth explanation of each topic involved in the chapter. Students studying in class 9 can access these solutions for free in PDF format.

All these solutions are prepared by expert teachers and updated for the current academic session. NCERT Solutions for Class 9 Science Chapter 10 Work and Energy help students to understand the fundamental concepts given in class 9 Science textbook. We have prepared the answers to all the questions in an easy and well-structured manner. It helps students to grasp the chapter easily.

NCERT Class 9 Science Work and Energy Intext Questions (Solved)

PAGE NO. 115

Question 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?

Answer: When a force F acts on an object to displace it through a distance S in its direction, then the work done W on the body by the force is given by:

Workdone = Force × Displacement W = F × S Where, F = 7 N S = 8 m Therefore, work done, W = 7 × 8 = 56 Nm = 56 J

PAGE NO. 116

Question 1: When do we say that work is done?

Answer: Work is done whenever the given conditions are satisfied:

• A force acts on the body.
• There is a displacement of the body caused by the applied force along the direction of the applied force.

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

Answer: When a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression: Workdone = Force × Displacement W = F × s 

Question 3: Define 1 J of work.

Answer: 1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force. 80

Question 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?

Answer: Work done by the bullocks is given by the expression: Workdone = Force × Displacement W = F × d Where, Applied force, F = 140 N Displacement, d = 15 m W = 140 × 15 = 2100 J Hence, 2100 J of work is done in ploughing the length of the field.

PAGE NO. 119

Question 1: What is the kinetic energy of an object?

Answer: Kinetic energy is the energy possessed by a body by the virtue of its motion. Every moving object possesses kinetic energy. A body uses kinetic energy to do work. Kinetic energy of the hammer is used in driving a nail into a log of wood, kinetic energy of air is used to run windmills, etc. 

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

Answer: If a body mass m is moving with a velocity v, then its kinetic energy  is given by the expression, 

Its SI unit is Joule (J).

Question 3: The kinetic energy of an object of mass, m moving with a velocity of 5 m s −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?

m = Mass of object   v = Velocity of the object in ms −1              Given that kinetic energy, E K = 25J

If the velocity of an object is doubled, then v = 5 × 2 = 10 ms −1 . Therefore, its kinetic energy becomes 4 times its original value, because it is proportional to the square of the velocity.  Hence, kinetic energy = 25 × 4 = 100 J.

If velocity is increased three times, then its kinetic energy becomes 9 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 9 = 225J.

PAGE NO.123

Question 1: What is power?

Answer: Power is 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 the expression, Power = Work/Time ⇒ Power = Energy/Time ⇒ P = W/T It is expressed in watt (W).

Question 2: Define 1 watt of power:

Answer: A body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 s, i.e.,1W=1J/1s

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

Answer: Given, Work done = Energy consumed by the lamp = 1000 J  Power is given by the expression, Power=Work done /Time ⇒ P = 1000/10 ⇒ P = 100 W

Question 4: Define average power.

Answer: A body can do different amount of work at different time intervals. Hence, it is better to find average power. Average power is obtained by dividing the total amount of work done in the total time taken to do this work.

Average Power =Total work done / Total time taken

CBSE Class 9 Science Work and Energy Exercise Questions and Answers

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’. (a) Suma is swimming in a pond. (b) A donkey is carrying a load on its back. (c) A wind-mill is lifting water from a well. (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.

(a) Suma is swimming in a pond.   Suma is doing work as she is able to move herself by applying force with the movement of her arms and legs in the water.

(b) A donkey is carrying a load on its back. Donkey is not doing any work (in the sense of physics) as the weight he is carrying (the direction of force) and displacement are perpendicular to each other.

(c) A windmill is lifting water from a well.   Windmill is lifting water from a well and doing work against gravity.

(d) A green plant is carrying out photosynthesis. No force and displacement are present here, so work done is zero.

(e) An engine is pulling a train. During the pulling a train, engine does the work against the friction, present between the railway track and wheels. 

(f) Food grains are getting dried in the sun.   During the drying the grains, there is no force as well as displacement is present. So, no work is done.

(g) A sailboat is moving due to wind energy. Work is done by the wind as it moves the sailboat towards the direction of the force (force of blowing air).

Question 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?

Answer: When the object moves upwards, the work done by gravity is negative (as the direction of gravitational force is towards the Earth’s centre) and when the object comes downwards, there is a positive work done. So, the total work down is zero throughout the motion.

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

Answer 3: 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 5ms –1 to 2ms –1 . Calculate the work done by the force.

Answer: Mass of the body = 20 kg Initial velocity = 5 m/s Final velocity = 2 m/s

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 down by the gravitational force acting on the body is zero because the direction of force is vertically downward and the displacement is horizontal i.e. 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?

Answer: The potential energy of freely falling object decreases and its kinetic energy increases (as its velocity increases) progressively. So, in this way the total mechanical energy (Kinetic energy + potential energy) remains constant. Thus, the law of conservation of energy is not violated.

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

Answer: The muscular energy of the cyclist is converted into kinetic (rotational) energy of wheels of the cycle which is further converted into kinetic energy to run the bicycle.

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: When we push the rock and fail to move it. Some of our energy is absorbed by the rock in the form of potential energy and the rest of our energy is goes to the environment through our muscles and the surface between the rock and out hand.

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

Answer: Energy consumed in one month = 250 unit = 250 kWh = 250 kW×h = 250×1000W×3600s = 900,000,000 J = 9 × 10 8  J.

Question 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.

Answer: We know that, potential energy = mgh Where, m = 40 kg g = 9.8 m/s 2 h = 5 m So, the potential energy = 40 × 9.8 × 5 J = 1960 J According to law of conservation of energy, the total mechanical energy (Kinetic and potential energy) of an object remains constant.  Therefore, when the object is half-way down, its potential energy become half the original energy and remaining half converted into kinetic energy. Hence, the kinetic energy = ½ (1960) J = 980 J

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

Answer: When a satellite moves around the Earth, the displacement in short interval is along the tangential direction and the force (gravitational force) is towards the centre of the Earth. Since, the force and displacement are perpendicular to each other, the work done by gravitational force is zero.  

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.

Answer: Yes, it is true. There may be displacement in the absence of force. We know that, F = ma, In the absence of force, F = 0, then ma = 0 

If a = 0, the object is either at rest or in a state of uniform motion in a straight line. In case the object is moving in a straight line, there must be displacement. So, in the absence of force, there may be displacement in the object. 

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 holding a bundle of hay get tired because his muscular energy is converting into thermal energy. There is no displacement at all, so he had no work as workdone = Force × displacement.

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

Answer: We know that, Energy = Power × time Here, Power = 1500 W Time = 10 hours = 10 × 60 × 60 seconds = 36000 seconds Therefore,  The energy used by heater  = Power × time = 1500 × 36000 J = 54000000 J = 5.4 × 10 7 J

Question 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?

Answer : In the given pendulum, there are three cases of points to be discussed.

At the points B and C, the pendulum bob is at its maximum height, so its potential energy is maximum and kinetic energy is zero. In this way the total mechanical energy remains constant. At the point A, the pendulum bob is at its lowest point, total potential energy is converted into kinetic energy. Now the kinetic energy is maximum and potential energy is zero. Once again the total mechanical energy remains constant

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

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

The kinetic energy of the car, when it comes to rest = 0 J Workdone on object  = change in kinetic energy = 208333.3 − 0 = 208333.3 J Hence, the work required to stop the car is 208333.3 J.

Question 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.

Answer: In the first case, the force and displacement are perpendicular to each other, so work done is zero. In the second case, the force and displacement are in the same direction, so the work done is positive. In the third case, the force and displacement are in the opposite direction, so the work done is negative.

Question 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?

Answer 19: If the resultant force acting on a body in different directions is zero, then the acceleration will be zero.   We know that,                               F = ma, In the net force is zero,             F = 0 ⇒ ma = 0  ⇒ a = 0  [as m ≠ 0]

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

Answer: The power of four devices = 4 × 500 W = 2000 W Time = 10 hours Therefore, the energy consumed = power × time = 2000 × 10 Wh = 20000 Wh = 20 kWh  = 20 units   [1 unit = 1 kWh]

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

Answer: When a freely falling body eventually stops on reaching the ground, its kinetic energy gets converted into heat energy (as the body and ground become warm due to collision), sound energy and into potential energy (due to change of shape or deformation). 

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• NCERT Exemplar for Class 9 Science Chapter 11 - Work and Energy
• Textbook Solutions

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Access NCERT Exemplar Solutions for Class 9 Science Chapter 11 - Work and Energy

Multiple choice questions  .

1. When a body falls freely towards the earth, then its total energy

(a) increases 

(b) decreases 

(c) remains constant  

(d) first increases and then decreases  

Ans: The correct answer is (c) remains constant 

When the body falls freely towards the earth, then the sum of potential energy and kinetic energy remains the same. And the total energy is equivalent to the sum of kinetic energy and potential energy. Therefore, the total energy remains constant.  

2. A car is accelerated on a leveled road and attains a velocity 4 times its initial velocity. In this process, the potential energy of the car  

(a) does not change  

(b) becomes twice to that of initial 

(c) becomes 4 times that of initial  

(d) becomes 16 times that of initial  

Ans: The correct answer is (a) does not change 

The potential energy of the system depends upon the height at which the object is situated. As the formula of calculating the potential energy is the product of the ‘mass of the object’ and gravity and its height. Since in this case, a car is on a leveled road and the height is zero. So the potential energy of the car does not change.  

3. In the case of negative work the angle between the force and displacement is  

(a) \[{0^o}\] 

(b) \[{45^o}\] 

(c ) \[{90^o}\] 

(d) \[{180^o}\] 

Ans: The correct answer is (d) \[{180^o}\] 

We are aware of the formula of the work done is: 

\[w = FS\cos \theta \] 

Here F is the force and ‘S’ is the displacement. 

So, from the above formula it is clear that the value of angle \[\cos {180^0}  = - 1\] 

Therefore, the force and displacement will take place in the opposite direction i.e. have an angle of \[{180^0}\] . and thus the work done is negative. 

4. An iron sphere of mass 10 kg has the same diameter as an aluminum sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a  tower. When they are 10 m above the ground, they have the same  

(a) acceleration  

(b) momenta  

(c) potential energy 

(d) kinetic energy 

Ans: The correct answer is (a) acceleration

Acceleration does not depend on the mass and velocity of the object rather it is due to gravity. while all the other options like momenta, potential energy, and kinetic energy all depend on mass, velocity, height, and other factors. 

5. A girl is carrying a school bag of 3 kg on her back and moves 200 m on a leveled road. The work done against the gravitational force will be (g \[ = 10m{s^{ - 2}}\] ) 

(a) \[6 \times {10^3}J\] 

(b) \[6J\] 

(c) \[0.6J\] 

(d) zero 

Ans: The correct answer is (d) zero 

The formula of the work done is \[w = FS\cos \theta \] 

In this case the work done by the gravitational force is perpendicular to the  displacement of the girl. So the \[\theta = {90^o}\] and the value of  \[\cos {90^o} = 0\] . hence , the work done is zero. 

6. Which one of the following is not the unit of energy? 

(a) joule  

(b) newton meter  

(c) kilowatt 

(d) kilowatt-hour 

Ans: The correct answer is (c) kilowatt 

Kilowatt is not the unit of energy as Kilowatt is the unit of power. Power is a change in work done divided by total time. 

While all other three i.e. Joule, Newton meter, and kilowatt-hour is the unit of energy.

7. The work done on an object does not depend upon the-:

(a) displacement 

(b) force applied 

(c) the angle between force and displacement 

(d) the initial velocity of the object  

Ans: The correct answer is (d) the initial velocity of the object 

Work done on an object depends on the force applied on the object and the displacement covered by the object and also on the angle formed between the force and the displacement. The work done does not depend upon the initial velocity of the object.  

8. Water stored in a dam possesses 

(a) no energy 

(b) electrical energy 

(c) kinetic energy 

(d) potential energy  

Ans: The correct answer is (d) potential energy 

The energy stored in the object due to its position is known as potential energy. and the water is stored in a dam and thus possesses potential energy. 

9. A body is falling from a height ‘h’. After it has fallen a height  \[\dfrac{h}{2}\] , it will possess 

(a) only potential energy 

(b) only kinetic energy 

(c) half potential energy and half kinetic energy 

(d) more kinetic and less potential energy

Ans: The correct answer is (c) half potential energy and half kinetic energy 

The kinetic energy of the body is maximum at the ground level and the potential energy of an object is maximum at the height, h. so when an object is at height  \[\dfrac{h}{2}\] i.e. half the distance so, it will have half potential energy and half kinetic energy. 

Short Answer Questions 

10. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies? 

Ans: Let’s assume that:

Initial velocity  \[ = v\]

When velocity is tripled  \[v' = 3v\]

Therefore the initial kinetic energy is , \[KE = \dfrac{1}{2}m{v^2}\]

Final kinetic energy will be , \[KE' = \dfrac{1}{2}mv{'^2}\]

Putting the value of  \[v' = 3v\], we will get :  \[KE' = \dfrac{1}{2}mv{'^2} = \dfrac{1}{2}m{(3v)^2} = 9(\dfrac{1}{2}m{v^2})\]

By comparing the two equations we get   \[KE:KE' = 1:9\]

11. Avinash can run with a speed of \[8m{s^{ - 1}}\] against the  frictional force of \[10N\] , and Kapil can move with a speed of \[3m{s^{  - 1}}\] against the frictional force of \[25N\]. Who is more powerful and why? 

Ans : To solve this question we will use the formula of the ‘Power’. We know that , Power \[ = \] Force \[ \times \] Displacement  

Power of Avinash , \[{P_A} = {F_A} \times {V_A} = 10 \times 8 =  80W\]

Now calculating the power of Kapil, \[{P_K} = {F_K} \times {V_K} =  25 \times 3 = 75W\] 

So, as we can see that Avinash is more powerful than Kapil. 

12. A boy is moving on a straight road against a frictional force of \[5N\].  After traveling a distance of \[1.5\] km he forgot the correct path at a  roundabout (Fig. \[11.1\] ) of radius \[100\] m. However, he moves on the circular path for one and half-cycles and then he moves forward up to \[2.0\]  km. Calculate the work done by him. 

Ans: In this case first we will not calculate the circumference, because we need a displacement and not a distance. 

Therefore, Displacement \[ = 1500m + 200m + 200m = 3700m\] 

Now, calculating the total work done , \[W = F \times S = 5 \times  3700m = 18500J\] 

13. Can any object have mechanical energy even if its momentum is zero?  Explain .

Ans: The answer is Yes. Mechanical energy is defined by both potential energy and kinetic energy. So if momentum is zero, this implies that velocity is zero and thus the kinetic energy is zero. But the object may have potential energy and thus an object can have mechanical energy even if its momentum is zero. 

14. Can any object have momentum even if its mechanical energy is zero?  Explain . 

Ans: The answer is zero. Mechanical energy is defined by both potential energy and kinetic energy. So, if mechanical energy is zero, this implies that kinetic energy is zero, thus velocity is zero and thus momentum is also zero. 

  15. The power of a motor pump is \[2\] kW. How much water per minute  the pump can raise to a height of \[10\] m? (Given g = 10 \[10m{s^{ - 2}}\] ) 

Ans:   We know the formula of ‘Power’ is:   \[P = \dfrac{W}{{\Delta t}}\]

Also, we can write , \[P = \dfrac{W}{{\Delta t}} = \dfrac{{mgh}}{{\Delta t}}\]

Now, putting the values in the formula we will get:

 \[2 = \dfrac{{m \times 10 \times 10}}{{60}}\]

\[m = \dfrac{{12000}}{{10}} = 1200kg\]

Therefore the answer is  \[  1200kg\].

16. The weight of a person on 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 he can jump on planet A?

Ans: In the question, it is given that the weight of a person on planet A is about half that on the earth so, this implies that also, the acceleration due to gravity will be half of that of earth. So he will be able to jump double the height that he can jump on the earth. Therefore he can jump \[0.8m\] on plant A. 

17. 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 the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body . 

Ans: We will use the equation \[{v^2} - {u^2} = 2as\] 

On rearranging it we will get \[s = \dfrac{{{v^2} - {u^2}}}{{2as}}\] Now we know that , work done , \[W = F \times S\] 

Also, \[F = m \times a\] 

So, work done by this force F will be : \[W = ma(\dfrac{{{v^2} - {u^2}}}{{2as}}) = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}\] 

\[W = (K{E_f}) - (K{E_i})\] 

18. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example. 

Ans: Yes this can be true in the case of circular motion. In a circular motion, the force will always be perpendicular to the displacement of the object. 

19. A ball is dropped from a height of \[10\] m. If the energy of the ball reduces by \[40\% \] after striking the ground, how high can the ball bounce back? (g \[10m{s^{ - 2}}\] ) 

Ans: work done is \[ = mgh = m \times 10 \times 10 = 100mJ\]

As, the energy is reduced by \[40\% \] , therefore the energy remaining is  \[60\% \] i.e. \[60mJ\] 

Now, putting the values, we will get: 

\[60m = m \times 10 \times h'\] 

\[h' = 6m\] 

20. If an electric iron of \[1200\] W is used for \[30\] minutes every day,  find electric energy consumed in the month of April? 

Ans: we know that formula to calculate Electrical energy is: 

Electrical energy \[ = \] Power \[ \times \] Time \[ \times \] Days Now, the values given in the question are – 

Power , \[P = \dfrac{{1200}}{{1000}}KW\] 

Time , \[t = \dfrac{{30}}{{60}}hr = 0.5hr\] 

Now putting the values, we get: 

Electrical energy \[ = \] Power \[ \times \] Time \[ \times \] Days \[E = 1.2 \times 0.5 \times 30\] 

\[E = 18KWh\] 

Long Answer Questions 

21. A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has larger kinetic energy? 

Ans: Let’s assume \[{p_1} = {m_1}{v_1}\] and \[{p_2} = {m_2}{v_2}\]. Also, we know that \[{p_1} = {p_2}\]

This means that \[{m_1}{v_1} = {m_2}{v_2}\].

Also, \[{m_1} < {m_2}\].

So, \[{v_1} > {v_2}\]

Now, kinetic energy of first body \[(K{E_1}) = \dfrac{1}{2}{m_1}{({v_1})^2} = \dfrac{1}{2}({m_1}{v_1}){v_1}\]

Kinetic energy of second body \[(K{E_2}) = \dfrac{1}{2}{m_2}{({v_2})^2} = \dfrac{1}{2}({m_2}{v_2}){v_2}\]

On dividing both the

equations to find the ratio, we get:

$\dfrac{{(K{E_1})}}{{(K{E_2})}}$

${\dfrac{1}{2}}({m_1}{v_1}){v_1}={\dfrac{1}{2}}({m_2}{v_2}){v_2}$

$=\dfrac{{{v_1}}}{{{v_2}}}$

We know that \[{v_1} > {v_2}\]. So, \[(K{E_1}) > (K{E_2})\]. 

22. An automobile engine propels a \[1000\] kg car (A) along a levelled  road at a speed of \[36km{h^{ - 1}}\] . Find the power of the opposing frictional force is \[100\] N. Now, suppose after traveling a distance of  \[200\] m, this car collides with another stationary car (B) of the same mass and comes to rest. Let its engine also stop at the same time. Now the car (B) 

starts moving on the same level road without getting its engine started.  Find the speed of the car (B) just after the collision 

Ans: We are given that mass of body A is equal to the mass of body B \[{m_A} - {m_b} = 1000kg\] 

Also, \[v = 36km{h^{ - 1}} = 10m{s^{ - 1}}\] 

And the frictional force is \[100\] N 

the engine of car A applies a force equal to the frictional force because car A is in uniform motion. 

Also the formula of power is: 

\[P = \dfrac{{Force \times dis\tan ce}}{{time}} = force \times  velocity\] 

On outing the values, we get:  

$ P = 100N \times 10m{s^{ - 1}}$

P = 1000W  

Also, we know that after the collision:

${m_A}{u_A} + {m_B}{u_B}={m_A}{v_A} + {m_B}{v_B} $

$  1000 \times 10 + 1000 \times 0 = 1000 \times 0 + 1000 \times {v_B} $

 $ {v_B} = 10m{s_{ - 1}}$

23. A girl having mass of \[35\] kg sits on a trolley of mass \[5\] kg. The  trolley is given an initial velocity of \[4m{s^{ - 1}}\] by applying a force.  The trolley comes to rest after traversing a distance of \[16m\] . (a) How  much work is done on the trolley? (b) How much work is done by the girl?

Ans: we know that. Mass of girl = \[35\] kg, mass of trolley = \[5\] kg, u =  \[4m{s^{ - 1}}\] , v = \[0\] and s = \[16m\] 

(a) First calculate the acceleration by the equation \[{v^2} - {u^2} = 2as\] 

\[a = \dfrac{{{v^2} - {u^2}}}{{2s}} = \dfrac{{0 - 16}}{{2 \times 16}}  = - 0.5m{s^{ - 1}}\] 

Also, 

$ W = F \times S $

$W = m \times a \times s$

$ W = 40 \times 0.5 \times 16 $

W = 320J  

(b) In this case force applied by the girl is zero, hence the work done is zero. 

24. 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 the men in lifting the box? (b) How much work do they do in just holding it? (c) Why do they get tired while holding it? (g \[ = 10m{s^{ - 2}}\]) 

Ans: (a) we know that  

$ F = 250kg \times 10m{s^{ - 2}}$

 F = 2500N  

 s = 1m  

 $ W = F \times S $

 $W = 2500N \times 1m$

  W = 2500Nm 

(b) In this case ‘displacement’ is zero, thus work done is zero. 

(c) In this case, the force applied is equal and opposite to the gravitational force experienced by the box, so its net displacement is zero but the muscular force is being applied, and thus they get tired while holding the box. 

25. 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 waterfalls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g \[ = 10m{s^{ - 2}}\]) 

Ans: We are given h= \[20\] m, and mass = \[2000\] × \[{10^3}\] kg =  \[2 \times {10^6}\] kg 

Also, we know the formula of the power as:   \[P = \dfrac{W}{{\Delta t}} = \dfrac{{mgh}}{{\Delta t}}\]

$ P = \dfrac{{2 \times {10}^5}} \times 10 \times{{ 20}}{{60}}W$

$ P = \dfrac{2}{3} \times {10^7}W $

26. How is the power related to the speed at which a body can be lifted?  How many kilograms will a man working at the power of \[100\] W, be able to lift at a constant speed of \[1m{s^{ - 1}}\] vertically? (g \[ =  10m{s^{ - 2}}\])

Ans: We know that   \[P = \dfrac{W}{{\Delta t}} = \dfrac{{mgh}}{{\Delta t}}\]

Also, we can write this in the form of:

$P = mg\dfrac{h}{{\Delta t}}$

${\text{  }}\dfrac{h}{{\Delta t}} = v = speed{\text{ }}$

On rearranging we will get:  

\[m = \dfrac{{power}}{{speed \times g}}\] 

Now on putting the values we will get: 

\[m = \dfrac{{100}}{{10 \times 1}} = 10kg\] 

27. Define watt. Express kilowatt in terms of joule per second. A \[150\] kg car engine develops \[500\] W for each kg. What force does it exert in moving the car at a speed of \[20m{s^{ - 1}}\]? 

Ans: A watt is a unit of power or radiant flux. One watt is the power of an agent that does work at the rate of \[J{s^{ - 1}}\] . 

Also we can write \[1\] Kilowatt \[ = 1000\] \[J{s^{ - 1}}\] formula of the force is: 

$force = \dfrac{{power}}{{velocity}}$

$force = \dfrac{{1500 \times 500}}{{20}}$

$ F = 3.75 \times {10^3}N$

F = 3750N   

28. Compare the power at which each of the following is moving upwards against the force of gravity? (given g \[ = 10m{s^{ - 2}}\]) 

Ans: (i) a butterfly of mass 1.0 g that flies upward at a rate of \[0.5m{s^{ - 1}}\] . (ii) a $250$ g squirrel climbing up on a tree at a rate of \[0.5m{s^{ - 1}}\] . Ans: (I) power \[ = \] mg \[ \times \] velocity, 

\[m = 1g = {10^{ - 3}}kg\] 

Therefore , power \[ = \] \[{10^{ - 3}} \times 10 \times 0.5\] \[P = 5 \times {10^{ - 3}}W\] 

(ii) m \[ = 250 \times {10^{ - 3}}W\]

$ P = 250 \times {10^{ - 3}} \times 10 \times 0.5W$

  P = 1.25W  Squirrel is climbing with more power than a butterfly.

NCERT Exemplar for  Chapter 11 (Science) of Class 9

Chapter 11 of Class 9 Science NCERT deals with the concepts and fundamentals of Work and Energy. The chapter gives an insight into certain relevant information that will develop the base of the students for their higher education and competitive exams. Some of the key points mentioned in the chapter are:

Work done on an object is referred to as the product of the magnitude of force applied on the object and the distance that the object moves due to application of force.

Work is measured in Joule (J).

If the displacement of the object on application of force is zero then the work done on the object will be zero.

When an object is capable of doing certain work then the object is said to possess energy. Therefore, work and energy have the same unit.

An object in motion possesses kinetic energy. This energy is half of the product of mass and square of final velocity.

The energy possessed by an object because of the change in the position or shape of the object is referred to as potential energy. 

The law of conservation of energy states that the energy can neither be created nor destroyed, it only changes from one form to another.

There are several forms in which energy can exist in nature, like, heat energy, kinetic energy, potential energy, chemical energy and more.

Power is another important concept dealt in the chapter. It is defined as the rate of doing work and is measured in watt (W).

FAQs on NCERT Exemplar for Class 9 Science Chapter 11 - Work and Energy

1. How will the science NCERT exemplar for class 9 chapter 11 help me?

The class 9 exemplar is well formulated by NCERT and is designed with an aim to help students excel in exams. Some of the ways in which NCERT exemplar can help class 9 students are:

It provides a variety of questions for students to practice.

It helps students understand concepts from an examination point of view.

It also makes students capable of dealing with advanced versions of work and energy that are taught in higher classes.

It also makes students understand the pattern of questions that are asked in exams by NCERT.

NCERT exemplar boosts students' preparation for exams.

2. What kind of questions are present in the Science NCERT Exemplar of class 9 for chapter 11?

Science NCERT exemplar provides a wide range of questions to students ranging from easy to difficult level. Class 9 chapter 11 of science present in NCERT exemplar contains multiple-choice questions (MCQs), short answer questions, and long answer questions. There are a total of 26 questions present in the NCERT exemplar, out of which, 9 are multiple-choice type questions, 11 are short answer questions and 8 are long answer questions. The questions given in NCERT Exemplar for class 9 science helps to prepare for the exams.

3. What are the topics covered in the science NCERT exemplar mainly in chapter 10?

There are different topics covered in the class 9 science chapter 11. All topics are important for class 9 science students. The science NCERT exemplar contains the most important topics of chapter 11, that is, work and energy. The topics included in the exemplar are:

Concept of work

Work done by a constant force

Energy, its forms, and unit

Kinetic energy and its associated numerical

Potential energy and its associated numerical

Laws of conservation of energy

Rate of doing work

4. On which parts of chapter 11 are the short answer questions and long answer questions in the science NCERT exemplar of class 9 mainly focuses on?

The multiple-choice questions in the NCERT exemplar mainly focus on factual questions on force, displacement, kinetic energy, acceleration, and SI units. Short answer questions focus on areas related to numerical that will enhance students' quantitative skills. Long answer questions of the exemplar are also numerical questions related to energy, work, and power that enhance not only quantitative but also logical skills of students. Students must read all topics carefully to understand the topics given in chapter 11 for class 9. 

5. What will be the output of studying chapter 11 of the class 9 science from the NCERT Exemplar?

By studying chapter 11 of class 9 from the NCERT Exemplar, that is, work and energy from the class 9 science syllabus, it provides students with an idea about the concept of work and energy. It also gives them insight into kinetic energy and possible energy and the formulas related to them. Another important outcome of the chapter will be the delivery of the most basic concept of the law of conservation of energy and the rate of doing work.

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.

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.

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.

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

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

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,

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),

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

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 )

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

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

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.

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

NCERT solutions for Class 9 Science  Chapter 11 Work and Energy helps you lay a good foundation for your exam preparation. Those students who refer the NCERT Solutions regularly are benefited with the comprehensive methodology of the topic, and also with the detailed step by step procedure, which will fetch them good marks in their examinations.

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

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

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, the 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 caused by the applied force on the direction of the applied 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 that the quantity of labor done by a force of one N on associate degree object that displaces it through a distance of one m within the direction of the applied 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= a hundred and forty × fifteen = 2100 J

Hence, 2100 J of labour is finished in tilling the length of the sector.

Exercise-11.3 Page: 152

1. What is the kinetic energy of an object?

The energy possessed by a body by the 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. Kinetic energy of the hammer is employed in driving a nail into a log of wood, mechanical energy of air is employed to run wind mills, etc.

3. 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).

4. 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?

Exercise-11.4 Page: 156

1. What is power?

Power is that the rate of doing work or the speed of transfer of energy. If W is that the quantity of work wiped out time t, then power is given by the expression,

It is expressed in watt (W).

2. Define 1 watt of power.

A body is claimed to possess power of one watt if it will work on 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

4. Define average power

The average Power of an agent could also be outlined because the total work done by it within the total time taken.

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.

(c) A wind-mill is lifting water from a well.

(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:

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

(a) Whereas swimming, Suma applies a force to push the water backwards. Therefore, Suma swims within the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, work is finished by Seema whereas swimming.

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

(c) A wind mill works against the gravity to elevate water. Hence, work is finished by the wind mill in lifting water from the well.

(d) During this case, there’s no displacement of the leaves of the plant. Therefore, the work done is zero.

(e) An engine applies force to tug the train. This permits the train to maneuver within the direction of force. Therefore, there’s a displacement within the train in the same direction. Hence, work is finished by the engine on the train.

(f) Food grains don’t move within the presence of alternative energy. Hence, the work done is zero during the method of food grains obtaining dried within the Sun.

(g)Wind energy applies a force on the sailing ship to push it within the forward direction. Therefore, there is a displacement within the boat in the direction of force. Hence, work is finished by wind on the boat.

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 within the initial and final positions/heights of the thing that is zero.

Gravity-related work is expressed as,

h= Vertical displacement = zero

W = m g × zero = 0 J

Consequently, the work done on the given object by gravity 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-energy and warmth energy. Hence, the transformation of energy within the given situation may 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.

Given data:

Initial velocity u=5 ms –1

Mass of the body = 20kg

Final velocity v = 2 ms –1

The initial kinetic energy

Ei = (1/2) mu 2 = (1/2) x 20 x (5 ms –1 ) 2 =250kgms -2

= 250Nm = 250J

Final kinetic energy Ef = (1/2) mv 2 = (1/2) x 20 x (2 ms –1 ) 2 =40kgms -2 = 40 Nm =40J

Work done = Change in kinetic energy

Work done = E f – E i

Work done =40J – 250J

Work done = -210J

Where negative sign indicates that force acts contrary to motion direction.

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, hat 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= mg × zero = 0

Therefore the work done on the body by gravity is therefore 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 once the body falls from a height, then its mechanical energy changes into kinetic energy increasingly. A decrease within the mechanical energy is capable a rise in the kinetic energy of the body. Throughout the method, total 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 is regenerate into heat and mechanical energy.

Kinetic energy provides 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 large rock, there’s no transfer of muscular energy to the stationary rock. Also, there’s no loss of energy as a result of muscular energy is transferred into 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 is up to one B.T.U. (kWh).

1 unit = one 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 metre the object has a potential energy of 2000 J.

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

P.E at Half way down= m× g×h

P.E= 40× 10 × 2.5= 1000J

Potential Energy at Half way down= 1000 joules.

According to law of conservation of energy:

Total potential energy= potential energy at Half way down+ kinetic energy at Half way of a down

2000 = 1000 + K.E at Half way down

K.E at Half way down= 2000- 1000= 1000 J

Kinetic energy at half way 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 2 conditions are satisfied:

→ a force acts on the body.

→ there’s a displacement of the body by the appliance of force in or opposite to the direction of force.

If the force direction is perpendicular to the displacement, the work performed is zero. When a satellite moves round the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Hence, the work done on the satellite by the planet 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, consider a uniformly moving object,

Suppose an object is moving with constant rate. The web force performing on its zero. But, there is a displacement on the motion of the article. Hence, there will be a displacement while not 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 fodder over his head, then there’s no displacement within the bundle of fodder. Although, force of gravity is functioning on the bundle, the person isn’t applying any force thereon. Hence, within 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?

With the help of the expression, energy consumed by an electric heater will be obtained,

Power rating of the heater,

P = 500 W = 1.5 power unit Time that the heater has operated,

T = ten h Work done = Energy consumed by the heater

Therefore, energy consumed = Power × Time

= 1.5 × 10 = 15 kWh

Hence, the energy consumed by the heater in 10h 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 oscillation pendulum.

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 solely P.E. 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 solely K.E. This method is perennial as long because the apparatus oscillates.

The bob doesn’t oscillate forever. It involves rest as a result of air resistance resists its motion. The apparatus loses its K.E. to beat this friction and stops once a while. The law of conservation of energy isn’t desecrated as a result of 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?

An object with mass ‘m’ moving with velocity ‘V’ has kinetic energy of 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?

The mass of the body = 1500kg

Velocity v = 60km/hr

The work required to stop the car = kinetic energy change of the car

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.

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

In this case, the direction of force functioning on the block is within the direction of displacement. Therefore, work done by force on the block are going to be positive.

In this case, the direction of force on the block is contrary to the direction of displacement. Therefore, work done by force on the block are going to 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 associate object might be zero even once many forces are working on it. This happens once all the forces get rid of one another i.e., the online force working on the thing is zero. For a uniformly moving object, the online force working on the thing 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.

Energy consumed by an electrical device will be obtained with the assistance of the expression for power,

Power rating of the device,

P = five hundred W = 0.50 power unit Time that the device runs,

T= ten h Work d consumed by the device thus, energy c × Time

= 0.50 × 10 = 5 kWh

Hence, the energy consumed by four equal rating devices in

10 h = 4 × 5 kWh = 20 kWh = 20 Units.

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

When the object falls freely towards the bottom, its mechanical energy decreases and K.E. will increase, because the object touches the bottom, all its mechanical energy gets reborn into K.E. Because the object hits the laborious ground, all its K.E. gets reborn into heat and sound energy. It may also deform the bottom relying upon the character of the ground and therefore the quantity of K.E. possessed by the thing.

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 more number of solved numericals which are repetitively asked in the finals. 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 Question (3 long, 2 short)
• Power -5 Question (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 one important topic that provides a basic foundation for all your future studies. Work and Energy are closely related terms which are quite often used in our daily life activities. Here in this topic, students learn more in detail about work, power, force, energy and how are they 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 in detailed explanation for all the complex topics.
• Provides completely solved solutions to all the questions present in the Class 9 Science NCERT textbooks.
• These solutions include few important questions at the end of every chapter
• The language used in NCERT Solutions for Class 9 Science is easy and simple to understand by 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.

Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 11

What are the topics covered under the chapter 11 of ncert solutions for class 9 science, what is the kinetic energy of an object in the chapter 11 of ncert solutions for class 9 science, can students rely on the ncert solutions for class 9 science chapter 11 from coolgyan’s.

NCERT Solutions for Class 9 Science 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 Science Important Questions

Students can read the important questions given below for Work and Energy Class 9 Science . All Work and Energy Class 9 Notes and questions with solutions have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT and KVS. You should read all notes provided by us and Class 9 Science Important Questions provided for all chapters to get better marks in examinations. Science Question Bank Class 9 is available on our website for free download in PDF.

Important Questions of Work and Energy Class 9

Question. A body is falling from a height h. After it has fallen a height 2 X h, 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

Question. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car (a) does not change (b) becomes twice to that of initial (c) becomes 4 times that of initial (d) becomes 16 times that of initial

Question. When a body falls freely towards the earth, then its total energy (a) increases (b) decreases (c) remains constant (d) first increases and then decreases

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

Question. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g =10 ms –2 ) (a) 6 ×10 3 J (b) 6 J (c) 0.6 J (d) zero

Question. Which one of the following is not the unit of energy? (a) joule (b) newton metre (c) kilowatt (d) kilowatt hour

Question. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same (a) acceleration (b) momenta (c) potential energy (d) kinetic energy

Question. In case of negative work the angle between the force and displacement is (a) 0 0 (b) 45 0 (c) 90 0 (d) 180 0

Question. The work done on an object does not depend upon the (a) displacement (b) force applied (c) angle between force and displacement (d) initial velocity of the object

Question. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies? Answer.  Initial velocity = v, then v’ = 3v Initial kinetic energy = 1/2mv 2 Final kinetic energy (K.E) = 1/2 mv ,2 = 1/2 m (3u) 2 = 9 (1/2mu 2 )

Question. Can any object have momentum even if its mechanical energy is zero? Explain. Answer. No. Since mechanical energy is zero, there is no potential energy and no kinetic energy. Kinetic energy being zero, velocity is zero. Hence, there will be no momentum.

Question. A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a round about (Fig. 11.1) of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him.

Answer.  Displacement = 1500 m + 200 m + 2000 m = 3700 m Work done = Force × displacement = 5 N × 3700 m = 18500 J (Note: We do not need to calculate the circumference because we need to take displacement and not distance)

Question. Can any object have mechanical energy even if its momentum is zero? Explain. Answer.  Yes, mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero. Hence, there is no kinetic energy but the object may possess potential energy.

Question. The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given g = 10 m s –2 ) Answer.

Question. Avinash can run with a speed of 8 ms –1 against the frictional force of 10 N, and Kapil can move with a speed of 3 ms –1 against the frictional force of 25 N. Who is more powerful and why? Answer. Power of Avinash P A = F A U A = 10 x 8 = 80W The power of kapil P K = F K U K = 25 x 3 = 75W So, Avinash is more powerful than kapil.

Question. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g= 10 m s –2 ) Ans. mgh = m×10×10 =100mJ. Energy is reduced by 40% then the remaining energy is 60mJ. Therefore, 60m = m×10× h ‘ or h ‘ = 6m

Question. 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 the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body. Answer.  u 2 −u 2 = 2a.s This give s = u 2 – u 2 / 2a F = ma We can write work done (W) by this force F as

Question. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example. Answer. Yes, it is possible, if an object is moving in a circular path. Because force is always acting perpendicular to the direction of displacement.

Question . If an electric iron of 1200 W is used for 30 minutes every day, find electric energy consumed in the month of April. Answer. P = 1200 / 1000 = 1.2 KW t = 30 / 50 = 0.5h E = Power × time × days = 1.2 × .5 × 30 = 18 kWh

Question. Compare the power at which each of the following is moving upwards against the force of gravity? (given g = 10 ms -2 ) (i) a butterfly of mass 1.0 g that flies upward at a rate of 0.5 ms –1 . (ii) a 250-g squirrel climbing up on a tree at a rate of 0.5 ms –1 . Answer.  (i) power = mg × velocity, g = 10 ms -2

Hence, the power with which the squirrel is climbing is much higher than that of a butterfly flying.

Question. A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy? Answer. P 1 = m 1 u 1 P 2 = m 2 u 2 But P1 = P 2 or m 1 u 2 = m 2 u 2 If m 1 < m 2 then u 1 < u 2

Question. 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.  Power is the rate of doing work. Kilowatt is the unit of power and kilowatt hour is the unit of energy. h= 20 m, and mass = 2000 × 10 3  kg = 2 × 10 6  kg

Question. A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 m s -1 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? Ans.  Mass of girl = 35 kg, mass of trolley = 5kg, u = 4 m/s, v =0 and s = 16m (a) Work = F × s = m × a × s We need to calculate acceleration as follows:

Using the above value of a; work done on trolley can be calculated as follows: W = mas = 40 × 0.5 × 16 = 320 J (b) In this case, the girl is not applying any force and so work done by girl = 0

Question. 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 the men in lifting the box? (b) How much work do they do in just holding it? (c) Why do they get tired while holding it? (g= 10 ms –2 ) Answer.  (a) F = 250 kg × g (g = 10ms -2 ) = 2500 N S = 1m W = F.s = 2500 Nm = 2500 J (b) Zero, as the box does not move at all, while holding it. (c) In order 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. So, they feel tired.

Question. The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high he can jump on the planet A? Answer.  Since, weight of the person on planet A is half that on the earth, acceleration due to gravity there, will be 1/2 that on the earth. Hence he can jump double the height with the same muscular force.                                                              Or The potential energy of the person will remain the same on the earth and on planet A. Thus, mg1h1 = mg 2 h 2

Question. How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of 1 ms –1  vertically? (g= 10 ms –2 ) Answer.

Force = power / velocity = 7.5 x 104 / 20 = 3.75 x 103 N Force = 3750 N.

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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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