CBSE Class 11 Physics Exemplar NCERT Solutions: Chapter 5 - Work, Energy and Power

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 11 PDF

This chapter provides NCERT Solutions for Class 11 Physics Exemplar, focusing on Work, Energy, and Power. It covers fundamental concepts such as the change in kinetic energy of a system, the work done by forces (including magnetic and frictional forces), and the conservation of energy. The solutions explain why certain quantities remain constant during free fall and inelastic collisions, and how work done is related to force and displacement. These solutions are designed to help students grasp the core principles of work, energy, and power, and their applications in various physical scenarios, aiding in effective exam revision.

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Board	CBSE
Class	Class 11
Subject	Physics Exemplar
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 5

Chapter summary

NCERT Solutions for Class 11 Physics Exemplar, Chapter 5, delves into the principles of Work, Energy, and Power. It addresses multiple-choice questions related to kinetic energy changes, work done by different forces, and the conservation of mechanical energy during free fall. The chapter also clarifies the conservation of linear momentum in inelastic collisions, providing clear explanations for each concept.

Learning outcomes

Understand the concept of work done by forces.
Analyze changes in kinetic energy of a system.
Explain the conservation of mechanical energy.
Apply the principle of conservation of linear momentum.
Differentiate between work done by different types of forces.
Solve problems involving work, energy, and power.

Topics covered

Paper topics

Work Done
Kinetic Energy
Magnetic Force and Work
Force and Motion
Work Done by Friction
Conservation of Mechanical Energy
Free Fall
Inelastic Collision
Linear Momentum
Work-Energy Theorem

Important topics

Conservation of Mechanical Energy
Work Done by Various Forces
Conservation of Linear Momentum
Kinetic Energy Changes
Work-Energy Theorem

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Questions and Solutions

Multiple Choice Questions (MCQs)

Q. 1 An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is, because

(a) the two magnetic forces are equal and opposite, so they produce no net effect

(b) the magnetic forces do not work on each particle

(d) the magnetic forces are necessarily negligible

Solution: The magnetic force exerted by one charged particle on another is always perpendicular to the velocity of the particle. According to the definition of work done (

W = \vec{F} \cdot \vec{d}

), if the force is perpendicular to the displacement (or velocity, in this context), the work done by that force is zero. Therefore, the magnetic forces do not contribute to the change in kinetic energy of the system, and they are ignored in this calculation. The correct option is (b).

Q. 2 A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is

(a) same as the same force law is involved in the two experiments

(b) less for the case of a positron, as the positron moves away more rapidly and the force on it weakens

(d) same as the work done by charged particle on the stationary proton

Solution: The electrostatic force between two charges is given by Coulomb's law,

F = k \frac{q_1 q_2}{r^2}

. The magnitude of the force between a proton and another proton is the same as the magnitude of the force between a proton and a positron, as both have the same magnitude of charge. However, a positron is significantly lighter than a proton. According to Newton's second law (

F = ma

), the lighter positron will experience a greater acceleration for the same force. Consequently, in the same amount of time, the positron will move through a much larger distance compared to the proton. Since work done is calculated as

W = F \times d

(when force is constant and in the direction of motion), and the force is the same but the distance moved by the positron is larger, the work done on the positron will be greater than the work done on the proton. Thus, the correct option is (c).

Q. 3 A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is

(a) constant and equal to mg in magnitude

(b) constant and greater than mg in magnitude

(d) at first greater than mg and later becomes equal to mg

Solution: When a man is squatting, he is not in a stable upright position. To maintain this position, the ground must exert a reaction force that not only balances his weight (mg) but also provides any necessary support against unbalanced forces, which might include slight horizontal components or forces due to the posture. As the man begins to stand up, he exerts a force on the ground to push himself upwards. Initially, to lift his body mass against gravity and accelerate it upwards, the ground's reaction force must be greater than his weight (mg). As he completes the motion and stands upright, the upward acceleration ceases, and the reaction force from the ground becomes equal to his weight (mg) to keep him stationary. Therefore, the reaction force is initially greater than mg and then becomes equal to mg. The correct option is (d).

Q. 4 A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is

(a) + 2000J

(b) - 200J

(d) - 20,000J

Solution: The force exerted by the road on the bicycle is 200 N, and it opposes the motion over a distance of 10 m. The work done by the road on the bicycle is

W_{on\_bicycle} = F \times d \times \cos(\theta)

. Since the force is opposed to the motion,

\theta = 180^\circ

, and

\cos(180^\circ) = -1

. So, the work done by the road on the bicycle is

W_{on\_bicycle} = 200 \, \text{N} \times 10 \, \text{m} \times (-1) = -2000 \, \text{J}

. This work done by friction causes the bicycle to stop. The question asks for the work done *by the cycle on the road*. According to Newton's third law, the bicycle exerts an equal and opposite force on the road. However, the road is stationary; it does not move. Since work is defined as force applied over a distance, and the road does not undergo any displacement, the work done by the bicycle on the road is zero. The correct option is (c).

Q. 5 A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?

(a) Kinetic energy

(b) Potential energy

(d) Total linear momentum

Solution: When a body is falling freely under gravity in a vacuum, the only force acting on it is gravity. Gravity is a conservative force. In the absence of non-conservative forces (like air resistance), the total mechanical energy of the system (body + Earth) remains constant. As the body falls, its potential energy decreases because its height decreases, and its kinetic energy increases because its speed increases. However, the sum of potential energy (PE) and kinetic energy (KE), which is the total mechanical energy, remains constant. Total linear momentum is not conserved because the gravitational force causes a change in momentum. Kinetic energy and potential energy individually change during the fall. Therefore, the correct option is (c).

Q. 6 During inelastic collision between two bodies, which of the following quantities always remain conserved?

(a) Total kinetic energy

(b) Total mechanical energy

(d) Speed of each body

Solution: In any type of collision, whether elastic or inelastic, the total linear momentum of the system is conserved, provided that no external forces act on the system. In an inelastic collision, some kinetic energy is lost or converted into other forms of energy such as heat, sound, or deformation. Therefore, total kinetic energy is not conserved. Total mechanical energy is also generally not conserved in inelastic collisions. The speed of individual bodies will change during the collision. Thus, the only quantity that always remains conserved during an inelastic collision (in the absence of external forces) is the total linear momentum of the system. The correct option is (c).

Common mistakes

Confusing work done by a force with the force itself.
Incorrectly assuming kinetic energy is conserved in all scenarios.
Not accounting for all forces when calculating net work done.
Misapplying conservation laws in inelastic collisions.

Revision tips

Review the definitions of work, kinetic energy, and potential energy.
Focus on understanding the conditions under which mechanical energy is conserved.
Practice problems involving different types of forces (gravitational, magnetic, frictional).
Pay close attention to the conservation of linear momentum in collisions.

Practice MCQs

Q1. An electron and a proton are moving under the influence of mutual forces. When calculating the change in the kinetic energy of the system, the magnetic force of one on another is ignored because:

Q2. A proton is kept at rest. A positively charged particle is released from rest at a distance 'd' in its field. If the experiment is repeated with a positron instead of a proton, in the same time 't', the work done on the two moving charged particles is:

Q3. A man squatting on the ground gets straight up and stands. The force of reaction of the ground on the man during this process is:

Q4. A bicyclist comes to a skidding stop in 10 m. The force on the bicycle due to the road is 200 N, directly opposed to the motion. The work done by the bicycle on the road is:

Q5. A body is falling freely under the action of gravity alone in a vacuum. Which of the following quantities remains constant during the fall?

Q6. During an inelastic collision between two bodies, which of the following quantities always remains conserved?

Frequently asked questions

What is the main focus of Chapter 5 of the CBSE Class 11 Physics Exemplar?

Chapter 5 focuses on the fundamental concepts of Work, Energy, and Power, including how work is done, the relationship between work and energy, and the conservation principles governing these quantities.

How do these NCERT Solutions help with understanding work done by magnetic forces?

The solutions explain that magnetic forces do no work on a moving charged particle because they are always perpendicular to the velocity, thus not changing the kinetic energy.

What quantity is conserved during an inelastic collision?

During an inelastic collision, the total linear momentum of the system is always conserved, assuming no external forces act on the system.

When does total mechanical energy remain constant?

Total mechanical energy remains constant when a body is falling freely under gravity in a vacuum, or in any situation where only conservative forces are doing work.

Are the solutions provided for Class 11 Physics Exemplar?

Yes, these are NCERT Solutions specifically for Chapter 5 of the Class 11 Physics Exemplar book.

How can these solutions aid in exam preparation?

These solutions offer clear explanations and step-by-step reasoning for complex problems, helping students solidify their understanding of key concepts and improve their problem-solving skills for exams.

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