case study questions class 11 physics work energy and power

Class 11 Physics Case Study Questions Chapter 6 Work, Energy, and Power

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In Class 11 Final Exams there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 11 Physics Chapter 6 Case Study and Passage-Based Question s with Answers were Prepared Based on the Latest Exam Pattern. Students can solve Class 11 Physics Case Study Questions Work, Energy, and Power  to know their preparation level.

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In CBSE Class 11 Physics Paper, 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, Energy, and Power Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 11 Physics  Chapter 6 Work, Energy, and Power

Case Study/Passage-Based Questions

Case Study 1: The kinetic energy possessed by an object of mass, m, and moving with a uniform velocity, v is

Kinetic energy is a scalar quantity. The kinetic energy of an object is a measure of the work and 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).

Work energy theorem:  The change in kinetic energy of a particle is equal to the work done on it by the net force. Mathematically

K f  – K i  = W

Where K i  and K f  are respectively the initial and final kinetic energies of the object. Work refers to the force and the displacement over which it acts. Work is done by a force on the body over a certain displacement.

1) Kinetic energy is

• a) Scalar quantity
• b) Vector quantity
• c) None of these

Answer: a) Scalar quantity

2) Which of the following has the same unit?

• a) Potential energy and work
• b) Kinetic energy and work
• c) Force and weight
• d) All of the above

Answer: d) All of the above

3) A car with an initial kinetic energy of 400 J is brought to a stop by applying brakes. How much work is done by the brakes? a) -400 J b) 0 J c) 400 J d) 800 J

Answer: a) -400 J

4) What happens to the kinetic energy of an object if its velocity is doubled? a) It remains the same. b) It becomes four times. c) It becomes half. d) It becomes twice.

Answer: b) It becomes four times.

5) According to the work-energy theorem, if a body’s kinetic energy decreases, the work done by the net force must be: a) Positive b) Negative c) Zero d) Infinite

Answer: b) Negative

What is the kinetic energy of an object with a mass of 10 kg and a velocity of 5 m/s? a) 125 J b) 250 J c) 500 J d) 50 J

Answer: a) 125 J

Case Study 2: The work-energy theorem states that – the change in the kinetic energy of a body is equal to the work done by the net force. In deriving the theorem, it is assumed that force is effective only in changing the KE. When the force and displacement are in the same direction, KE increases and work done is positive. When the force and displacement are in opposite directions, KE decreases and work done is negative. When the body is in uniform motion, KE does not change and work done by centripetal force is zero.

(i) A body of mass 10 kg initially at rest, acquires a velocity of 10 m/s. The work done is: (a) -500J (b) 500J (c) 50J (d) – 50J

Answer: (a) -500J

(ii) How much work must be done by a force on a 50 kg body in order to accelerate from rest to 20 m/s in 10 sec? (a) 10 3 J (b) 10 4 J (c) 2 X 10 3 J (d) 4 X 10 4 J

Answer: (b) 104 J

(iii) A gun of mass M fires a bullet of mass m with maximum speed v. The KE of gun will be? (a) ½ mv2 (b) 1/2 Mv2 (c) more than ½ mv2 (d) less than½ mv2

Answer: (d) less than½ mv2

(iv) An unloaded car moving with velocity v on a frictionless road can be stopped in a distance s. If the passengers add 40% to its weight and the breaking force remains the same then the stopping distance will be: (a) 1.4 s (b) 1.5 s (c) 1.6 s (d) 1.8 s

Answer: (a) 1.4 s

(v) A block of mass 10 kg is moving in the x direction with a constant speed of 10 m/s. It is subjected to a retarding force F = – 0.1 xj/m during its travel from x=20m to x=30m. Final KE will be (a) 250J (b) 275J (c) 450J (d) 475J

Answer: (d) 475J

Hope the information shed above regarding Case Study and Passage Based Questions for Class 11 Physics Chapter 6 Work, Energy, and Power with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 11 Physics Work, Energy, and Power 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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CBSE Case Study Questions Class 11 Physics PDF Download

Are you a Class 11 Physics student looking to enhance your understanding and prepare effectively for your exams? Look no further! In this comprehensive guide, we present a curated collection of CBSE Case Study Questions Class 11 Physics that will help you grasp the core concepts of Physics while reinforcing your problem-solving skills.

CBSE 11th Standard CBSE Physics question papers, important notes, study materials, Previous Year Questions, Syllabus, and exam patterns. Free 11th Standard CBSE Physics books and syllabus online. Important keywords, Case Study Questions, and Solutions.

Class 11 Physics Case Study Questions

CBSE Class 11 Physics question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 11 Physics students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 11 Physics

• Chapter 1: Physical World
• Chapter 2: Units and Measurements
• Chapter 3: Motion in a Straight Line
• Chapter 4: Motion in a Plane
• Chapter 5: Laws of Motion
• Chapter 6: Work, Energy, and Power
• Chapter 7: System of Particles and Rotational Motion
• Chapter 8: Gravitation
• Chapter 9: Mechanical Properties of Solids
• Chapter 10: Mechanical Properties of Fluids
• Chapter 11: Thermal Properties of Matter
• Chapter 12: Thermodynamics
• Chapter 13: Kinetic Theory
• Chapter 14: Oscillations
• Chapter 15: Waves

Class 11 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 11 Physics examinations. Our expert faculty for standard 11 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 11 students understand the concepts and also easy-to-learn solutions.

Class 11 Books for Boards

Why Case Study Questions Matter

Case study questions are an invaluable resource for Class 11 Physics students. Unlike traditional textbook exercises, these questions simulate real-life scenarios, challenging students to apply theoretical knowledge to practical situations. This approach fosters critical thinking and helps students build a deep understanding of the subject matter.

Let’s delve into the different topics covered in this collection of case study questions:

1. Motion and Gravitation

In this section, we explore questions related to motion, velocity, acceleration, and the force of gravity. These questions are designed to test your grasp of the fundamental principles governing motion and gravitation.

2. Work, Energy, and Power

This set of questions delves into the concepts of work, energy, and power. You will encounter scenarios that require you to calculate work done, potential and kinetic energy, and power in various contexts.

3. Mechanical Properties of Solids and Fluids

This section presents case study questions about the mechanical properties of solids and fluids. From stress and strain calculations to understanding the behavior of fluids in different situations, these questions cover a wide range of applications.

4. Thermodynamics

Thermodynamics can be a challenging topic, but fear not! This part of the guide offers case study questions that will clarify the laws of thermodynamics, heat transfer, and thermal expansion, among other concepts.

5. Oscillations and Waves

Get ready to explore questions related to oscillations, simple harmonic motion, and wave characteristics. These questions will deepen your understanding of wave propagation and the behavior of oscillatory systems.

6. Kinetic Theory and Laws of Motion

Kinetic theory and the laws of motion can be complex, but with our case study questions, you’ll find yourself mastering these topics effortlessly.

Discover a wide array of questions dealing with light, lenses, and mirrors. This section will improve your problem-solving skills in optics and enhance your ability to analyze optical phenomena.

8. Electrical Effects of Current

Electricity and circuits are fundamental to physics. The case study questions in this section will challenge you to apply Ohm’s law, Kirchhoff’s laws, and other principles in various electrical circuits.

9. Magnetic Effects of Current

Delve into the fascinating world of magnets and magnetic fields. This set of questions will strengthen your understanding of magnetic effects and their applications.

10. Electromagnetic Induction

The final section covers electromagnetic induction, Faraday’s law, and Lenz’s law. You’ll be presented with scenarios that test your ability to predict induced electromotive forces and analyze electromagnetic phenomena.

In conclusion, mastering Class 11 Physics requires a thorough understanding of fundamental concepts and their practical applications. The case study questions provided in this guide will undoubtedly assist you in achieving a deeper comprehension of the subject.

Remember, practice is key! Regularly attempt these case study questions to strengthen your problem-solving abilities and boost your confidence for the exams. Happy studying, and may you excel in your Physics journey!

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Class 11th Physics - Work, Energy and Power Case Study Questions and Answers 2022 - 2023

By QB365 on 08 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Physics Subject - Work, Energy and Power, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Work, energy and power case study questions with answer key.

11th Standard CBSE

Final Semester - June 2015

Mechanical energy exists in two forms: Kinetic energy and Potential energy. Kinetic energy is the energy possesed by a body by virtue of motion. Potential energy is the energy possessed by the body by virtue of its position or configuration. These two forms of energy are interconvertible. If no other form of energy is involved in a process, the sum of kinetic energy and potential energy always remains constant. (i) State law of conservation of mechanical energy. (ii) State two particles having mass m 1  and m 2 , both have equal linear momenta. What is the ratio of their kinetic energies? (iii) Two particles of masses m 1  and m 2 have equal kinetic energies. What is the ratio of their linear momenta? (iv) A particle of mass m has half the kinetic energy of another particle of mass m/2. If the speed of the heavier particle is increased by 2 ms -1 , its new kinetic energy equals the original kinetic energy of the lighter particle. What is the ratio of the original speeds of the lighter and heavier Particle? (v) A uniform rod of mass m and length I is made to stand vertically on one end. What is the potential energy of the rod in this position? (vi) Give an example where a force does work on a body but fails to change its K-E. (vii) Does K-E depend upon the direction of motion involved? Can it be negative? Does its value depend on frame of refrence?

Work is said to be done by a force acting on a body, provided the body is displaced actually in any direction except in a direction perpendicular to the direction of the force-mathematically,  \(W=\bar{F} \cdot \bar{s}=F s \cos \theta\)  whereas energy is the capacity of a body to do the work and Power is the rate at which the body do the work. \(P=\frac{\mathrm{W}}{t}=\frac{\overline{\mathrm{F}} \cdot \bar{s}}{t}=\overline{\mathrm{F}} \cdot \bar{v}\) Both, work and energy are measured in Joule while power is measured in watt. (i) A box is pushed through 4.0 m across a floor offering 100 N resistance. Determine the work done by the applied force. (ii) In the above question, determine the work done by the resistive force and by the gravity. (iii) A truck draws a tractor of mass 1000 kg at a steady rate of 20 ms -1  on a level road. The tension in the coupling is 2000 N. What is the power spent on the tractor? (iv) Determine the work done on the tractor in one minute?

In a conservative force field, we can find the component of force from the potential energy at a point in the field. A positive force means repulsion and a negative force means attraction. From the given potential energy function U (r) we can find the equilibrium position where force is zero. Suppose the potential energy at a distance r from centre of the field is given as \( \mathrm{U}(r)=\frac{A}{r^{2}}-\frac{B}{r} \) Where A and B are positive constants. (i) What should be the nature of the field as per conclusion drawn from the form of given potential energy? (ii) Determine the work done to move the particle from equilibrium to infinity. (iii) If K.E of a body becomes 4 times of its initial value, then what would be new linear momentum? (iv) Determine the percentage change in K.E of a body if momentum of a body increases by 0.01 % (v) What does it meant by unstable equilibrium of a particle? Write condition for unstable equilibrium. (vi) What are conservative forces?

*****************************************

Work, energy and power case study questions with answer key answer keys.

(i) Total mechanical energy of a system is conserved if the forces doing work on it are conservative i.e., Individually kinetic energy K and the potential energy V(x) may vary from point to point, but their sum is constant throughout. (ii) Since  \(K=\frac{1}{2} m v^{2}=\frac{p^{2}}{2 m} ; \frac{K_{1}}{K_{2}}=\frac{m_{2}}{m_{1}}\) (iii) Since Linear momentum  \(p=\sqrt{2 m \mathrm{~K}}\) \(\frac{p_{1}}{p_{2}}=\sqrt{\frac{m_{1}}{m_{2}}}\) (iv) If v and v' be the original speed of the heavier and the lighter particles respectively. \(\frac{1}{2} m v^{2}=\frac{1}{2} \times\left\{\frac{1}{2}\left(\frac{m}{2}\right) v^{\prime 2}\right\}\) \(\therefore \ v^{2}=\frac{v^{\prime 2}}{4} \text { or } v^{\prime}=2 v\) (v) Potential energy in the vertical position = work done is raising if from horizontal position to vertical position. Since the entire mass of the rod is concentrated at centre of mass which is raised through a height  \(\frac{h}{2}\)  therefore workdone = mgh =  \(m g \frac{l}{2}\) (vi) When a body is dragged on a rough horizontal surface with a constant velocity work is done against friction but K-E = constant. (vii) No, Kinetic energy does not depend on the direction of motion and it K.E cannot be negative. The magnitude of K.E depends upon the frame of Refrence.

(i) W = Fs cos \(\theta\)  = 100 x 4 cos 0° = 400 J (ii) Resistive force opposes the applied force. Box moves at 180 o  to the resistive force. Therefore W = Fs .cos 180 = -400 J. Since motion is along horizontal and gravity is along the vertical, therefore workdone by gravity is W = Fs cos 90 = zero. (iii) Force applied = tension in coupling = 2000 N As P = Fv cos \(\theta\)  = 2000 x 20 cos 0 o  = 40000W = 40 kW (iv) Work done = Power x time = 40 kW x 60 s = 2400 kJ

(i) Workdone  \( W=\bar{F} \cdot \bar{r} \) \( =\overline{\mathrm{F}} \cdot\left(\bar{r}_{2}-\bar{r}_{1}\right) \) \( =(\hat{i}+3 \hat{j}+\hat{k}) \cdot[(5 \hat{i}+6 \hat{j}+9 \hat{k})-(3 \hat{i}+2 \hat{j}-4 \hat{k})] \) \( =(\hat{i}+3 \hat{j}+\hat{k}) \cdot(2 \hat{i}+4 \hat{j}+13 \hat{k})=27 \mathrm{~J} \)   (ii) Net displacement from the time body is projected to the time it hits the ground is = h (vertically downwards) \(\therefore\)  work done W = mgh (iii) For time  \( t=\frac{u \sin \theta}{g} \)  body attains the maximum height  \( h_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g} \) therefore work done  \( m g h_{\max }=\frac{m u^{2} \sin ^{2} \theta}{2} \)   (iv) Since during the time  \( t=\frac{2 u \sin \theta}{g} \text { i.e. } \)   the time of flight of projectile, displacement is zero, therefore workdone by gravitational force is zero. (v) Workdone = Force x displacement W = area enclosed by F-x curve from given figure \( w=3 \times 3+\frac{1}{2} \times 3 \times 3=13.5 \mathrm{~J} \)   (vi) Velocity is maximum when K.E. is maximum for minimum P.E \( \frac{d v}{d x}=0 \Rightarrow x^{2}-x=0 \)   \( \Rightarrow \quad x=\pm 1 \) \( \text { P.E. }=\frac{1}{4}-\frac{1}{2}=\frac{-1}{4} \) \( \mathrm{K} \cdot \mathrm{E}_{\max }+\mathrm{P.E}_{\min }=2 \) \( \mathrm{K} \cdot \mathrm{E}_{\max }=2+\frac{1}{4}=\frac{9}{4} \) \( \Rightarrow \quad \frac{1}{2} M v_{\max }^{2}=\frac{9}{4} \) \( \Rightarrow \quad \frac{1}{2} \times 1 \times v_{\max }^{2}=\frac{9}{4} \) \( \Rightarrow \quad v_{\max }=\frac{3}{\sqrt{2}} \mathrm{~m} / \mathrm{s} \)

(i) The field is surely conservative (ii) Work required can be calculated from P.E at the equilibrium position \( W=\frac{A}{r_{0}^{2}}-\frac{B}{r_{0}} \) where r 0 - equilibrium position At equilibrium  \( r_{0}, \frac{d U}{d r}=0 \Rightarrow r_{0}=\frac{2 A}{B} \) Therefore  \( W=\frac{B^{2}}{4 A} . \)   (iii) Since K.E = p 2 /2M, when K.E becomes 4 times, p 2 is 4 times therefore linear momentum P becomes 2 times. (iv) Since  \( \Delta P=M \Delta V=0.01 \% \) \(\therefore\)   \(\triangle\)  V = 0.01% Now  \( \Delta E=\frac{1}{2} m(\Delta V)^{2} \)   \(\therefore\)   \(\triangle\) E = 2 x 0.0 I = 0.02% (v) In this a particle when displaced from equilibrium position, tends to move away from equilibrium position. potential energy V = maximum and  \( \frac{d^{2} V}{d x^{2}} \)   = negative. (vi) When work done by against a force in displacing a particle does not depend upon the path along which particle is displaced, then such a force is known as conservative force. These forces keep the K.E constant and work done will be zero when particle is displaced in closed path.

(i) From the conservation of X and Y components of linear momentum mu = mv 1  cos 30° + mv 2 cos 30° \( \Rightarrow \ u=\left(v_{1}+v_{2}\right) \frac{\sqrt{3}}{2} \)   and 0 = mv 1  sin 30 - mV 2 sin 30 \(\Rightarrow\)  v 1  = v 2 \( v_{1}=\frac{u}{\sqrt{3}} \)   (ii) Total K.E before collision is  \( \mathrm{K}_{i}=\frac{1}{2} m u^{2} \)   and after collision  \( \mathrm{K}_{f}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2} \)   \( =\frac{1}{2} m\left(v_{1}^{2}+v_{2}^{2}\right)=\frac{1}{2} m\left(2 v_{1}^{2}\right)=\frac{m u^{2}}{3} \) \( \frac{\mathrm{K}_{f}}{\mathrm{~K}_{i}}=\frac{2}{3} \)   (iii)  \( e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=0 \)   (iv) As the collision is inelastic, body loses some energy so that K.E of ball does not remain the same. However, total energy and total momentum of ball and earth system remain the same.

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• NCERT Solutions for Class 11 Physics Chapter 6 - Work Energy And Power
• NCERT Solutions

Download NCERT Solutions for Class 11 Physics Chapter 6: Free PDF

NCERT Solutions for Class 11 Physics Chapter 6 have been prepared for students to provide them with in-depth knowledge of Chapter 6 Work, Energy, and Power.

NCERT Solution of Work, Energy, and Power Class 11 also comes in handy for students who are preparing for competitive exams like JEE. These solutions efficiently cover all the essential concepts within the chapter. Students often find Physics Ch 6 Class 11 Work, Energy, and Power to be extremely difficult. But you’ll be able to effortlessly understand the chapter if the difficult concepts are explained methodically.

Would you like to view a summarized version of this chapter? Check out the 'Chapter at a glance' section below the PDF of NCERT Solutions.

Work, Energy and Power Chapter at a Glance - Class 11 NCERT Solutions

1. Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities. The work done by the friction or viscous force on a moving body is negative. 

2. A force is conservative if (i) work done by it on an object is path independent and depends only on the initial and final position, or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position. 

3. The gravitational potential energy of a particle of mass m at a height x about the earth’s surface is U(x) = m g x where the variation of g with height is ignored. 

4. The elastic potential energy of a spring of force constant k and extension x is $U(x)=\dfrac{1}{2}kx^{2}$

5. The potential energy of a body subjected to a conservative force is always undetermined upto a constant. For example, the point where the potential energy is zero is a matter of choice. For the gravitational potential energy mgh, the zero of the potential energy is chosen to be the ground.  For the spring potential energy kx 2 /2, the zero of the potential energy is the equilibrium position of the oscillating mass. 

6. For a conservative force in one dimension, we may define a potential energy function U(x) such that

$F(x)=-\dfrac{dU(x)}{dx}$ or $U_{i}-U_{f}=\int_{x_{i}}^{x_{f}}F(x)dx$

7. For equilibrium $F=-\dfrac{dU}{dx}=0$

8. The work – energy theorem states that the change in kinetic energy of a body is the work done by the net force on the body. 

K f – K i = W net

9. The work done by a force can be calculated sometimes even if the exact nature of the force is not known. This is calculated with the help of work energy theorem by using change in kinetic energy

10. The WE theorem holds in all inertial frames. It can also be applied in non  internal frames provided we include the pseudo forces in the calculation of the net force acting on the body under consideration.

11. Every force encountered in mechanics does not have an associated potential energy. For example, work done by friction over a closed path is not zero and no potential energy can be associated with friction. 

12. The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces that act on the body are conservative.

13. Linear Momentum

The total linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. $\vec{P}=M\vec{v}_{cm}$

14. Impulse

Impulse of a force $\vec{F}$ acting on a body for the time interval t=t 1 to t=t 2 as: $I=\int_{t_{1}}^{t_{2}}F.dt$ And also, 

$\vec{I}=m(\vec{v_{2}}-\vec{v_{1}})=\Delta \vec{P}$

= Change in momentum due to force $\vec{F}$ .

15. Conservation of Linear Momentum

According to law of conservation of linear momentum, total linear momentum of a system of particles remain constant or conserved in the absence of any external force.

i.e, When $\vec{F}_{ext}=0$

$\Rightarrow \dfrac{d\vec{p}}{dt}=0$

$\Rightarrow \vec{p}$ constant

i.e. $\vec{p}_{initial}=\vec{p}_{final}$

Also, for n number of particles

$\vec{p}_{1}+\vec{p}_{2}+\vec{p}_{3}...\vec{p}_{n}=constant$

For collision of two bodies, the total momentum before collision remains the same as the total momentum after the collision.

i.e., $m_{1}\vec{u}_{1}+m_{2}\vec{u}_{2}=m_{1}\vec{v}_{1}+m_{2}\vec{v}_{2}$ 

Recoil velocity of gun is calculated by, $v_{2}=-\dfrac{m_{1}v_{1}}{m_{2}}$

where, m 2 = mass of the gun, m 1 = mass of bullet and v 1 = velocity of the bullet.

16. Collision

It is an isolated event, in which two or more colliding bodies exert strong forces on each other for a short duration of time.

It is mainly of two types: elastic and inelastic collision.

For every type of collision, linear momentum of colliding body or system is conserved.

i.e $m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

where, m 1 and m 2 = masses of the body which undergo collision.

u 1 = initial velocity of the body of mass m 1 ,

u 2 = initial velocity of the body of mass m 2 ,

v 1 = final velocity of the body of mass m 1 , and 

v 2 = final velocity of the body of mass m 2 .

But kinetic energy of the colliding body and system is conserved in elastic collision only.

17. Coefficient of Restitution (e) 

It is the ratio of relative velocity of separation after collision to the relative velocity of approach before collision. It is expressed as $e=\dfrac{\left | v_{2}-v_{1} \right |}{\left |u_{1}-u_{2}  \right |}$ where $0\leq e\leq 1$

(i) For perfectly inelastic collision, e = 0.

(ii) For perfectly elastic collision, e = 1 and for inelastic collision 0 < e < 1.

(iii) For other collisions, it can be 0 < e < 1.

18. Head-on Collision

For bodies with masses m 1 and m 2 respectively following are the important relations for head-on collision.

(i) When collision is elastic, final velocities for m 1 i.e, $v_{1}=\dfrac{(m_{1}-m_{2})}{m_{1}+m_{2}}u_{1}+\dfrac{2m_{2}u_{2}}{m_{1}+m2}$ and for m 2 , $v_{2}=\dfrac{2m_{1}u1}{m_{1}+m_{2}}+\dfrac{(m_{2}-m_{1})}{m_{1}+m_{2}}$

(ii) When collision is inelastic Final velocities for m 1 ,

$v_{1}=\left ( \dfrac{m_{1}-em_{2}}{m_{1}+m_{2}} \right )u_{1}+\left ( \dfrac{(1+e)m_{2}}{m_{1}+m_{2}} \right )u_{2}$

and for m 2 ,

$v_{2}=\left ( \dfrac{(1+e)m_{1}}{m_{1}+m_{2}} \right )u_{1}+\left ( \dfrac{m_{2}-em_{1}}{m_{1}+m_{2}} \right )u_{2}$

If after collision, approaching bodies move with a common velocity, i.e. e = 0 (get stuck with one another). then collision is said to be perfectly inelastic.

19. For perfectly elastic oblique collision 

Along X-axis, $m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}cos\theta +m_{2}v_{2}cos\phi$

Along Y-axis, $0=m_{1}v_{2}sin\theta -m_{2}v_{2}sin\phi$

If two bodies of equal masses undergo perfect elastic oblique collision then scattering angle $\theta +\phi =\dfrac{\pi }{2}$ and $u_{1}^{2}=v_{1}^{2}+v_{2}^{2}$.

20. Rebounding of a Ball on collision with Floor.

Speed of the ball after nth rebound, $v_{n}=e^{n}v_{0}=e^{n}\sqrt{2gh}$

Height covered by the ball after nth rebound, $h_{n}=e^{2n}h$

Total distance s covered  by the ball before it stops bouncing , $s=h\left ( \dfrac{1+e^{2}}{1-e^{2}} \right )$  

where, h = height of the ball dropped from ground and e = coefficient of restitution

Access NCERT Solutions for Class 11 Physics Chapter 6 – Work Energy and Power

1. The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.

Ans: Positive. Work done is the dot product of force and displacement. This means that the work will be positive if the projection of one of them on the other is in the same direction as the other. In other words, if the two vectors make an acute angle, then the work done is positive. Here, the bucket is being lifted upwards, which is in the same direction of the pull-force being applied. Hence the work done is positive.

work done by gravitational force in the above case,

Ans: Negative. The gravitational force is acting downward, and the displacement is upwards, which means the angle these make \[\sim 180{}^\circ \](obtuse). This means that the dot product of the two vectors would be negative.

work done by friction on a body sliding down an inclined plane,

Ans: Negative.

The direction of frictional force is opposite to the direction of motion; hence the work done by the frictional force is negative in this case.

work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,

Ans: Positive. The frictional force is acting in the direction opposite to the motion of the body. However, to maintain the constant velocity, extra force is applied in the direction of the motion. So this means that the work done by this extra force is positive.

work done by the resistive force of air on a vibrating pendulum in bringing it to rest. 

The resistive force always acts in the direction opposite to the motion of the body (pendulum). Thus the work done by this force is also negative.

2. A body of mass \[\mathbf{2}\text{ }\mathbf{kg}\] initially at rest moves under the action of an applied horizontal force of \[7N\] on a table with coefficient of kinetic friction \[=0.1\]. Compute the

work done by the applied force in \[10s\]

Ans: Given the mass of the body, \[m=2kg\]

The force applied on the body, \[F=7N\]

The given coefficient of kinetic friction, \[\mu =0.1\]

The initial velocity of the body, \[u=0\]

Time, \[t=10s\]

The acceleration produced in the body by the applied force is given by Newton’s second law

of motion as:

\[a'={}^{F}/{}_{m}\text{=}{}^{7}/{}_{2}\text{=}3.5\text{ }m{{s}^{-2}}\] 

And the frictional force is given as: \[f=\mu mg=0.1\times 2\times 9.8=1.96N\]

The acceleration produced by the frictional force:

\[a''=1.96/2=0.98\text{ }m{{s}^{-2}}\]

Total acceleration of the body is the sum of the two accelerations: \[a'+a''=3.5+0.98\text{ }m{{s}^{-2}}\]

Using the total acceleration, the distance traveled by the body is given by the equation of motion:

\[s=ut+{}^{1}/{}_{2}a{{t}^{2}}=0+{}^{1}/{}_{2}\times 2.52\times {{10}^{2}}=126m\]

Work done by the applied force (F),

\[{{W}_{a}}=F.s=7\times 126=882J\].

Hence the work done by the applied force is 882J.

work done by friction in \[10s\], 

Ans: The work done by the frictional force (f), 

\[{{W}_{f}}=f.s=-1.96\times 126=-247J\].

work done by the net force on the body in \[10s\], 

Ans: The net force acting on the body\[=7+\left( 1.96 \right)=5.04N\]

Hence the work done by the net force in the 10s, is equal to the product of the force and the distance travelled by the body, \[{{W}_{Net}}=5.04\times 126=635J\] .

change in kinetic energy of the body in \[10s\], and interpret your results

Ans: Using the first equation of motion, final velocity can be calculated as:

\[v=u+at=0+2.52\times 10=25.2\text{ }m/s\]

The change in kinetic energy is

\[={}^{1}/{}_{2}m{{v}^{2}}-{}^{1}/{}_{2}m{{u}^{2}}={}^{1}/{}_{2}m({{v}^{2}}-{{u}^{2}})=635J\]

Hence the change in the kinetic energy is obtained as 635J.

3. Given in figure are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

The kinetic energy (T) of a body is given as the difference in the total energy of a system (E) and the potential energy (V):

\[T=E-V\] 

Kinetic energy of a body is dependent on its mass and the square of its velocity. Neither of these quantities can be negative, hence Kinetic energy is also a positive quantity. 

Therefore, the particle cannot exist in a region where T becomes negative or when \[V>E\].

For distance more than a, i.e., for \[x>a\], the Kinetic energy becomes negative, hence the case is not possible.

The minimum energy required is equal to the maximum potential energy shown i.e., E should be at least equal to \[{{V}_{0}}\]

All throughout, the total energy E is smaller than the potential energy V. Hence the kinetic energy is negative, which isn’t possible in a physical case. Hence a particle can’t be found for the given energy. The minimum energy required is equal to the maximum potential energy shown i.e., the E should be at least equal to \[{{V}_{3}}\]

For distance \[{x < a}\] and \[{x>b}\],  the kinetic energy becomes negative. Hence the scenario is not possible. However, for \[b>x>a\], the kinetic energy is either positive or zero. This is possible. 

But for the particle to stay anywhere in the system, the minimum energy required is equal to the maximum potential energy is shown, i.e., E should be at least equal to \[{{V}_{0}}\].

For \[b/2<x<a/2;\,\text{ }a/2<x<b/2;\]  the potential energy of the particle becomes greater than the total energy. Hence the particle can’t stay here. 

However, it can stay everywhere else. In case the particle is needed in this region, then the minimum total energy in this case would be \[{{V}_{0}}\].

4. The potential energy function for a particle executing linear simple harmonic motion is given by\[V(x)=k{}^{{{x}^{2}}}/{}_{2}\], where k is the force constant of the oscillator. For \[k=0.5N/m\], the graph of \[V(x)\]versus x is shown in the given Fig. Show that a particle of total energy \[1J\]moving under this potential must ‘turn back’ when it reaches \[x=\pm 2m\].

Ans: As given, the total energy of the particle, \[E=\text{ }1\text{ }J\] 

The potential energy of the particle \[V(x)=k{}^{{{x}^{2}}}/{}_{2}\]

The force constant of the oscillator is \[k=0.5N{{m}^{-1}}\]

The total energy is: \[E=V+T\] where T is the Kinetic energy of the particle. According to the conservation law, this energy does not change throughout the motion.

\[\Rightarrow 1={}^{1}/{}_{2}m{{v}^{2}}+{}^{1}/{}_{2}k{{x}^{2}}\]

When the pendulum ‘turns back’, velocity (and hence T) becomes zero. 

\[\Rightarrow 1={}^{1}/{}_{2}k{{x}^{2}}\]

\[\Rightarrow x=\pm 2\]

Hence, the particle turns back when it reaches \[x=\pm 2\].

5. Answer the following:

The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?

Ans: Rocket.

Total energy is given by the relation:

$TE=PE+KE=\frac{1}{2}m{{v}^{2}}+mgh$

Now, when the casing of a rocket burns due to friction, the mass of the rocket gets reduced, which further causes the total energy to reduce. Clearly, it is at the cost of the rocket that the heat needed for burning is generated.

Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?

Ans: Gravitational force is a conserved force, which means that the work done by this force is independent of the path taken by the object. This means that if an object comes back to a place where it started from, i.e. when the total displacement is zero (i.e., a closed path), the work done is zero.

An artificial satellite orbiting the earth in a very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?

Ans: The atmospheric resistance costs the satellite its total energy. This means that the total energy keeps decreasing. So even when the kinetic energy increases due to the increase in the speed, the potential energy decreases much faster, leading to an overall drop in the energy. 

So basically, the energy of the satellite is given as the sum of its kinetic energy, its potential energy, and the energy it loses to the atmospheric resistance.

In (i), the man walks \[\mathbf{2m}\] carrying a mass of \[15kg\] on his hands. In (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley and a mass of \[15kg\] hangs at its other end. In which case is the work done greater? 

Ans: In the case where the person is carrying the weight horizontally, which means he has applied the force (F) in the vertically upward direction, but the displacement (s) is horizontal. The work done (W) is given by \[W=F.s=Fs\cos \theta \] and the \[\theta \] is the angle between the two vectors. Since \[\theta =90{}^\circ \]the work becomes zero. 

Similarly, in the second case, the person is pulling the rope horizontally and the motion of the weight is upwards, and hence the two vectors are perpendicular to each other. Hence the work is zero.

6. Underline the correct alternative:

 When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.

Ans: decreases.

The body moves towards the lower potential hence decreasing the potential energy of the body. 

 Work done by a body against friction always results in a loss of its kinetic/potential energy.

Ans: Kinetic energy.

Friction usually reduces the velocity of the body and hence reduces the Kinetic energy as well.

 The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.

Ans: External force.

Internal forces cannot produce any change in the total momentum of a body. It’s a fact that the sum of all the internal forces is zero. Hence, the total momentum of a many- particle system is proportional to the external forces acting on the system.

In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies

Ans: Total linear momentum.

The momentum can neither be created nor destroyed. However the energies can change their forms.

7. State if each of the following statements is true or false. Give reasons for your answer.

 In an elastic collision of two bodies, the momentum and energy of each body is conserved.

Law of conservation of momentum states that the total momentum of the system is conserved when there is no net external force applied on the system. The momentum of one body can be transferred to other body in the system.

 Total energy of a system is always conserved, no matter what internal and external forces on the body are present.

Any external force on a body can change the total energy of the body.

 Work done in the motion of a body over a closed loop is zero for every force in nature.

Total energy over a closed-loop motion is only zero if the force involved is conservative. Eg. The friction would do zero work in a closed-loop.

 In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

In inelastic collisions, the total energy is usually lost as sound or heat or vibrations which means that the initial kinetic energy is more than the final kinetic energy

8. Answer carefully, with reasons:

In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?

At the point of contact, the moving billiard ball drops its velocity to zero and is still in the process of transferring its momentum and energy to the next ball. At this moment, the ball, which came to a sudden stop after hitting the next ball, has transferred its kinetic energy to the potential energy (of vibration), which is then transferred to the next ball as its kinetic energy.

Is the total linear momentum conserved during the short time of an elastic collision of two balls?

In an elastic collision, the total linear momentum is always conserved at every point in time.

What are the answers to (a) and (b) for an inelastic collision?

Ans: No, Yes.

There is a definite loss of kinetic energy in inelastic collisions, which happens during the collision. This is another reason why the total kinetic energy is not conserved.

The law of conservation of momentum is applicable to both elastic and inelastic collisions. Hence the total linear momentum is always conserved throughout the collision.

If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Ans: Elastic.

If the potential energy is only dependent on the separation distance, this means that the induced force is independent of the path that is taken by the balls. Hence the force is conservative. This means that the collisions here are also going to be elastic.

9. A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to 

\[(i)\,{{t}^{1/2}}\,\,(ii)\,t\,\,(iii)\,{{t}^{3/2}}\,\,(iv)\,{{t}^{2}}\]

Ans: It is given that the body is accelerated with a constant value-\[a\] hence the corresponding force experienced is \[F=ma\].

Let’s say that the initial velocity of the body is \[u\].

Thus, the velocity- \[v\] of the body at any given point in time- \[t\] can be written as \[v-u=a{{t}^{{}}}\]

And the power is

\[P=F.v=(ma)v\]

\[\Rightarrow P=ma(at+u)=m{{a}^{2}}t+mau\]

The mass, acceleration, and initial velocity are constant.

\[\Rightarrow P\propto t\]

Hence it can be noted that power is linearly dependent on t.

10.  A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to 

\[(i)\,{{t}^{1/2}}\,\,(ii)\,t\,\,(iii)\,{{t}^{3/2}}\,\,(iv)\,{{t}^{2}}\] 

Ans: In this case, the power is constant:

Since the mass of the body doesn’t change, the acceleration and the velocity are inversely proportional.

\[\Rightarrow a=\frac{F}{mv}\]

\[\frac{dv}{dt}=a\]

Equating the two accelerations

\[\frac{dv}{dt}=a=\frac{F}{mv}\]

\[\Rightarrow \frac{m}{F}\int{vdv}=\int{dt}\]

\[\Rightarrow \frac{m}{F}{{v}^{2}}=t+C\]

\[\Rightarrow v\propto {{t}^{1/2}}\]

Hence the option (i) is valid.

11. A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \[F=(-\hat{i}+2\hat{j}+3\hat{k})N\]

Where i, j, k are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?

Ans: Force exerted on the body, \[F=(-\hat{i}+2\hat{j}+3\hat{k})N\] .

Displacement, \[s=4\hat{k}m\] .

Hence the work done, \[W=F.s=(-\hat{i}+2\hat{j}+3\hat{k}).(4\hat{k})=12J\] .

Therefore it is concluded that the work done is \[12J\].

12. An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy \[\mathbf{10}\text{ }\mathbf{keV}\], and the second with \[\mathbf{100}\text{ }\mathbf{keV}\].Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass \[{{m}_{e}}=9.11\times {{10}^{-31}}kg\], proton mass \[{{m}_{p}}=1.67\times {{10}^{-27}}kg\],\[\mathbf{eV}=\text{ }\mathbf{1}.\mathbf{60}\times \mathbf{1}{{\mathbf{0}}^{-19}}J\]) .

Ans: The given mass of the electron, \[{{m}_{e}}=9.11\times {{10}^{-31}}kg\] and that of the proton is \[{{m}_{p}}=1.67\times {{10}^{-27}}kg\].

The given kinetic energy of the electron is \[{{T}_{e}}=10keV=1.60\times {{10}^{-15}}J\] , and the kinetic energy of the proton is \[{{T}_{p}}=100keV=1.60\times {{10}^{-14}}J\].

The velocity of the electron is hence obtained as: \[{{v}_{e}}={{\left( \frac{2{{T}_{e}}}{{{m}_{e}}} \right)}^{1/2}}=5.93\times {{10}^{7}}m/s\]

Similarly, the velocity of the proton can be obtained as   \[{{v}_{p}}={{\left( \frac{2{{T}_{p}}}{{{m}_{p}}} \right)}^{1/2}}=4.38\times {{10}^{6}}m/s\]

The ratio of their velocities is \[\frac{{{v}_{e}}}{{{v}_{p}}}=\frac{13.54}{1}\]

Hence the electron moves 13.54 times faster.

13. A rain drop of radius \[\mathbf{2}\text{ }\mathbf{mm}\] falls from a height of \[\mathbf{500}\text{ }\mathbf{m}\] above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is \[10m/s\]?

Ans: The given radius of the raindrop, \[r=2\times {{10}^{-3}}m\]

The volume of the drop is\[V=(4/3)\pi {{r}^{3}}\]

And hence the mass is \[m=\rho V=\rho (4/3)\pi {{r}^{3}}\]

Therefore the gravitational force is \[F=mg=\rho (4/3)\pi {{r}^{3}}g\]

Thus the work by gravitation in the first half of the journey of the drop: \[{{W}_{1}}=F.s=0.082J\] and here \[s=250m\]

This is equal to the amount of work done in the second half of the journey, \[{{W}_{2}}=0.082J\]

If there is no resistance force, then the total energy of the raindrop will remain the same. Hence the total energy on the top is: \[{{E}_{T}}=mgh+0=0.164J\].

The resistive force makes the drop have a velocity of \[10m/s\]. Hence the total energy at the ground:

\[{{E}_{G}}=(1/2)m{{v}^{2}}=1.675\times {{10}^{-3}}J\]

Therefore the energy lost in the resistance is the difference between the energies: \[={{E}_{T}}-{{E}_{G}}=-0.162J\]

Hence the energy due to the resistive force is -0.162J.

14. A molecule in a gas container hits a horizontal wall with speed \[200m/s\] and angle \[{{30}^{o}}\] with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Ans: The momentum of any system remains conserved as long as there aren’t any external forces acting on the system. Similarly, the momentum of the system in the gas container remains conserved irrespective of the collision being elastic or inelastic.

As the gas molecules strike the stationary wall of the container, they rebound at the same speed. Thus the velocity of the wall remains zero. This means that the collision is elastic. Hence, the total kinetic energy of the molecule remains conserved during the collision. 

15. A pump on the ground floor of a building can pump up water to fill a tank of volume \[30{{m}^{3}}\] in \[15\min \]. If the tank is \[40m\] above the ground, and the efficiency of the pump is \[30\%\], how much electric power is consumed by the pump?

Ans: Volume that can be filled in 15 minutes: \[V=30{{m}^{3}}\]

And the time of operation, \[t=15\min =900s\].

The tank is at \[h=40m\]height

And the given pump is of the efficiency \[\eta =30\%=0.3\] .

The average power required for lifting the water to this height is:

\[P=\frac{E}{t}=\frac{mgh}{t}=13.067\times {{10}^{3}}W\]

This is the required power output, so we need a pump whose power output needs to be equal to this. In other words, the \[30\%\]of the pump's output should be equal to the required power.

\[P=30\%\,{{P}_{pump}}\]

\[\Rightarrow {{P}_{pump}}=\frac{100P}{30}=43.6\times {{10}^{3}}W=43.6kW\]

Hence we need a pump that consumes \[43.6kW\].

16. Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V . If the collision is elastic, which of the following figure is a possible result after collision?

Ans : In all the cases of collisions where there is no external force, the linear momentum of the system is unchanged. In the case of elastic collision, the kinetic energy is also conserved.

Let’s say that all the ball bearings have mass- \[m\], so the total kinetic energy before collision:

\[{{T}_{i}}=\frac{1}{2}m{{V}^{2}}+\frac{1}{2}m{{0}^{2}}+\frac{1}{2}m{{0}^{2}}=\frac{1}{2}m{{V}^{2}}\]

Now, one needs to check that the final kinetic energy is equal to the initial one.

The final kinetic energy \[{{T}_{f}}=\frac{1}{2}m{{0}^{2}}+\frac{1}{2}m{{\left( \frac{V}{2} \right)}^{2}}+\frac{1}{2}m{{\left( \frac{V}{2} \right)}^{2}}=\frac{1}{4}m{{V}^{2}}\].

The final kinetic energy is\[{{T}_{f}}=\frac{1}{2}m{{0}^{2}}+\frac{1}{2}m{{0}^{2}}+\frac{1}{2}m{{V}^{2}}=\frac{1}{2}m{{V}^{2}}\].

The final kinetic energy is \[{{T}_{f}}=\frac{1}{2}m{{\left( \frac{V}{3} \right)}^{2}}+\frac{1}{2}m{{\left( \frac{V}{3} \right)}^{2}}+\frac{1}{2}m{{\left( \frac{V}{3} \right)}^{2}}=\frac{1}{6}m{{V}^{2}}\]

Hence it is clearly seen that the kinetic energy is conserved only in case (ii).

17. The bob A of a pendulum released from\[{{30}^{o}}\] to the vertical hits another bob B of the same mass at rest on a table as shown in figure. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Ans: If the collision is elastic, then the bob-A will transfer all its momentum to the other bob. After the collision, bob A will come to rest, and bob B will move with the initial velocity of A in the horizontal direction. Thus the bob A will not rise after the collision. 

18. The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance? 

Ans: As provided, the length of the pendulum, \[l=1.5m\] 

The mass of the bob \[=m\]

Energy dissipated \[=\text{ }5%\]

The law of conservation of momentum states the linear momentum of a system is unchanged in the absence of external force.

In the horizontal position

The PE of the bob is \[{{E}_{P}}=mgl\,\], while the KE is \[{{E}_{K}}=0\]and hence the total energy is \[E=mgl\]

While at the lowermost point or the mean position, the potential energy is \[{{E}_{P}}=0\] , the KE is \[{{E}_{K}}=\frac{1}{2}m{{v}^{2}}\] , and the total energy is \[E=\frac{1}{2}m{{v}^{2}}\].

But the bob loses 5% of its energy. 

Hence the total energy at the lowermost point is 95% of the total energy at the higher/horizontal position

\[\Rightarrow \frac{1}{2}m{{v}^{2}}=\frac{95}{100}mgl\]

\[\Rightarrow v=5.28m/s\].

Hence the bob arrives with 5.28m/s at the lowest point.

19. A trolley of mass \[300kg\] carrying a sandbag of \[25kg\] is moving uniformly with a speed of \[27km/h\] on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of. \[\mathbf{0}.\mathbf{05}\text{ }\mathbf{kg}/\mathbf{s}\]. What is the speed of the trolley after the entire sand bag is empty?

Ans: As the trolley carrying the sand bag is moving without friction, which means, there is no friction which can act as an external force on the system. Hence the law of conservation of momentum is applicable.

The given initial mass of the bag \[{{m}_{s}}(t=0)=25kg\] and of the trolley is\[{{m}_{Tr}}=300kg\], hence the total initial mass is \[{{M}_{Tot}}(t=0)=325kg\].  And the mass drop rate is \[\frac{dM}{dt}=0.05kg/s\].

The initial velocity of the trolley is \[v(t=0)=\,27km/h=7.5m/s\]

The momentum at any given time \[t\]is \[p=(325-0.05t)v(t)\] and at \[t=0\], the momentum \[p(t=0)=325v(t=0)=325\times 27=8775kg\,\,m/s\]

The time at which the sandbag empties, \[{{t}_{final}}=\frac{25}{0.05}=500s\]

Thus the momentum at \[{{t}_{final}}\]can be written as  

\[8775=(325-0.05\times 500).v({{t}_{final}}=500)\]

\[\Rightarrow 8775=300.v({{t}_{final}})\]

\[\Rightarrow v({{t}_{final}})=29.25m/s\]

Hence the final velocity when the bag is emptied is \[29.25m/s\].

20. A body of mass \[m=0.5kg\,\] travels in a straight line with velocity \[v=a{{x}^{3/2}}\,\] where \[a=5{{m}^{1/2}}{{s}^{-1}}\]. What is the work done by the net force during its displacement from \[x=0\] to \[x=2\]?

Ans: The given mass of the body is \[m=0.5kg\] and is moving with velocity\[v=a{{x}^{3/2}}\]where \[a=5{{m}^{1/2}}{{s}^{-1}}\].

The kinetic energy at \[x=0\] is \[{{E}_{K}}(x=0)=\frac{1}{2}0.5\times 0\] and at \[x=2\] is \[{{E}_{K}}(x=2)=\frac{1}{2}0.5\times {{(5\times {{2}^{3/2}})}^{2}}=50J\].

The difference in these two values is equal to the work done.

\[W={{E}_{K}}(x=2)-{{E}_{K}}(x=0)=50J\]

Hence the work done is \[50J\].

21. The blades of a windmill sweep out a circle of area \[=A\]

If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?

Ans: As provided, the circular area swept by the windmill \[=A\]

And the velocity of the wind is \[=v\]

And the density of the air is \[=\rho \]

The volume flow rate of wind through the windmill is\[=Av\]

And hence the mass flow rate is \[m/t=\rho Av\]

Hence the mass flowing in time t is \[m=\rho Avt\].

What is the kinetic energy of the air? 

Ans: Using this mass the kinetic energy is obtained as

\[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\rho A{{v}^{2}}t={{E}_{wind}}\]

This is also equal to the energy that the wind carries with itself.

Assume that the windmill converts \[25%\] of the wind’s energy into electrical energy, and that \[A=30{{m}^{2}}\], \[v=36km/h\] and the density of air is \[\rho =1.2kg/{{m}^{3}}\]. What is the electrical power produced?

Ans: The area of the circle swept is \[A=30{{m}^{2}}\]

And the given velocity is \[v=36km/h=10m/s\]

The density is \[\rho =1.2kg/{{m}^{3}}\]

Since the windmill converts only 25% of wind energy to electricity.

 \[{{E}_{electric}}=25%{{E}_{wind}}=\frac{1}{8}\rho A{{v}^{2}}t\]

Hence the electric power is obtained as \[{{P}_{electric}}={{E}_{electric}}/t=4.5kW\].

The electric power output from the windmill is 4.5kW.

22. A person trying to lose weight (dieter) lifts a \[10kg\] mass, one thousand times, to a height of \[0.5m\] each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

How much work does she do against the gravitational force?

Ans: Mass of the dumbbell \[m=10kg\]

The height up to which it is lifted \[h=0.5m\]

Hence the work done with every lift is\[mgh\]

Number of times the dumbbell is lifted\[n=1000\]

Hence the total work done against gravity is  \[=n\,mgh=49kJ\].

Fat supplies \[3.8\times {{10}^{7}}J\] of energy per kilogram which is converted to mechanical energy with a \[20%\] efficiency rate. How much fat will the dieter use up?

Ans: Energy supplied with every kilogram of fat is\[3.8\times {{10}^{7}}J\]

The efficiency is\[20%\]

Hence the mechanical energy supplied by the body

\[=\frac{20}{100}3.8\times {{10}^{7}}J=7.6\times {{10}^{6}}J\]

And the equivalent fat loss is \[6.45\times {{10}^{-3}}kg\]

Hence the body loses \[6.45\times {{10}^{-3}}kg\] of fat.

23. A family uses \[8kW\]of power. 

Direct solar energy is incident on the horizontal surface at an average rate of \[200W\]per square meter. If \[20%\] of this energy can be converted to useful electrical energy, how large an area is needed to supply \[8kW\]? 

Ans: The given power requirement of a family is\[P=8kW=8\times {{10}^{3}}W\]

The solar energy per sq mts \[{{P}_{solar}}/A=200W\]

Efficiency in converting solar to electrical energy \[\eta =20%\]

Let the area needed to generate the required electricity be \[A\]

Total solar power is  \[{{P}_{solar}}=({{P}_{solar}}/A).A\]

And the expected electrical power output can be written as\[P=20%({{P}_{solar}})=\frac{20}{100}200\times A\]

\[\Rightarrow A=200{{m}^{2}}\]

Hence the area required is \[A=200{{m}^{2}}\].

Compare this area to that of the roof of a typical house.

Ans: A typical house can have a roof of the dimensions \[15\times 15\,{{m}^{2}}\], hence the area is \[\ge 200{{m}^{2}}\]. And this is more than the required area.

24. A bullet of mass \[0.012kg\] and horizontal speed \[70m/s\]strikes a block of wood of mass \[0.4kg\]and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Ans: As provided, the mass of the bullet is\[m=0.012kg\]

And the initial speed of the bullet before hitting the block , \[{{u}_{b}}=70m/s\]

Mass of the wooden block, \[M=0.4kg\]

The initial speed of the stationary wooden block, \[{{u}_{B}}=0\]

Let the final speed of the system of the bullet and the block be \[{{v}_{b+B}}\]

According to the law of conservation of momentum: 

\[m{{u}_{\mathbf{b}}}+M{{u}_{B}}=(m+M){{v}_{b+B}}\]

\[\Rightarrow {{v}_{b+B}}=\frac{(0.012\times 70+0.4\times 0)}{0.012+0.4}=2.04m/s\]

For the system of the bullet and the wooden block when they are stuck together: 

The total mass of the system, \[{{M}_{b+B}}=0.412kg\]

And the resultant velocity of the system \[{{v}_{b+B}}=2.04m/s\]

Height up to which the system rise\[=h\]

Applying the law of conservation of energy to this system: 

The potential energy at the highest point is equal to the Kinetic energy at the lowest point 

\[{{M}_{b+B}}gh=(1/2){{M}_{b+B}}{{v}_{b+B}}^{2}\]

\[\Rightarrow h=\frac{{{v}_{b+B}}^{2}}{2g}=0.2123m\]

Hence the wooden block rises up to the height of \[0.2123m\]

The energy lost in the form of heat is the drop in the energy of the bullet.

\[=\frac{1}{2}m{{u}_{b}}^{2}-\frac{1}{2}{{M}_{b+B}}{{v}_{b+B}}^{2}=29.4-0.857=28.54J\]

Hence the heat produced is \[28.54J\].

25. Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (figure). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. 

Given \[{{\theta }_{1}}={{30}^{o}}\] , \[{{\theta }_{2}}={{60}^{o}}\] and \[h=10m\] , what are the speeds and times taken by the two stones?

Ans: AB and AC are both smooth planes. 

Intuitively it can be said that, as height of both the planes is the same, therefore, north the stones will reach the bottom with same speed. But since the side AB is longer, the stone will take more time to reach down.

The mathematics for the same is as follows

The potential energy at A is supposed to be equal to the kinetic energies of the stones when they reach B and C

\[mgh=\frac{1}{2}m{{v}_{1}}^{2}=\frac{1}{2}m{{v}_{2}}^{2}\]

\[\Rightarrow {{v}_{1}}={{v}_{2}}\]

The accelerations along the sides are given as

\[{{a}_{1}}=g\sin {{\theta }_{1}}\]

\[{{a}_{2}}=g\sin {{\theta }_{2}}\]

As \[{{\theta }_{2}}>{{\theta }_{1}}\], \[\Rightarrow {{a}_{2}}>{{a}_{1}}\]

Now since \[v=u+at=0+at\]

\[\Rightarrow {{t}_{2}}<{{t}_{1}}\]

Hence the second stone will reach sooner than the first stone, which attains the same speed while reaching down but will take less time.

26. A \[1\text{ }kg\]block situated on a rough incline is connected to a spring of spring constant \[100N/m\]as shown in figure. The block is released from rest with the spring in the upstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Ans: AS provided in the question, the mass of the block, \[M=\text{ }1\text{ }kg\]

Spring constant, \[k=100N/m\]

Displacement in the block along the plane, \[x=0.1\,m\]

At equilibrium, the normal reaction is \[R=Mg\cos {{37}^{o}}\]

The friction force acting opposite to \[Mg\sin {{37}^{o}}\]is \[f=\mu R=\mu Mg\cos {{37}^{o}}\]

Here the coefficient of friction is \[\mu \]

The net force acting on the block is 0, and its expression in terms of its components is: \[Mg\sin {{37}^{o}}-f=Mg(\sin {{37}^{o}}-\mu \cos {{37}^{o}})\]

Also, the work done is equal to the potential energy of the spring 

\[Mg(\sin {{37}^{o}}-\mu \cos {{37}^{o}})x=(1/2)k{{x}^{2}}\]

Hence the required coefficient of friction is

\[\Rightarrow \mu =\frac{\left( \sin {{37}^{o}}-(1/2)\frac{kx}{Mg} \right)}{\cos {{37}^{o}}}=0.115\]

Therefore the coefficient of friction is \[0.115\].

27. A bolt of mass \[0.3kg\]falls from the ceiling of an elevator moving down with an uniform speed of \[7m/s\] . It hits the floor of the elevator (length of the elevator = \[3m\]) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Ans: The given bolt has a mass of \[m=0.3kg\]

The elevator moves with speed \[v=7m/s\]

And the height of the elevator is \[h=3m\]

The relative velocity of the bolt wrt lift is zero; hence at the time of impact, the change in potential energy is converted to heat.

\[\Delta PE=mgh=0.3\times 9.8\times 3=8.82J\]

The heat produced is dependent on the relative velocity of the bolt wrt the elevator, and not on anything else. Hence the heat generated remains the same even if the elevator is moving.

28. A trolley of mass \[200kg\]moves with a uniform speed of \[v=36km\]on a frictionless track. A child of mass \[20kg\,\,\]runs on the trolley from one end to the other (10 m away) with a speed of \[4m/s\]relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Ans: As provided, the trolley weighs \[M=200kg\]  and is \[l=10m\] long

It moves with a velocity of \[v=36km/h=10m/s\]

And the mass of the boy is \[m=20kg\]

The initial momentum of trolley+boy is \[(M+m)v=2200kg\,m/s\]

Now let the final velocity be \[v'\] 

And the boy’s final velocity wrt ground is\[v'-4\]

Hence the final momentum is

\[Mv'+m(v'-4)=220v'-80\]

Since there is no external force, the final momentum is equal to the initial momentum

\[2200=220v'-80\]

\[\Rightarrow v'=10.36m/s\]

The boy’s speed is \[v''=4m/s\,\,\]

The time taken to run across the trolley \[t=10/4=2.5s\,\,\]

Hence the distance moved by the trolley is\[v''t=25.9m\].

29. Which of the following potential energy curves in figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Ans: The gravitational potential energy of a system of two masses is inversely related to the separation between the two of them. 

However, In the given case, the potential energy of the system of the two balls will decrease as they come closer to each other. It will become zero when the two balls touch each other, i.e., when the distance between the two is 2R. 

The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions, and hence they can’t describe the elastic collisions. 

30. Consider the decay of a free neutron at rest:\[n\to {{p}^{+}}+{{e}^{-}}\]

Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in decay the of a neutron or a nucleus (figure).

Ans: The given neutron decay at rest can be written as 

\[n\to {{p}^{+}}+{{e}^{-}}\]

According to Einstein's mass-energy relation, the mass lost (\[\Delta m\]) will be taken by the ejected electron as energy\[\Delta m{{c}^{2}}\]. The mass defect (\[\Delta m\]) is the difference in the mass of the neutron and the sum of the proton masses and electron.

\[\Delta m={{m}_{neutron}}-({{m}_{{{p}^{+}}}}+{{m}_{{{e}^{-}}}})\]

The speed of light is denoted by \[c\]

Hence the equation misses showing the change in the mass or the energy and where it went. Adding an energetic neutrino \[\nu \] on the RHS would solve the issue.

A Glimpse of Class 11 Physics Work, Energy, and Power NCERT Solutions

Although you are familiar with the three terms: Work, Energy Power, learning about the relationship between these three is made easy with the help of NCERT Class 11 Physics Work, Energy Power solutions.

The first two subtopics explain the scalar product and the notions of work and kinetic energy. Additionally, the first part of Work, Energy and Power includes the work-energy theorem, the concept of kinetic energy, potential energy, conservation of mechanical energy, and much more.

NCERT solutions Class 11th Physics Chapter 6 provides you with step-by-step solutions of Mathematical problems and detailed theoretical representation to help you grasp these concepts.

NCERT solution of Physics Class 11 Chapter 6 in the next part explains the concept of power and collisions by providing lucid diagrammatic representations.

Topics Covered in Class 11 Physics Chapter 6 - Work, Energy, and Power 

Below are some of the key concepts discussed in this chapter.

The work-energy theorem

Kinetic energy

Work done by a variable force

The work-energy theorem for a variable force

The concept of potential energy

The conservation of mechanical energy

The potential energy of a spring

Law of conservation of energy

NCERT Solutions For Class 11 Physics Chapter 6 PDF Download

NCERT solutions for Class 11 Physics Chapter 6 Work, Energy and Power is now available for download in PDF format. The Class 11 Physics Chapter 6 NCERT solutions are prepared by our subject-matter experts who have many years of experience in the field of physics.

The comprehensive, step-by-step Class 11 Physics Ch 6 NCERT solutions strictly adhere to the CBSE guidelines and hence, act as an extremely reliable resource. Work, Energy and Power Class 11 NCERT PDF also includes illustrations to help you understand the complicated problems better.

NCERT solutions for Class 11 Physics Work, Energy and Power PDF mainly focus on complex problems based on the above-mentioned topics. But it takes a smart approach in providing you with in-depth theoretical knowledge by incorporating questions that require you to choose the correct alternative.

Work, Energy and Power Class 11 NCERT PDF has proved to be extremely helpful for students to prepare for their examinations in a short period by highlighting all essential theories and mathematical problems.

Download the NCERT Physics Class 11 Chapter 6 PDF today through the link given on this page. Study anywhere and everywhere from your favourite device. Scoring the highest marks and standing out in class has never been easier.

Key Features of NCERT Solutions for Class 11 Physics Chapter 6 - Work, Energy and Power

These NCERT Solutions are prepared in order to help students in quickly finding important concepts from Work, Energy, and Power.

All concepts are explained in a detailed manner.

NCERT Solutions are clear and easy to understand.

They are prepared by subject experts as per the syllabus curated by experts.

These NCERT Solutions on Work, Energy, and Power help students in developing strong conceptual foundations for students, which is important in the final stages of preparation for board and competitive exams.

These NCERT Solutions are available in PDF format and can be downloaded for free.

Benefits of NCERT Solutions for Class 11 Physics Chapter 6

NCERT solutions for Class 11 Physics Chapter 6 Work, Energy and Power have been able to help the students in many ways.

Here are some of the benefits of NCERT Solutions of Work, Energy and Power Class 11:

Provides Step-by-Step Solutions: Physics NCERT solutions Class 11 Work, Energy and Power provides step-by-step solutions to complex problems that are essential for your class tests and final examinations. It covers the entire exercise given in your book.

Quality of Answers: CBSE Class 11 Physics Chapter 6 NCERT solutions are prepared by our team of experts abiding by all the instructions given by CBSE. All answers are uniquely designed to cover the entire topic, explaining to you all the important concepts along with illustrations when necessary.

Available for Free: You can access the Work, Energy and Power Class 11 NCERT solutions without paying any charges.

Useful for Competitive Exams: The extensive study material Physics NCERT solutions Class 11 Work, Energy and Power is valid and extremely useful even for competitive exams.

Requires Less and Qualitative Time to Prepare: NCERT solutions Physics Class 11 Chapter 6 helps you to grasp the complex topics like power, collisions, potential energy quicker by providing well-structured answers.

Important Questions from Work, Energy, and Power (Short, Long, and Practice Questions)

Short answer type questions.

1.  A spring is kept compressed by pressing its ends together lightly. It is then placed in strong acid and released. What happens to its stored potential energy?

2. Why are the clock pendulums made of invar, a material of low value of the coefficient of linear expansion?

3. How would a thermometer be different if glass expanded more with increasing temperature than mercury?

Long Answer Type Questions

1. An object of mass 0.4kg moving with a velocity of 4m/s collides with another object of mass 0.6kg moving in the same direction with a velocity of 2m/s. If the collision is perfectly inelastic, what is the loss of K.E. due to impact?

2. A ball is dropped on the floor from a height of 2cm. After the collision, it rises up to a height of 1.5m. Assuming that 40% of mechanical energy lost goes to thermal energy into the ball. Calculate the rise in temperature of the ball in the collision. The specific heat capacity of the ball is 800J/k. Take g = 10m/s 2 .

3. If the volume of a block of metal changes by 0.12% when it is heated to 200C. What is the coefficient of linear expansion of the metal?

Practice Questions

1. If one Mole of a monatomic gas is mixed with 3 moles of a diatomic gas. What is the molecular-specific heat of the mixture at constant volume?

2. A stone of mass 5 kg falls from the top of a cliff 30 m high and buries itself one metre deep into the sand. Find the average resistance offered and the time taken to penetrate into the sand.

3. A pump on the ground floor of a building can pump up water to fill a tank of volume 30m 3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

The given N CERT solutions are based on the latest NCERT curriculum. These solutions provide the students to understand the physics concepts in an easy way. All the solutions are clearly solved and presented in a step-by-step process. Students can rely on these solutions as they are prepared by highly skilled subject experts. You can download NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy, and Power pdf to access it offline anytime. This pdf helps you during  the exam preparation and is extremely helpful in scoring good marks in the exam. This PDF covers all the important types of the problems in the chapter. 

FAQs on NCERT Solutions for Class 11 Physics Chapter 6 - Work Energy And Power

1. What does Class 11 Physics Chapter 6 of NCERT book depict?

Class 11 Physics Chapter 6 of NCERT book deals with Work, Energy, and Power. With the help of this chapter, students will smoothly understand the relation between these three physical quantities. Scalar product and its distributive laws have also been taught in this chapter to form the foundation of further complex concepts.

Various other topics related to Work, Energy, and Power have been discussed in this chapter such as concepts of Kinetic energy, work done by variable force, work-energy theorem for a variable force, potential energy, conservation of mechanical energy, potential energy of a spring. Numerical problems have also been explained in a very detailed manner using theoretical and mathematical expressions.

Keeping in mind that energy comes in various forms such as heat, chemical energy, electrical energy, laws of conservation of energy is also explained in a detailed manner. Similarly, the principle of conservation of energy is also included in this chapter.

2. What are the topics/ subtopics covered under this chapter of Class 11 Physics?

There are a total of 12 topics and sub-topics in Chapter 6 of Class 11 Physics titled Work, Energy and Power. Check the topics and sub-topics given below.

Ex 6.1 - Introduction

Ex 6.2 - Notions of work and kinetic energy: The work-energy theorem

Ex 6.3 - Work

Ex 6.4 - Kinetic energy

Ex 6.5 - Work done by a variable force

Ex 6.6 - The work-energy theorem for a variable force

Ex 6.7 - The concept of potential energy

Ex 6.8 - The conservation of mechanical energy

Ex 6.9 - The potential energy of a spring

Ex 6.10 - Various forms of energy: the law of conservation of energy

Ex 6.11 - Power

Ex 6.12 - Collisions

3. How many questions are contained in the exercises of Class 11 Physics Chapter 6 of NCERT book?

A total of 30 questions are there in the exercises of Class 11 Physics Chapter 6 of NCERT book. In Addition, answers to all these 30 questions are being provided in the NCERT Solutions by Vedantu. These answers are accumulated by the expert teachers of India's leading ed-tech site in an accurate and logical manner.

Being a part of Unit IV of NCERT textbook, Work, Energy and Power that holds 17 marks of weightage in the final Class 11 Physics exam along with Unit V and VI; Motion of System of Particles and Rigid Body and Gravitation respectively.

4. Why should one opt for the NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy, and Power offered by Vedantu?

There are so many advantages of referring to the NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and power offered by Vedantu. These solutions include all the important topics with in-depth and detailed explanations that aim to help the students of Class 11 understand the concepts in a better way. These solutions help the students to build a strong foundation about Work Energy and power.

The NCERT Solutions for Class 11 Physics Chapter 6 pdf available provided by Vedantu are explained in a logical order so that the students can easily understand all the concepts and hold a strong command over those. Going through the solutions provided here will also help the students to understand how they should approach and solve all the problems that they may face during the actual exam.

5. Where can I find numericals for Class 11 Physics Chapter 6 work, energy and power?

Students can solve all numericals given in Class 11 physics NCERT textbook. Students can find solutions to all numericals given in the textbook on Vedantu’s website. Students can understand the method of solving numericals to score high marks in Class 11 physics. All numericals are solved using simple strategies that can help students to understand the key points for solving numericals easily in the exam. All numericals are solved by expert physics teachers for easy understanding of the students. 

6. What is the triple point of water?

The triple point of any substance refers to the combination of temperature and pressure at which a substance can exist in all three states: solid, liquid, and gas. The triple point of water is given a value of 273.16 K and a vapour pressure of 611.66 Pascal. At this point, water can exist in all three states: vapour, liquid, and ice by making small changes in pressure and temperature.

7. Which chapter is Work Energy and Power in Class 11 Physics?

Work, Energy, and Power is Chapter 6 of Class 11 Physics. It is an important Chapter as it covers a wide range of important concepts from the exam point of view. Students can get NCERT Solutions for Chapter 6 Class 11 on the Vedantu website and the app. Vedantu gives all Solutions in PDF format. Students can download the PDF file free of cost and study the notes prepared by expert teachers easily just by visiting the website. 

8. How can I download NCERT Solutions Class 11 Physics Chapter 6 Work Energy And Power?

NCERT Solutions Class 11 Physics Chapter 6 Work Energy And Power are given on the internet. Students can download the NCERT Solutions Class 11 Physics Chapter 6 from the Vedantu website.

Visit the page-NCERT Solutions Class 11 Physics Chapter 6.

You download the given PDF file on your computer.

Save the file on your computer.

You can use the saved file anytime for offline study. 

NCERT solutions can help students to prepare for their exams and score good marks. 

9. What is the formula of Kinetic Energy?

The formula of Kinetic Energy $\dfrac{1}{2}MV^2$. You can find the complete explanation of the chapter on Vedantu. It will help make the chapter easier. Chapter 6 is an interesting chapter as long as you understand the concepts thoroughly. 

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NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power PDF Download

The NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power contains a vast collection of questions to practise. By practising these questions of Chapter 6 Work, Energy and Power, students can strengthen their problem solving skills as well as reasoning skills; these skills can be used in further chapters and real-life problems. These questions of Chapter 6 Work, Energy and Power can also help students to build a strong foundation for the chapter. 

NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power PDF

The Chapter 6 Work, Energy and Power Chapter 6 Work, Energy and Power can be a bit confusing if students are not able to understand the concepts properly. For that purpose students need to start practising Chapter 6 Work, Energy and Power questions from the NCERT textbook. After practising those questions, students can prefer referring to the NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power PDF which is available in the Selfstudys website. 

Exercise Wise NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

In the NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power, a plethora of exercises are given so that students can understand the concepts as well as can solve confusions then and there. By practising exercise wise questions, students can learn to approach different questions of Chapter 6 Work, Energy and Power in different and creative ways. 

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The questions in the NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power are arranged according to each and every formula. By practising the questions of Chapter 6 Work, Energy and Power formula wise, students can become more confident while applying the formulas. By applying the right formulas in the right questions of Chapter 6 Work, Energy and Power, students can score good marks in those questions.

Where can Students Find the NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power?

Students can find the NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power, steps to attempt are clearly explained below: 

• Visit the Selfstudys website. 

• Bring the arrow towards NCERT Books & Solutionss which can be seen in the navigation bar. 
• A drop down menu will appear, select NCERT Solutionss from the list. 

• A new page will appear, select Class 11 from the list of classes. 

• Now select the subject Physics from the list. 
• Again a new page will appear, select the chapter Chapter 6 Work, Energy and Power. 

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MCQ Questions for Class 11 Physics Chapter 6 Work, Energy and Power with Answers

We have compiled the NCERT MCQ Questions for Class 11 Physics Chapter 6 Work, Energy and Power with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 11 Physics with Answers on a daily basis and score well in exams. Refer to the Work, Energy and Power Class 11 MCQs Questions with Answers here along with a detailed explanation.

Work, Energy and Power Class 11 MCQs Questions with Answers

Multiple Choice Type Questions

Question 1. A bullet is fired horizontally and gets embedded in a block kept on a table. If the table is frictionless, then (a) only momentum is conserved. (b) only potential energy is conserved. (c) only K.E. is conserved. (d) both (a) and (b).

Answer: (d) both (a) and (b).

Question 2. A shell is fired from a canon with a velocity v ms -1 at an angle 6 with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal mass. One of the pieces retraces its path to the canon and the speed of the other piece (in ms -1 ) just after the explosion is (a) v cos θ (b) 3v cos θ (c) 2v cos θ (d) \(\frac {v}{2}\) cos θ

Answer: (b) 3v cos θ

Question 3. The work performed on an object does not depend upon (a) the displacement. (b) the force applied. (c) the angle at which the force is applied to the displacement (d) initial velocity of the object.

Answer: (d) initial velocity of the object.

Question 4. A ball is dropped from a height of 15 m. It gets embedded in sand by 10 mm and then stops. Which of the following is conserved? (a) Temperature (b) Momentum (c) Kinetic energy (d) Both (a) and (c)

Answer: (b) Momentum

Question 5. A chain of length L and mass M is held on a smooth table with its \(\frac {1}{n}\) th part hanging over the edge. The work done in pulling the chain is directly proportional to (a) n -3 (b) √n (c) n (d) n -2

Answer: (d) n -2

Question 6. A pump delivers water at the rate of V cubic meter per second. By what factor its power should be raised so that it delivers water at the rate of nV cubic metre per second. (a) n³ (b) √n (c) n (d) n²

Answer: (a) n³

Question 7. Liquid of density ρ flows along a horizontal pipe of uniform area of cross-section ‘a’ with a velocity v through a right angled bend. What force should be applied to the bend to hold it in equilibrium? (a) \(\frac {av^2ρ}{2}\) (b) \(\frac {av^2ρ}{√2}\) (c) 2av²ρ (d) √2 av²ρ

Answer: (c) 2av²ρ

Question 8. A body of mass M accelerates uniformly from rest to a velocity v in time t. What is the instantaneous power delivered to the body at time T. (a) \(\frac {mv}{t}\)T (b) \(\frac {mv^2}{t}\)T (c) \(\frac {m^2v^2}{t^2}\)T (d) \(\frac {mv^2}{t^2}\)T

Answer: (d) \(\frac {mv^2}{t^2}\)T

Question 9. A man weighing 50 kgf carries a load of 10 kgf to the top of the building in 5 minutes. The work done by him is 10 5 J. If he carries the same load in 10 minutes, the work done by him will be: (a) 10 5 J (b) 5 × 10 5 J (c) 12 × 10 5 J (d) 2.5 × 10 5 J

Answer: (a) 10 5 J

Question 10. If v be the instantaneous velocity of a body dropped from the top of a tower, when it is located at a height h, then which of the following remains constant? (a) gh + \(\frac {1}{2}\) v² (b) gh – \(\frac {1}{2}\) v² (c) gh + v² (d) gh – v²

Answer: (a) gh + \(\frac {1}{2}\) v²

Question 11. A ball B 1 of mass m is moving with a velocity v along north. It collides with another ball B 2 of same mass moving with a velocity v along east. After the collision, both the balls stick together and move along north east. The velocity of the combination is (a) \(\frac {v}{√2}\) (b) √2v (c) 2v (d) v

Answer: (a) \(\frac {v}{√2}\)

Question 12. A ball B 1 of mass m moving with velocity u, collides head on with another ball B 2 of the same mass at rest. Given that the coefficient of restitution is e. Then the ratio of velocities of two balls after collision will be (a) \(\frac {1+e}{1-e}\) (b) e (c) \(\frac {1+e}{2}\) (d) \(\frac {1-e}{2}\)

Answer: (a) \(\frac {1+e}{1-e}\)

Question 13. A vehicle of mass m is moving on a rough horizontal road with a momentum p. If the coefficient of friction between the tyres and the road be µ, then the stopping distance is given by (a) \(\frac {p}{2µm^2g}\) (b) \(\frac {p}{2µmg}\) (c) \(\frac {p^2}{2µmg}\) (d) \(\frac {p^2}{2µm^2g}\)

Answer: (d) \(\frac {p^2}{2µm^2g}\)

Question 14. A uniform chain of length l and mass m is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g be the acceleration due to gravity, then the work required to pull the hanging part on the table is : (a) \(\frac {mg l}{3}\) (b) \(\frac {mgl}{9}\) (c) \(\frac {mgl}{18}\) (d) mg l

Answer: (c) \(\frac {mgl}{18}\)

Answer: (d) \(\frac {2mg(h+x)}{x^2}\)

Question 16. The slope of the potential energy versus position vector gives (a) momentum (b) force (c) work done (d) power

Answer: (b) force

Question 17. The slope of the kinetic energy versus position vector gives the time rate of change of: (a) momentum (b) force (c) work done (d) power

Answer: (a) momentum

Question 18. A bus weighing 100 quintals moves on a rough road with a constant speed of 72 kmh -1 . The friction of the road is 9% of its weight and that of air is 1 % of its weight. What is the power of the engine? Take g = 10 ms -2 . (a) 100 kW (b) 150 kW (c) 200 kW (d) 50 kW

Answer: (c) 200 kW

Question 19. A metre rod is pivoted at its end and stands vertically. If it is displaced through 60° with the vertical, what will be the ratio of its potential energy in this position to its maximum potential energy? (a) 0.25 (b) 0.75 (c) 0.80 (d) 0.50

Answer: (d) 0.50

Question 20. A bullet is fired into a block of sand and its velocity decreases by 50% when it penetrates through 9 cm. What will be the total distance penetrated by the bullet? (a) 9 cm (b) 10 cm (c) 12 cm (d) 18 cm

Answer: (c) 12 cm

Fill in the blanks

Question 1. When a body is dropped from a certain height on the ground, the work is done on the body ……………. and is ……………….

Answer: by the force of gravity, positive

Question 2. When it is lifted from the ground to a certain height, the work is done on the body ……………. and is ……………

Answer: against the force of gravity, negative

Question 3. The statement that inertial mass of a body is equal to the gravitational mass is called ……………

Answer: principle of equivalence

Question 4. When a vertical force equal to the weight of a body is applied to it, then its acceleration is ……………..

Answer: zero.

Question 5. When work done is zero, then the speed of a body is ……………..

Answer: uniform

Question 6. Coefficient of restitution is one for ……………… collision.

Answer: perfectly elastic collision

Question 7. The coefficient of restitution is zero (e = 0) for ……………… collision which means that ……………….. of the colliding body is lost and is changed to other forms like ………………. or …………………..

Answer: perfectly inelastic, kinetic energy, sound energy, or heat energy

Question 8. The mass of a body increases with the increase in the velocity of the particle according to the relation ………………..

Answer: m = \(\frac {m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

Question 9. Mass of the particle becomes infinite if it moves with the ……………….. and the acceleration produced in the body by a given force F becomes ………………

Answer: velocity of light c = 3 × 10 8 ms -1 , zero (∵ a = \(\frac {F}{m}\) = \(\frac {F}{∞}\) = 0)

Question 10. E = mc² shows that mass and energy are …………….

Answer: inter convertible

Energy is transferred from one form to another. Name the transformations of energy in the following in the form ……………….. to ……………….. Question 11. lifting of a book by a boy ………………..

Answer: Lifting of a book by a boy transforms energy from chemical to gravitational.

Question 12. burning of candle ………………..

Answer: Burning of candle transforms energy from chemical to light and heat.

Question 13. a moving truck on a road ………………..

Answer: A moving truck on a road transforms energy from chemical to kinetic energy of motion.

Question 14. emission of light by Sun ………………..

Answer: Emission of light by Sun transforms energy from nuclear to electromagnetic energy.

Question 15. The work done against friction is ……………….. proportional to the force with which the surfaces are pressed together.

Answer: directly

Question 16. If the momentum of a body is doubled then its K.E. becomes ………………..

Answer: four times

Question 17. The sum of K.E. and P.E. for an isolated system is always ………………..

Answer: constant

Question 18. If the range of a projectile be R, then its K.E. is minimum when horizontal distance covered by it is ………………..

Answer: \(\frac {R}{2}\)

Question 19. ………………. is not conserved during an inelastic collision.

Answer: K.E.

Question 20. When a bullet striking a block gets embedded in it, then the nature of collision is ……………….

Answer: perfectly inelastic

Question 21. ………………. is a non-conservative force.

Answer: friction

Question 22. For a perfectly inelastic collision, the coefficient of restitution is ……………….

Answer: zero

Question 23. The work done …………….. updn the time taken in doing it.

Answer: does not depend

Question 24. The work done by the conservative force 1s always ……………..

Answer: positive

Question 25. The P.E. gained by the body raised from the surface of Earth by a height equal to radius of Earth is ……………..

Answer: mg \(\frac {R}{2}\)

Question 26. K.E. of a projectile is …………….. at the highest point of its trajectory.

Answer: minimum

Question 27. The work done on a body by a resultant external force is always ………………. to the change in its K.E.

Answer: equal

Question 28. A body is suspended by vertical string. The work done on the body by the tension in the string is …………….

Question 29. Power is ………………. of force and velocity.

Answer: scalar product

Question 30. Work is ………………. of force and displacement.

Question 31. 1 kwh is the energy consumed by an appliance of power I kw in ………….

Answer: 1 hour

Question 32. The work done by a conservative force moving an object on a closed path is …………….

Question 33. In a perfectly inelastic collision, the two bodies ………………… after collision.

Answer: stick together

Question 34. For an inelastic collision, the coefficient of restitution lies between ………………

Question 35. Kinetic energy increases during a ……………….

Answer: super elastic collision

Question 36. ………………. energy results from the chemical bounding between the atoms.

Answer: Chemical

Question 37. ………………. energy results due to the separation between two objects in the gravitational field.

Answer: Gravitational

Question 38. ………………… energy between two nucleons is due to nuclear force.

Answer: Nuclear

Question 39. Chemical, gravitational and nuclear energies are the ……………….. for different types of forces in nature.

Answer: Potential energies

True/False Type Questions

1. Which of the following statement is True/False? (a) Work done can be positive, negative and zero. (b) A teacher sitting in a chair is dong work. (c) Work is the cross product of force and displacement. (d) Work done by a coolie in carrying a bag on his head is zero.

Answer: (a) True (b) False (c) False (d) True

2. Which of the following statement is True/False? (a) The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is extremely small. (b) In an inelastic collision, the linear momentum is conserved but not the kinetic energy. (c) The collision is said to be elastic if two bodies stick together after collision. (d) The unit of energy is increased by 16 times if the unit of force and length be each increased by 4 times.

Answer: (a) True (b) True (c) False (d) True

3. Tell which of the following statement is True/False? (a) Momentum is conserved in elastic collision but not in inelastic collision. (b) Total kinetic energy is conserved in elastic collision but momentum is not conserved in elastic collision. (c) Both K.E. and momentum are conserved in all types of collision. (d) Total energy, K.E. and the momentum is conserved in elastic collision.

Answer: (a) False (b) False (c) False (d) True

4. Which of the following statement is True/False? (a) Both momentum and energy are conserved in an inelastic collision. (b) Neither momentum nor energy is conserved in an inelastic collision. (c) Momentum is conserved but not the K.E. in an inelastic collision. (d) Momentum is not conserved but the K.E. is conserved in an inelastic collision.

Answer: (a) False (b) False (c) True (d) False

5. Which of the following statement is True/False? There will be an increase in potential energy of the system if work is done upon the system by (a) a conservative force. (b) a non-conservative force. (c) any conservative or non-conservative force.

Answer: (a) True (b) False (c) False

6. Which of the following statement is True/False? (a) The gain in the P.E. of an object of mass m raised from the surface of earth to a height equal to the radius R is \(\frac {mgR}{2}\). (b) 1 eV = 1.6 × 10 -19 J (c) Power is the cross-product of force and velocity. (d) Work is the dot product of force and displacement.

7. Which of the following statement is True/False? (a) A teacher rubbing the blockboard by a duster is doing no work. (b) A light and a heavy body have equal K.E. of translation. The heavier body has larger momentum. (c) A bullet is fired from a rifle. If the rifle recoils freely, then the K.E. of the rifle is greater than that of the bullet. (d) The energy produced when 10 gram of coal is burnt is 9 × 10 4 J.

8. Tell which one of the following statement is True/False? (a) A truck and car having same K.E. cover equal distances before stopping after applying equal retarding force through brakes. (b) 1 kg m = 9.8 J (c) 1 J = 10 7 erg (d) For an elastic collision, the velocity of separation after collision is always equal to the velocity of approach before collision.

Answer: (a) True (b) True (c) True (d) True

9. Tell which of the following statement is True/False? (a) Friction is non-conservative force. (b) Gravity is a conservative force. (c) No work is done against gravity while moving an object along horizontal. (d) Kilowatt-hour is the unit of energy.

10. Tell which of the following statements are True/False? (a) The forces involved during the elastic collision are conservative in nature. (b) K.E. lost in an inelastic collision appears in some other form of energy such as heat, sound etc. (c) Mechanical energy is not converted into any other form of energy in an elastic collision. (d) Coefficient of restitution is zero (e = 0) for a perfectly inelastic collosion.

Match Type Questions

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MCQ Questions for Class 11 Physics Chapter 6 Work Energy and Power

• Last modified on: 2 years ago
• Reading Time: 12 Minutes

Q.1. The magnitude of work done by a force (a) depends on frame of reference (b) does not depend on frame of reference (c) cannot be calculated in non-inertial frames. (d) both (a) and (b)

Q.2. No work is done if (a) displacement is zero (b) force is zero (c) force and displacement are mutually perpendicular (d) All of these

Q.3. When the force retards the motion of body, the work done is (a) zero (b) negative (c) positive (d) Positive or negative depending upon the magnitude of force and displacement

Q.4. Work is always done on a body when (a) a force acts on it (b) it moves through a certain distance (c) it experiences an increase in energy through a mechanical influence (d) None of these

Q.5. A boy carrying a box on his head is walking on a level road from one place to another is doing no work. This statement is (a) correct (b) incorrect (c) partly correct (d) cannot say

Q.6. In which of the following work is being not done? (a) Shopping in the supermarket (b) Standing with a basket of fruit on the head (c) Climbing a tree (d) Pushing a wheel barrow

Q.7. According to work-energy theorem, the work done by the net force on a particle is equal to the change in its (a) kinetic energy (b) potential energy (c) linear momentum (d) angular momentum

Q.8. If a light body and heavy body have same kinetic energy, then which one has greater linear momentum? (a) Lighter body (b) Heavier body (c) Both have same momentum (d) Can’t be predicted

Q.9. The coefficient of restitution e for a perfectly inelastic collision is (a) 1 (b) 0 (c) infinity (d) –1

Q.10. A light and a heavy body have equal momentum. Which one has greater K.E.? (a) The lighter body (b) The heavier body (c) Both have equal K.E. (d) Data given is incomplete

Q.11. The temperature at the bottom of a high water fall is higher than that at the top because (a) by itself heat flows from higher to lower temperature (b) the difference in height causes a difference in pressure (c) thermal energy is transformed into mechanical energy (d) mechanical energy is transformed into thermal energy.

Q.12. A bullet is fired and gets embedded in block kept on table. If table is frictionless, then (a) kinetic energy gets conserved (b) potential energy gets conserved (c) momentum gets conserved (d) both (a) and (c)

Q.13. A man pushes a wall and fails to displace it, he does (a) negative work (b) positive but not maximum work (c) no work at all (d) maximum positive work

Q.14. Unit of energy is (a) kwh (b) joule (c) electron volt (d) All of these

Q.15. Which of the following is not a conservative force? (a) Gravitational force (b) Frictional force (c) Spring force (d) None of these

Q.16. The speed of an object of mass m dropped from an inclined plane (frictionless), at the bottom of the plane, depends on: (a) height of the plane above the ground (b) angle of inclination of the plane (c) mass of the object (d) All of these

Q.17. A particle is taken round a circle by application of force. The work done by the force is (a) positive non–zero (b) negative non–zero (c) Zero (d) None of the above

Q.18. The coefficient of restitution e for a perfectly elastic collision is (a) 1 (b) 0 (c) infinity (d) –1

Q.19. Four particles given, have same momentum. Which has maximum kinetic energy (a) Proton (b) Electron (c) Deutron (d) Alpha-particles

Q.20. If two particles are brought near one another, the potential energy of the system will (a) increase (b) decrease (c) remains the same (d) equal to the K.E

Q.21. When two spheres of equal masses undergo glancing elastic collision with one of them at rest, after collision they will move (a) opposite to one another (b) in the same direction (c) together (d) at right angle to each other

Q.22. In an inelastic collision (a) momentum is not conserved (b) momentum is conserved but kinetic energy is not conserved (c) both momentum and kinetic energy are conserved (d) neither momentum nor kinetic energy is conserved

Q.23. The potential energy of a system increases if work is done (a) upon the system by a non conservative force (b) by the system against a conservative force (c) by the system against a non conservative force (d) upon the system by a conservative force

Q.24. In an elastic collision, what is conserved ? (a) Kinetic energy (b) Momentum (c) Both (a) and (b) (d) Neither (a) nor (b)

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