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

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

Edited By Vishal kumar | Updated on Jun 25, 2025 12:40 AM IST

Curious about how work gets done or where energy goes? The NCERT Solutions for Class 9 Science Chapter 10 Work and Energy explain these everyday ideas in a simple, easy to follow way. These solutions will not only help you understand the basics better but also enhance your exam preparation. This Chapter is very important for your exam as well as for Olympiads, you can expect at least one question in your exam from this chapter. These NCERT solutions are created by expert faculty based on latest CBSE syllabus.

The NCERT Solutions for Class 9 Science Chapter 10 Work and Energy are provided to help students prepare effectively for their CBSE exams and build a solid foundation in Physics. These NCERT Solutions for Class 9 include Exercise Questions, Additional Hot Questions, and cover all the important topics from the chapter. You will also find a list of Important Formulae and a smart Approach to solve the Questions.

This Story also Contains
  1. NCERT Solutions for Class 9 Science Chapter 10: Work and Energy
  2. Work and Energy Class 9 NCERT: Solved Exercise Questions
  3. Class 9 Science NCERT Chapter 10: Higher Order Thinking Skills (HOTS) Questions
  4. Class 9 Science Chapter 10 Work and Energy: Topics
  5. NCERT Solutions for Class 9 Chapter 10 Science: Important Formulas
  6. Approach to Solve Questions of Work and Energy Class 9
  7. NCERT Solutions for Class 9 Science Chapter-wise
NCERT Solutions for Class 9 Science Chapter 10 - Work and Energy
NCERT Solutions for Class 9 Science Chapter 10 - Work and Energy

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

These solutions help strengthen problem-solving skills and prepare students for exams effectively. Download NCERT Solutions for class 9 science chapter 10 from the below.

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Work and Energy Class 9 NCERT: Solved Exercise Questions

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

• Suma is swimming in a pond.

• A donkey is carrying a load on its back.

• A windmill is lifting water from a well.

• A green plant is carrying out photosynthesis.

• An engine is pulling a train.

• Food grains are getting dried in the sun.

• A sailboat is moving due to wind energy

Answer:

(i) Work done by Suma is negative as the force and displacement are in the opposite direction.

(ii) Work done is zero as the gravity on the load is acting vertically downward, whereas its displacement is in a horizontal direction.

(iii) Work done is positive as both force and displacement are in an upward direction.

(iv) Work done is zero as there is no displacement involved.

(v) Work done is positive as force is acting in the direction of the motion.

(vi) Work done is zero as there is no displacement of the grains.

(vii) Work done by wind force is positive as it supports the motion of 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?

Answer:

We can see that there is no net displacement in the vertical direction. It has only horizontal displacement. Thus, work done by vertical forces on the stone will be zero.

The force of gravity is acting vertically downward, thus the work done by it is zero.

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

Answer:

The chemical energy stored in a battery is transformed into the heat energy which glows the filament of the bulb. Further, the heat energy is converted into light energy.

4. Certain force acting on a 20kg mass changes its velocity from 5ms1 to 2ms1 . Calculate the work done by the force.

Answer:

By the equations of motion, we can write :

v2 = u2 + 2as

s = v2  u22a

= 22  522a = 212a m

The work done is :

W = F.s

= 20a×212a

= 210 J

Thus work done is - 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

Answer:

The displacement of the object is horizontally on the table. We know that the gravitational force is acting in a downward direction. There is no displacement vertically.

Thus the work done by the gravitational force is zero.


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

Answer:

No, there is no violation of the law of conservation of energy. At the maximum height, the energy is in the form of potential energy. When the object reaches the ground, its potential energy decreases, whereas its kinetic energy is increasing (as the velocity of the object is increasing). Thus, there is no loss of total energy (energy transformation may take place ).

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

Answer:

The muscular energy of a person is transformed in the form of mechanical energy which helps to rotate the wheel of bicycles.

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:

Since the rock doesn't move thus net displacement is zero. As a result, the work done by the force is zero.

The energy that we apply on the rock gets transformed in the form of heat.

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

Answer:

We know that 1 unit of energy is given by : 1 unit = 1 KWh.

Also, 1 KWh = 3.6×106 J

Thus 250 units in joule is given by :

E = 250×3.6×106 = 9×108 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.

Answer:

The potential energy of an object of mass m and at the height at h is given as = mgh

The potential energy of the given object is :

P.E. = mgh = 40×10×5 = 2000 J

The potential energy is being converted into the K.E..

Thus at half-way kinetic energy of the object is :

K.E. = P.E.2 = 20002 = 1000 J


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

Answer:

The work done by the gravitational force is zero. This is because the satellite is moving in a circular orbit. Thus the direction of displacement of the satellite is perpendicular to the force of gravity. Hence work done 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.

Answer:

This depends on the initial state of the object. If the object is in motion initially, then it will continue to be in this state as no external force is acting. But if the object is at rest initially then the object can't move without external 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.

Answer:

Since there was no displacement thus the work done by man is zero . This work should not be considered relatable to the term 'work' we use in daily life.

Due to upholding the weight against gravity led the man to get tired.

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

Answer:

The relation between energy and power is given by :

Energy = Power × Time

Thus the energy used in 10 hours is :

E = 1500×10 = 15000 Wh = 15 KWh

Hence the energy used by the heater 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?

Answer :

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In the above figure, point B is the mean position about which the bob rotates.

When the bob is released from point C, it attains some velocity while moving down (up to B) and deaccelerated and stops at point A.

Thus point A and point C are the maximum height points. And the velocity of the bob at point B will be maximum.

The total energy at point A and point C is only the potential energy as its velocity at these points is zero. And at point B as the height of bob is zero thus the total energy is just the kinetic energy.

Thus in this manner, the conservation of energy takes place (by transforming in some other form.).

It eventually comes to rest due to the air resistance. It deacceleration the motion of bob. (as it is a frictional force.)

There is no violation of the energy conservation law as some amount of energy is converted in the form of heat.

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?

Answer:

At this moment, the energy of the object is :

K.E. = 12mv2

Thus in bringing the object to rest the work needed is : = K.E.

W = 12mv2 J

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

Answer:

Firstly convert the velocity in SI unit.

v = 60×518 = 503 m/s

Thus the work done to stop the car is equal to the kinetic energy of the car.

W = 12mv2 = 12×1500×(503)2

= 208333.33 J or 208.33 KJ

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.

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Answer:

(i) In first case, the force acting on the object is perpendicular to the displacement of the body. Thus the work done by the force is zero.

(ii) In this, the force is in direction of the displacement so the work done is positive.

(iii) In this case, the direction of force and displacement are opposite to each other. Thus the work done by the force is 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?

Answer:

Yes. If all the external forces acting on the body balances each other then the net force acting on the object is zero. Thus the acceleration produced is zero even if forces are acting on the body. The necessary condition for the zero acceleration is that the net force acting on the body should be zero.

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

Answer:

The energy consumed by 1 device is given by :

E = 500×10 = 5000 Wh

Thus the energy consumed by 4 devices is = 4×5000 = 20 KWh

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

Answer:

When an object is falling from a height, there is an increase in the kinetic energy of the object. Thus just before hitting the ground the kinetic energy of the object is very high. But after hitting the ground its velocity comes to zero and thus the kinetic energy becomes zero. The kinetic energy is transformed into other forms of energy such as heat, sound or some deformation on the ground.

Class 9 Science NCERT Chapter 10: Higher Order Thinking Skills (HOTS) Questions

Q.1 A car of mass 1000 kg accelerates from 0 to 20 m/s in 10 seconds. What power is developed by the car's engine?

Answer:

Power developed by the engine of the car P,

P= force × velocity =(1/2)× mass × velocity 2/ time =12×1000×202=4000 W

Q.2 A car with a constant speed is moving on a straight road. What is the nature of the work done by the engine?

Answer:

Since the car is moving at a constant speed, there is no change in its kinetic energy. Therefore, the net work done on the car by all the forces acting on it is zero, including the work done by the engine.

Q.3 A uniform force of 20 N is applied to a box, displacing it by 5 m. What is the nature of the work done by the force?

Answer:

Work done is given by the equation W = Fd for constant force, where W is the work done, F is the force applied, and d is the distance traveled in the direction of the force. As the force and displacement are in the same direction, the work done is positive.

Q.4 An engine of a car of mass m = 1000 Kg changes its velocity from 5 m/s to 25 m/s in 5 minutes. The power (in KW) of the engine is

Answer:

Power expressed as the rate of change of kinetic Energy -
Power = Work done / time = change in K.E / time

P=12m[(25)2(5)2]5×60=12×1000×6005×60=1000 W

P=1KW

Q.5 A truck of mass 30,000 kg moves up an inclined plane of slope 1 in 100 at a speed of 30 kmph. The power of the truck is (given g = 10 ms2)

Answer:

P=fVP=fvsinθP=mgsinθV=30000×10×1100×30×518
P=25 kW

Class 9 Science Chapter 10 Work and Energy: Topics

This chapter explains how work is defined in science, the different forms of energy, and how energy is conserved and transformed in daily life. The main topics covered in this chapter are listed below:

10.1 Work

10.1.2 Scientific Conception Of Work

10.1.3 Work Done By A Constant Force

10.2 Energy

10.2.1 Forms Of Energy

10.2.4 Potential Energy Of An Object At A Height

10.2.5 Are Various Energy Forms Interconvertible?

10.2.6 Law Of Conservation Of Energy

10.3 Rate Of Doing Work

NCERT Solutions for Class 9 Chapter 10 Science: Important Formulas

Here are the important formulas from Class 9 Science Chapter 10 Work and Energy. These will help you solve problems more easily and understand the chapter better. Go through them often so you can remember them well during exams.

Work (W)

 Work done = Force × Displacement ×cosθW=F×d×cosθ

Kinetic Energy (KE)

Kinetic energy of a moving body =1/2× mass × (velocity) 2
KE=1/2mv2


Potential Energy (PE)

Potential energy due to height = mass × gravity × height

PE=mgh


Mechanical Energy (ME)


Total energy = Kinetic Energy + Potential Energy

ME=KE+PE


Work-Energy Theorem

Work done = Change in kinetic energy

W=ΔKE


Power (P)

Power = Work done / Time taken

P=W/t

Approach to Solve Questions of Work and Energy Class 9

To solve questions from Class 9 Science Chapter 10: Work and Energy, start by carefully reading the question to identify the given values such as force, distance, mass, velocity, or time. Next, recall and choose the correct formula based on what the question is asking for example, use W=F×d×cosθ for work done, KE=12mv2 for kinetic energy, PE=mgh for potential energy, or P=Wt for power. Make sure all quantities are in standard SI units mass in kilograms, distance in meters, and time in seconds. After that, substitute the values into the formula and solve step by step. Finally, write your answer clearly with the correct unit like joules (J) for energy and work or watts (W) for power.

NCERT Solutions for Class 9 Science Chapter-wise

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NCERT Solutions for Class 9 - Subject Wise

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Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. What are the main topics to be covered to understand the Work And Energy Class 9 Science chapter 10 NCERT solutions?

The main topics covered in the NCERT syllabus of Class 9  chapter 10 are work, calculation of work, unit of work, energy, unit of energy,  kinetic energy, potential energy, work energy theorem, concept of conservation of energy, power and commercial unit of energy

2. Is the chapter helpful for exams like NSEJS?

Yes, definitely. Almost every year there are questions from the class 9 chapters of NCERT. More than one question can be expected from the chapter Work And Energy for NSEJS. In the some of the previous year papers more than one questions were asked.

3. Whay should I solve Work And Energy Class 9 NCERT Science Chapter 10 questions?

Solving the NCERT questions will give knowledge of how to apply the concepts studied to solve a numerical question and will give conceptual clarity. To solve more problems refer to NCERT Exemplar questions for Class 9 chapter 10 Work And Energy.

4. Are NCERT Solutions for Class 9 Science Chapter 10 sufficient for exam preparation?

While NCERT Solutions for Class 9 Science Chapter 10 are a good resource for exam preparation, it is recommended that students also practice solving other reference books and previous year's question papers to get a better understanding of the concepts and exam patterns.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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