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

Last updated date: 23rd May 2024
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NCERT Exemplar for Class 9 Science - Work and Energy - Free PDF Download

Free PDF download of NCERT Exemplar for Class 9 Science Chapter 11 - Work and Energy solved by expert Science teachers on Vedantu as per NCERT (CBSE) Book guidelines. All Chapter 11 - Work and Energy exercise questions with solutions to help you to revise complete syllabus and score more marks in your examinations.

Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 9 Science , Maths solutions and solutions of other subjects. You can also download NCERT Solutions for Class 9 Maths to help you to revise the complete syllabus and score more marks in your examinations.

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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. 

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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:




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 \[100\]

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 \[{v^2} - {u^2} = 2as\] 

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


$ 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.