case study on work energy and power class 11

Case Study Questions Class 11 Physics Work, Energy And Power

Case study questions class 11 physics chapter 6 work, energy and power, cbse case study questions class 11 physics work, energy and power, case study – 2, case study – 3, case study – 4.

The impact and deformation during collision may generate heat and sound. Part of the initial kinetic energy is transformed into other forms of energy. A useful way to visualize the deformation during collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains its original shape without loss in energy, then the initial kinetic energy is equal to the final kinetic energy but the kinetic energy during the collision time Δt is not constant. Such a collision is called an elastic collision. On the other hand the deformation may not be relieved and the two bodies could move together after the collision. A collision in which the two particles move together after the collision is called a completely inelastic collision. The intermediate case where the deformation is partly relieved and some of the initial kinetic energy is lost is more common and is appropriately called an inelastic collision. If the initial velocities and final velocities of both the bodies are along the same straight line, then it is called a one-dimensional collision, or head-on collision.

Answer key – 4

Case Study – 5

4) The instantaneous power is defined as the limiting value of the average power as time interval approaches zero.

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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 11 Physics Case Study Questions

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Looking for complete and comprehensive case study questions for class 11 Physics? myCBSEguide is just a click away! With extensive study materials, sample papers, case study questions and mock tests, myCBSEguide is your one-stop solution for class 11 Physics exam preparation needs. So, what are you waiting for? Log on to myCBSEguide and get started today!

What is the purpose of physics?

Physics is the study of the fundamental principles governing the natural world. It is a vital part of the scientific enterprise, providing the foundation on which other sciences are built. Physics is essential for understanding how the world works, from the smallest particles to the largest structures in the Universe. In class 11 Physics, students are introduced to the basic concepts of physics and learn about the fundamental principles governing the natural world. Class 11 Physics concepts are essential for understanding the world around us and for further study in physics and other sciences.

What are case study questions in physics?

In physics, case study questions are intended to evaluate a student’s ability to apply theoretical principles to real-life situations. These questions usually ask the student to assess data from a specific experiment or setting in order to discover what physical principles are at play. Problem-solving and critical-thinking skills are developed through case study questions, which are an important aspect of physics education.

CBSE Case Study Questions in Class 11 Physics

CBSE Class 11 Physics question paper pattern includes case study questions. Class 11 Physics case study questions assess a student’s ability to apply physics principles to real-world environments. The questions are usually focused on a situation provided in the Class 11 Physics question paper, and they demand the student to answer the problem using their physics knowledge. Class 11 Physics case study questions are an important aspect of the CBSE physics curriculum. Class 11 Physics case study questions are a useful way to assess a student’s expertise in the subject.

Sample Class 11 Physics Case Study Questions

Expert educators at myCBSEguide have created a collection of Class 11 physics case study questions. The samples of Class 11 physics case study questions are given below. Class 11 physics case study questions are designed to test your understanding of the concepts and principles of physics. They are not meant to be easy, but they should be done if you have a good grasp of the subject. So, take a look at the questions and see how you fare. Good luck!

Class 11 Physics Case Study Question 1

Read the case study given below and answer any four subparts: Potential energy is the energy stored within an object, due to the object’s position, arrangement or state. Potential energy is one of the two main forms of energy, along with kinetic energy. Potential energy depends on the force acting on the two objects.

• kinetic energy
• potential energy
• mechanical energy
• none of these
• potential energy decreases
• potential energy increases
• kinetic energy decreases
• kinetic energy increases
• only when spring is stretched
• only when spring is compressed
• both a and b
• 5  ×  10 4  J
• 5  ×  10 5  J

Answer Key:

Class 11 Physics Case Study Question 2

• distance between body
• source of heat
• all of the above
• convection and radiation
• (b) convection
• (d) all of the above
• (a) convection
• (a) increase
• (c) radiation

  Class 11 Physics Case Study Question 3

• internal energy.
• 1 +(T 2 /T 1 )
• (T 1 /T 2 )+1
• (T 1  /T 2 )- 1
• 1 – (T 2  / T 1 )
• increase or decrease depending upon temperature ratio
• first increase and then decrease
• (d) 1- (T 2 / T 1 )
• (b) increase
• (c) constant

Class 11 Physics Case Study Question 4 

• It is far away from the surface of the earth
• Its surface temperature is 10°C
• The r.m.s. velocity of all the gas molecules is more than the escape velocity of the moon’s surface
• The escape velocity of the moon’s surface is more than the r.m.s velocity of all molecules
• T(H 2 ) = T(N 2 )
• T(H 2 ) < T(N 2 )
• T(H 2 ) > T(N 2 )

The given samples of Class 11 Physics case study questions will help Class 11 Physics students to get an idea on how to solve it. These Class 11 Physics case study questions are based on the topics covered in the Class 11 Physics syllabus and are designed to test the student’s conceptual understanding. The questions are of varying difficulty levels and cover a wide range of topics. By solving these Class 11 Physics case study questions, students will be able to develop their problem-solving skills and improve their understanding of the concepts.

Examining Class 11 Physics syllabus

Senior Secondary school education is a transitional step from general education to a discipline-based curriculum concentration. The current curriculum of Class 11 Physics takes into account the rigour and complexity of the disciplinary approach, as well as the learners’ comprehension level. Class 11 Physics syllabus has also been carefully crafted to be similar to international norms.

The following are some of the Class 11 Physics syllabus’s most notable features:

• Emphasis is placed on gaining a fundamental conceptual knowledge of the material.
• Use of SI units, symbols, naming of physical quantities, and formulations in accordance with international standards are emphasised.
• For enhanced learning, provide logical sequencing of subject matter units and suitable placement of concepts with their links.
• Eliminating overlapping concepts/content within the field and between disciplines to reduce the curricular load.
• Process skills, problem-solving ability, and the application of Physics principles are all encouraged.

CBSE Class 11 Physics (Code No. 042)

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Class 11 Physics Case Study Questions PDF Download

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Class 11 Physics Case Study Questions are available here. You can read these Case Study questions by chapter for your final physics exam. Subject matter specialists and seasoned teachers created these quizzes. You can verify the right response to each question by referring to the answer key, which is also provided. To achieve high marks on your Board exams, practice these questions.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

We are providing Case Study questions for Class 11 Physics based on the Latest syllabus. There is a total of 14 chapters included in the CBSE Class 11 physics exams. Students can practice these questions for concept clarity and score better marks in their exams.

Table of Contents

Class 11th PHYSICS: Chapterwise Case Study Question & Solution

Case study questions play a crucial role in the Class 11 Physics curriculum. They are designed to assess your understanding of various concepts and principles in real-life scenarios. These questions help you apply theoretical knowledge to practical situations, enhancing your problem-solving skills.

Case Study-Based Questions for Class 11 Physics

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

Class 11 Physics MCQ Questions

Before the exams, students in class 11 should review crucial Physics Case Study issues. They will gain a better understanding of the kinds of Case Study questions that may be offered in Physics exams for Grade 11. These questions were created by our highly qualified faculty for standard 11 Physics based on the questions that appeared most frequently in last year’s exams. The solutions have been written in a way that will make them simple to grasp and will aid students in grade 11 in understanding the topics.

Class 11 Books for Boards

Class 11 Physics Syllabus 2024

Unit I: Physical World and Measurement 08 Periods

Chapter–2: Units and Measurements

Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. significant figures. Dimensions of physical quantities, dimensional analysis and its applications.

Unit II: Kinematics 24 Periods

Chapter–3: Motion in a Straight Line

The frame of reference, Motion in a straight line, Elementary concepts of differentiation and integration for describing motion, uniform and non-uniform motion, and instantaneous velocity, uniformly accelerated motion, velocity-time and position-time graphs. Relations for uniformly accelerated motion (graphical treatment).

Chapter–4: Motion in a Plane

Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors. Motion in a plane, cases of uniform velocity and uniform acceleration projectile motion, uniform circular motion.

Unit III: Laws of Motion 14 Periods

Chapter–5: Laws of Motion

Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication.

Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).

Unit IV: Work, Energy and Power 14 Periods

Chapter–6: Work, Energy and Power

Work done by a constant force and a variable force; kinetic energy, workenergy theorem, power. Notion of potential energy, potential energy of a spring, conservative forces: non- conservative forces, motion in a vertical circle; elastic and inelastic collisions in one and two dimensions.

Unit V: Motion of System of Particles and Rigid Body 18   Periods

Chapter–7: System of Particles and Rotational Motion

Centre of mass of a two-particle system, momentum conservation and Centre of mass motion. Centre of mass of a rigid body; centre of mass of a uniform rod. Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications. Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions. Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation).

Unit VI: Gravitation 12 Periods

Chapter–8: Gravitation

Kepler’s laws of planetary motion, universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite.

Unit VII: Properties of Bulk Matter 24 Periods

Chapter–9: Mechanical Properties of Solids

Elasticity, Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus, shear modulus of rigidity (qualitative idea only), Poisson’s ratio; elastic energy.

Chapter–10: Mechanical Properties of Fluids

Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure. Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem and its simple applications. Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise.

Chapter–11: Thermal Properties of Matter

Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv – calorimetry; change of state – latent heat capacity. Heat transfer-conduction, convection and radiation, thermal conductivity, qualitative ideas of Blackbody radiation, Wein’s displacement Law, Stefan’s law .

Unit VIII: Thermodynamics 12 Periods

Chapter–12: Thermodynamics

Thermal equilibrium and definition of temperature zeroth law of thermodynamics, heat, work and internal energy. First law of thermodynamics, Second law of thermodynamics: gaseous state of matter, change of condition of gaseous state -isothermal, adiabatic, reversible, irreversible, and cyclic processes.

Unit IX:   Behavior of Perfect Gases and Kinetic Theory of Gases 08   Periods

Chapter–13: Kinetic Theory

Equation of state of a perfect gas, work done in compressing a gas. Kinetic theory of gases – assumptions, concept of pressure. Kinetic interpretation of temperature; rms speed of gas molecules; degrees of freedom, law of equi-partition of energy (statement only) and application to specific heat capacities of gases; concept of mean free path, Avogadro’s number.

Unit X: Oscillations and Waves 26 Periods

Chapter–14: Oscillations

Periodic motion – time period, frequency, displacement as a function of time, periodic functions and their application. Simple harmonic motion (S.H.M) and its equations of motion; phase; oscillations of a loaded spring- restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation of expression for its time period.

Chapter–15: Waves

Wave motion: Transverse and longitudinal waves, speed of traveling wave, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, Beats.

FAQs about Class 11 Physics Case Studies

What is the best website for a  case   study  of physics  class   11 .

studyrate.in is the best website for Class 11 Physics Case Study Questions for Board Exams. Here you can find various types of Study Materials, Ebooks, Notes, and much more free of cost.

How do you write a case study question for Class 11?

The CBSE will ask two Case Study Questions in the CBSE Class 11th Maths Question Paper. Question numbers 15 and 16 will be case-based questions where 5 MCQs will be asked based on a paragraph.

Are the case study questions based on the latest syllabus?

Yes, the case study questions are curated to align with the latest Class 11 Physics syllabus.

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Work Energy and Power CBSE Class 11 Physics Revision Notes Chapter 6

• Work:-  Work done W is defined as the dot product of force F and displacement s .

Work done by the force is positive if the angle between force and displacement is acute (0 ° <θ<90 ° ) as cos θ is positive. This signifies, when the force and displacement are in same direction, work done is positive. This work is said to be done upon the body.

Work done by the force is negative if the angle between force and displacement is obtuse (90 ° <θ<180 ° )  as cos θ is negative. This signifies, when the force and displacement are in opposite direction, work done is negative. This work is said to be done by the body.

Work done by a variable force:-

If applied force F is not a constant force, then work done by this force in moving the body from position A to B will be,

Here ds is the small displacement.

Units:   The unit of work done in S.I is joule (J) and in C.G.S system is erg.

1J = 1 N.m , 1 erg = 1 dyn.cm

Relation between Joule and erg:- 1 J = 10 7 erg

Power: - The rate at which work is done is called power and is defined as,

P = W/t = F.s/v = F.v

Here s is the distance and v is the speed.

Instantaneous power in terms of mechanical energy:-  P  = dE / dt

Units: The unit of power in S.I system is J / s (watt) and in C.G.S system is erg / s .

1) Energy is the ability of the body to do some work. The unit of energy is same as that of work.

2) Kinetic Energy (K):- It is defined as,

K = ½ mv 2

Here m is the mass of the body and v is the speed of the body.

Potential Energy ( U ):- Potential energy of a body is defined as,  U = mgh

Here, m is the mass of the body, g is the free fall acceleration (acceleration due to gravity) and h is the height.

Gravitational Potential Energy:- An object’s gravitational potential energy U is its mass m times the acceleration due to gravity g times its height h above a zero level.

In symbol’s,

Relation between Kinetic Energy ( K ) and momentum  ( p ):-

K = p 2 /2 m

If two bodies of different masses have same momentum, body with a greater mass shall have lesser kinetic energy.

If two bodies of different mass have same kinetic energy, body with a greater mass shall have greater momentum.

For two bodies having same mass, the body having greater momentum shall have greater kinetic energy.

Work energy Theorem:- It states that work done on the body or by the body is equal to the net change in its kinetic energy .

For constant force,

W = ½ mv 2 – ½ mu 2

= Final K.E – Initial K.E

For variable force,

Law of conservation of energy:- It states that, “Energy can neither be created nor destroyed. It can be converted from one form to another. The sum of total energy, in this universe, is always same”.

The sum of the kinetic and potential energies of an object is called mechanical energy. So, E = K + U

In accordance to law of conservation of energy, the total mechanical energy of the system always remains constant.

So, mgh + ½ mv 2 = constant

In an isolated system, the total energy E total of the system is constant.

So, E = U + K = constant

Or, U i + K i = U f + K f

Or, ? U = -? K

Speed of particle v in a central force field:

v = √2/ m [ E - U (x)]

Conservation of linear momentum:-

So, p f = p i

Coefficient of restitution (e) :- It is defined as the ratio between magnitude of impulse during period of restitution to that during period of deformation.

e = relative velocity after collision / relative velocity before collision

= v 2 – v 1 / u 1 – u 2

Case (i) For perfectly elastic collision, e = 1. Thus, v 2 – v 1 = u 1 – u 2 . This signifies the relative velocities of two bodies before and after collision are same.

Case (ii) For inelastic collision, e<1. Thus, v 2 – v 1 < u 1 – u 2 . This signifies, the value of e shall depend upon the extent of loss of kinetic energy during collision.

Case (iii) For perfectly inelastic collision, e = 0. Thus, v 2 – v 1 =0, or v 2 = v 1 . This signifies the two bodies shall move together with same velocity. Therefore, there shall be no separation between them.

Elastic collision:- In an elastic collision, both the momentum and kinetic energy conserved.

One dimensional elastic collision:-

After collision, the velocity of two body will be ,

v 1 = ( m 1 - m 2 / m 1 + m 2 ) u 1 + (2 m 2 / m 1 + m 2 ) u 2

v 2 = ( m 2 - m 1 / m 1 + m 2 ) u 2 + (2 m 1 / m 1 + m 2 ) u 1

When both the colliding bodies are of the same mass, i.e., m 1 = m 2 , then,

v 1 = u 2 and v 2 = u 1

When the body B of mass m 2 is initially at rest, i.e., u 2 = 0, then,

v 1 = ( m 1 - m 2 / m 1 + m 2 ) u 1 and v 2 = (2 m 1 / m 1 + m 2 ) u 1

(a) When  m 2 <<m 1 , then, v 1 = u 1 and v 2 = 2u 1

(b) When  m 2 =m 1 , then, v 1 =0  and v 2 = u 1

(c) When  m 2 >>m 1 , then, v 1 = -u 1 and v 2 will be very small.

Inelastic collision:- In an inelastic collision, only the quantity momentum is conserved but not kinetic energy.

v = ( m 1 u 1 + m 2 u 2 ) /( m 1 + m 2 )

loss in kinetic energy, E = ½ m 1 u 1 2 + ½ m 2 u 2 2 - ½ ( m 1 + m 2 ) v 2

E = ½ ( m 1 u 1 2 + m 2 u 2 2 ) – ½ [( m 1 u 1 + m 2 u 2 )/( m 1 + m 2 )] 2

= m 1 m 2 ( u 1 - u 2 ) 2 / 2( m 1 + m 2 )

Points to be Notice:-

(i) The maximum transfer energy occurs if m 1 = m 2

(ii) If K i is the initial kinetic energy and K f is the final kinetic energy of mass m 1 , the fractional decrease in kinetic energy is given by,

K i – K f / K i = 1- v 1 2 / u 2 1

Further, if m 2 = nm 1 and u 2 = 0, then,

K i – K f / K i = 4 n /(1+ n ) 2

Conservation Equation:

(i)  Momentum – m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2

(ii) Energy – ½ m 1 u 1 2 + ½ m 2 u 2 2 = ½ m 1 v 1 2 + ½ m 2 v 2 2

Conservative force ( F ): - Conservative force is equal to the negative gradient of potential V of the field of that force. This force is also called central force.

So,  F = - ( dV / dr )

The line integral of a conservative force around a closed path is always zero.

Spring potential energy ( E s ):- It is defined as,

E s = ½ kx 2

Here k is the spring constant and x is the elongation.

Equilibrium Conditions:

(a) Condition for equilibrium, dU / dx = 0

(b) For stable equilibrium,

U ( x ) = minimum,

dU / dx = 0,

d 2 U / dx 2 = +ve

(c) For unstable equilibrium,

U ( x ) = maximum

dU / dx = 0

d 2 U / dx 2 = -ve

(d) For neutral equilibrium,

U ( x ) = constant

dU / dx = 0 

d 2 U / dx 2 = 0

UNITS AND DIMENSIONS OF WORK, POWER AND ENERGY

Work and Energy are measured in the same units. Power, being the rate at which work is done, is measured in a different unit.

Conversions between Different Systems of Units

1 Joule = 1 Newton ´ 1 m = 10 5 dyne ´ 10 2 cm = 10 7 erg

1 watt = 1 Joule/ sec = 10 7 erg/sec.

1 kwh  = 10 3 watt ´ 1 hr  = 10 3 watt ´ 3600 sec 

= 3.6 ´ 10 6 Joule

1HP = 746 watt.

1 MW = 10 6 watt.

1 cal = 1 calorie = 4.2 Joule

1eV = "e" Joule  = 1.6 ´ 10 -19 Joule

(e = magnitude of charge on the electron in coulombs)

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

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power are part of Class 11 Physics NCERT Solutions . Here we have given NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power.

Topics and Subtopics in  NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power :

QUESTIONS FROM TEXTBOOK

Question 6. 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: (a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket, (b) Work done by gravitational force in the above case, (c) Work done by friction on a body sliding down an inclined plane, (d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest. Answer:  Work done, W = T.S = Fs cos θ (a) Work done ‘positive’, because force is acting in the direction of displacement i.e., θ = 0°. (b) Work done is negative, because force is acting against the displacement i.e., θ = 180°. (c) Work done is negative, because force of friction is acting against the displacement i.e., θ= 180°. (d) Work done is positive, because body moves in the direction of applied force i.e., θ= 0°. (e) Work done is negative, because the resistive force of air opposes the motion i.e., θ = 180°.

Question 6. 2. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) Work done by the applied force in 10 s (b) Work done by friction in 10 s (c) Work done by the net force on the body in 10 s (d) Change in kinetic energy of the body in 10 s and interpret your results. Answer:  (a) We know that U k = frictional force/normal reaction frictional force = U k x normal reaction = 0.1 x 2 kg wt = 0.1 x 2 x 9.8 N = 1.96 N net effective force = (7 – 1.96) N = 5.04 N acceleration = 5.04/2 ms -2 = 2.52 ms -2 distance, s=1/2x 2.52 x 10 x 10 = 126 m work done by applied force = 7 x 126 J = 882 J (b) Work done by friction = 1.96 x 126 = -246.96 J (c) Work done by net force = 5.04 x 126 = 635.04 J (d) Change in the kinetic energy of the body = work done by the net force in 10 seconds = 635.04 J (This is in accordance with work-energy theorem).

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Question 6. 6. Point out the correct alternative: (a) When a conservative force does positive work on a body, the potential energy of the body  increases/decreases/remains unaltered. (b)Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces of the system. (d) 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. Answer: (a) Potential energy of the body decreases because the body in this case goes closer to the centre of the force. (b) Kinetic energy, because friction does its work against the motion. (c) Internal forces can not change the total or net momentum of a system. Hence the rate of change of total momentum of many particle system is proportional to the external force on the system. (d) 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.

Question 6. 7. State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on  the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. Answer:  (a) False, the total momentum and total energy of the system are conserved. (b) False, the external force on the system may increase or decrease the total energy of the system. (c) False, the work done during the motion of a body over a closed loop is zero only when body is moving under the action of a conservative force (such as gravitational or electrostatic force). Friction is not a conservative force hence work done by force of friction (or work done on the body against friction) is not zero over a closed loop. (d) True, usually in an inelastic collision the final kinetic energy is always less than the initial kinetic energy of the system.

Question 6. 8. Answer carefully, with reasons: (a) 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)? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls? (c) What are the answers to (a) and (b) for an inelastic collision? (d) 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). Answer:   (a) In this case total kinetic energy is not conserved because when the bodies are in contact dining elastic collision even, the kinetic energy is converted into potential energy. (b) Yes, because total momentum conserves as per law of conservation of momentum. (c) The answers remain unchanged. (d) It is a case of elastic collision because in this case the forces will be of conservative nature.

Question 6. 9.  A body is initially at rest. It undergoes a 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 Answer:  (ii) From v = u + at v = 0 + at = at As power, p = F x  v .’. p = (ma) x at = ma 2 t Since m and a are constants, therefore, p α t.

Question 6. 13. A raindrop of radius 2 mm falls from a height of 500 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 10 ms -1 ? Answer:  Here, r = 2 mm = 2 x 10 -3  m. Distance moved in each half of the journey, S=500/2= 250 m. Density of water, p = 10 3 kg/ m 3 Mass of rain drop = volume of drop x density m =4/3 π r 2 x ρ =4/3 x 22/7 (2 x 10 -3 ) 3 x 10 3 = 3.35 x 10 -5 kg .-. W = mg x s = 3.35 x 10 -5 x 9.8 x 250 = 0.082 J Note: Whether the drop moves with decreasing acceleration or with uniform speed, work done by the gravitational force on the drop remains the same. If there was no resistive forces, energy of drop on reaching the ground. E 1 = mgh = 3.35 x 10 -5 x 9.8 x 500 = 0.164 J Actual energy, E 2 = 1/2mv 2 = 1/2 x 3.35 x 10 -5 (10) 2 = 1.675 x 10 -3 J Work done by the resistive forces, W =E 1 – E 2 = 0.164 – 1.675 x 10 -3 W = 0.1623 joule.

Question 6. 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 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump? Answer:   Here, volume of water = 30 m 3  ; t = 15 min = 15 x 60 = 900s h = 40 m ; n= 30% As the density of water = p = 10 3 kg m -3 Mass of water pumped, m = volume x density = 30 x 10 3 kg Actual power consumed or output power p 0 = W/t = mgh/t =>p 0 =(30 x 10 3 x 9.8 x 40)/900=13070 watt If pi is input power (required), then as η=p 0 /p i => p i =p 0 /η = 13070/(30/100)=43567 W =43.56 KW

Question 6. 19. A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a friction less track. After a while, sand starts leaking out of a hole on the trolley’s floor at the rate of 0.05 kg s -1 . What is the speed of the trolley after the entire sand bag is empty? Answer:   The system of trolley and sandbag is moving with a uniform speed. Clearly, the system is not being acted upon by an external force. If the sand leaks out, even then no external force acts. So there shall be no change in the speed of the trolley.

Question 6. 20. A particle of mass 0.5 kg travels in a straight line with velocity u = 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 m? Answer:  Here m = 0.5 kg u=a x 3/2 , a = 5 m -1/2 s -1 . Initial velocity at x = 0, v 1 = a x 0 = 0 Final velocity at x = 2, v 2 = a (2) 3/2  = 5 x (2) 3/2 Work done = increase in K.E = 1/2 m(v 2 2 -v 1 2 ) = 1/2 x 0.5[(5 x (2) 3/2 ) 2 – 0] = 50 J.

Question 6. 22. A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated, (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 x 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? Answer:  Here, m = 10 kg, h = 0.5 m, n = 1000 (a) work done against gravitational force. W = n(mgh) = 1000 x (10 x 9.8 x 0.5) = 49000J. (b) Mechanical energy supplied by 1 kg of fat = 3.8 x 10 7 x20/100 = 0.76 x10 7 J/kg .-. Fat used up by the dieter =1kg/(0.76 x 10 7 ) x 49000 = 6.45 x 10 -3 kg

Question 6. 23. A family uses 8 kW of power, (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house. Answer:   (a) Power used by family, p = 8 KW = 8000 W As only 20% of solar energy can be converted to useful electrical energy, hence, power 8000 W to be supplied by solar energy = 8000 W/20 = 40000 W As solar energy is incident at a rate of 200 Wm -2 , hence the area needed A=4000 W/200 Wm -2 =200 m 2 (b) The area needed is camparable to roof area of a large sized house.

Question 6. 27. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms -1 . It hits the floor of the elevator (length of elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary? Answer:   P.E. of bolt = mgh = 0.3 x 9.8 x 3 = 8.82 J The bolt does not rebound. So the whole of the energy is converted into heat. Since the value of acceleration due to gravity is the same in all inertial system, therefore the answer will not change even if the elevator is stationary.

Question 6. 28. A trolley of mass 200 kg moves with a uniform speed of 36 km h -1 on a friction less track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 ms -1 relative to the trolley in a direction opposite to the trolley’s 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? Answer:   Let there be an observer travelling parallel to the trolley with the same speed. He will observe the initial momentum of the trolley of mass M and child of mass m as zero. When the child jumps in opposite direction, he will observe the increase in the velocity of the trolley by Δv. Let u be the velocity of the child. He will observe child landing at velocity (u – Δu) Therefore, initial momentum = 0 Final momentum = MΔ v – m (u – Δv) Hence, MΔ v – m (u – Δv) = 0 Whence Δv =mu/ M + m Putting values Δv =4 x 20/ 20 + 220 = ms -1 .-. Final speed of trolley is 10.36 ms -1 . The child take 2.5 s to run on the trolley. Therefore, the trolley moves a distance = 2.5 x 10.36 m = 25.9 m.

NCERT Solutions for Class 11 Physics All Chapters

• 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

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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.

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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?

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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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CBSE Class 11 Physics- Chapter 6 Work Energy and power- Study Materials

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Work, Energy and Power : Notes and Study Materials -pdf

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Class 11 Physics Chapter 6 Work Energy and Power

Topics and Subtopics in  Class 11 Physics Chapter 6 Work Energy and Power :

Work, Energy and Power Class 11 Notes Physics Chapter 6

• Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of applied force. It is measured by the product of the force and the distance moved in the direction of the force, i.e., W = F-S • If an object undergoes a displacement ‘S’ along a straight line while acted on a force F that makes an angle 0 with S as shown. The work done W by the agent is the product of the component of force in the direction of displacement and the magnitude of displacement.

• If we plot a graph between force applied and the displacement, then work done can be obtained by finding the area under the F-s graph. • If a spring is stretched or compressed by a small distance from its unstretched configuration, the spring will exert a force on the block given by F = -kx, where x is compression or elongation in spring, k is a constant called spring constant whose value depends inversely on unstretched length and the nature of material of spring. The negative sign indicates that the direction of the spring force is opposite to x, the displacement of the free-end.

• Energy The energy of a body is its capacity to do work. Anything which is able to do work is said to possess energy. Energy is measured in the same unit as that of work, namely, Joule. Mechanical energy is of two types: Kinetic energy and Potential energy. • Kinetic Energy The energy possessed by a body by virtue of its motion is known as its kinetic energy. For an object of mass m and having a velocity v, the kinetic energy is given by: K.E. or K = 1/2 mv   2 • Potential Energy The energy possessed by a body by virtue of its position or condition is known as its potential energy. There are two common forms of potential energy: gravitational and elastic. —> Gravitational potential energy of a body is the energy possessed by the body by virtue of its position above the surface of the earth. It is given by (U)P.E. = mgh where m —> mass of a body g —> acceleration due to gravity on the surface of earth. h —> height through which the body is raised. —> When an elastic body is displaced from its equilibrium position, work is needed to be done against the restoring elastic force. The work done is stored up in the body in the form of its elastic potential energy. If an elastic spring is stretched (or compressed) by a distance Y from its equilibrium position, then its elastic potential energy is given by U= 1/2 kx 2 where, k —> force constant of given spring • Work-Energy Theorem According to work-energy theorem, the work done by a force on a body is equal to the change in kinetic energy of the body.

• The Law of Conservation of Energy According to the law of conservation of energy, the total energy of an isolated system does not change. Energy may be transformed from one form to another but the total energy of an isolated system remains constant. • Energy can neither be created, nor destroyed. • Besides mechanical energy, the energy may manifest itself in many other forms. Some of these forms are: thermal energy, electrical energy, chemical energy, visual light energy, nuclear energy etc. • Equivalence of Mass and Energy According to Einstein, mass and energy are inter-convertible. That is, mass can be converted into energy and energy can be converted into mass.

• Collision Collision is defined as an isolated event in which two or more colliding bodies exert relatively strong forces on each other for a relatively short time. Collision between particles have been divided broadly into two types. (i) Elastic collisions (ii) Inelastic collisions • Elastic Collision A collision between two particles or bodies is said to be elastic if both the linear momentum and the kinetic energy of the system remain conserved. Example: Collisions between atomic particles, atoms, marble balls and billiard balls. • Inelastic Collision A collision is said to be inelastic if the linear momentum of the system remains conserved but its kinetic energy is not conserved. Example: When we drop a ball of wet putty on to the floor then the collision between ball and floor is an inelastic collision. • Collision is said to be one dimensional, if the colliding particles, move along the same straight line path both before as well as after the collision. • In one dimensional elastic collision, the relative velocity of approach before collision is equal to. the relative velocity of separation after collision.

•  Coefficient of Restitution or Coefficient of Resilience Coefficient of restitution is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision.

• Elastic and Inelastic Collisions in Two Dimensions

• Non-conservative Forces A force is said to be non-conservative if the work done in moving from one point to another depends upon the the path followed. Let W, be the work done in moving from A to B following the path 1. W2 through the path 2 and W3 through the path 3. Fig. (i).

Examples of non-conservative forces are : (i) Force of friction (ii) Viscus force Low of conservation of energy holds goods for both conservative and non-conservative forces.

CBSE Class 11 Physics Chapter-6 Important Questions

1 Marks Questions

1.If two bodies stick together after collision will the collision be elastic or inelastic?   

Ans:  Inelastic collision.

2.When an air bubble rises in water, what happens to its potential energy?

Ans:  Potential energy of an air bubble decreases because work is done by upthrust on the bubble.

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

Ans:  The loss in potential energy appears as kinetic energy of the molecules of the cid.

4.Define triple point of water?

Ans.  Triple point of water represents the values of pressure and temperature at which water co-exists in equilibrium in all the three states of matter.

5.State Dulong and petit law? 

Ans. Acc. to this law, the specific heat of all the solids is constant at room temperature and is equal to 3R.

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

Ans. The clock pendulums are made of Inver because it has low value of α (co-efficient of linear expansion) i.e. for a small change in temperature, the length of pendulum will not change much.

7.Why is mercury used in making thermometers?

Ans. Mercury is used in making thermometers because it has wide and useful temperature range and has a uniform rate of expansion.

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

Ans. If glass expanded more with increasing temperature than mercury, the scale of the thermometer would be upside down.

9.Show the variation of specific heat at constant pressure with temperature?

10.Two thermometers are constructed in the same way except that one has a spherical bulb and the other an elongated cylindrical bulb. Which one will response quickly to temperature change?

Ans. The thermometer with cylindrical bulb will respond quickly to temperature changes because the surface area of cylindrical bulb is greater than the of spherical bulb.

11.  A body constrained to move along the  z -axis of a coordinate system is subject to a constant force F given by 

Where are unit vectors along the axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the axis?

Ans.  Force exerted on the body, 

Displacement,  s = m

Work done,  W =  

Hence, 12 J of work is done by the force on the body.

12. A molecule in a gas container hits a horizontal wall with speed  and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Ans.   Yes; Collision is elastic

The momentum of the gas molecule remains conserved whether the collision is elastic or inelastic.

The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the container, rebounding with the same speed.

It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.

13.  The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Ans.  Bob A will not rise at all

In an elastic collision between two equal masses in which one is stationary, while the other is moving with some velocity, the stationary mass acquires the same velocity, while the moving mass immediately comes to rest after collision. In this case, a complete transfer of momentum takes place from the moving mass to the stationary mass.

Hence, bob A of mass  m , after colliding with bob B of equal mass, will come to rest, while bob B will move with the velocity of bob A at the instant of collision.

Ans.  The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sandbag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton’s first law of motion. Hence, the speed of the trolley will remain 27 km/h.

15.  Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here  r  is the distance between centres of the balls.

Ans.  (i), (ii), (iii), (iv), and (vi)

The potential energy of a system of two masses is inversely proportional to the separation between them. 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 (i.e., V( r ) = 0) when the two balls touch each other, i.e., at  r  = 2R, where R is the radius of each billiard ball. The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions. Hence, they do not describe the elastic collisions between them.

2 Marks Questions

1.A body is moving along Z – axis of a co – ordinate system is subjected to a constant force F is given by 

Where are unit vector 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 4m along the  Z – axis?

Ans:  W = 12 J

2.A ball is dropped from the height h 1  and if rebounces to a height h 2 . Find the value of coefficient of restitution?

Ans: Velocity of approach 

(Ball drops form height h 1 )

Velocity of separation 

(Ball rebounds to height h 2 )

Coefficient of restitution

3.State and prove work energy theorem analytically?

Ans: It states that work done by force acting on a body is equal to the change produced in its kinetic energy.

If  force is applied to move an object through a distance dS

Hence W = Kf – Ki Where Kf and Ki are final and initial kinetic energy.

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

Ans: m 1  = 0.4kg,    u 1  = 4m/s, m 2  = 0.6kg   u 2  = 2m/s.

Total K.E. be fore collision

Since collision is perfectly inelastic

Total K.E. after collision

Loss in K.E. =Ki – Kf = 4.4 – 3.92 = 0.48J

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NCERT Solutions for Class 11th: Ch 6 Work, Energy And Power Physics

Ncert solutions for class 11th: ch 6 work, energy and power physics science.

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Work, Energy and Power Class 11 Physics | Notes

Work is said to be done if a body gets displaced in the direction of applied force. It is a scalar quantity. Its unit is Joule (J) in S.I. and erg in C.G.S. It is also defined as the dot product of force ($\overrightarrow{F}$) and displacement ($\overrightarrow{S}$),

i.e. W =  $\overrightarrow{F}$.$\overrightarrow{S}$

$\therefore $  W = FS Cos$\theta $ where $\theta $ is the angle between $\overrightarrow{F}$and $\overrightarrow{S}$.

Special cases:

1. When $\theta $ = 0º,

w = FScos$\theta $ = FScos0º = FS (+ve)

The positive sign shows that the work is done in the direction of applied force. 

2. When $\theta $ = 180º 

w = FScos180º = –FS (–ve)

The negative sign shows that the work is done in the direction opposite to applied force.

3. When $\theta $ = 90º 

w = FScos90º = 0 (+ve)

The +ve sign shows that the work is done in the direction of applied force. 

Energy 

The capacity of doing work is known as energy. It’s unit is Joule (J) in S.I. and erg in C.G.S. 

1J = 10 7 erg

Types of energy 

1. kinetic energy (k.e.).

It is the energy possessed by a body due to its motion. It is given by K.E. = $\frac{1}{2}$ mv 2 .

2. Potential Energy (P.E.)

It is the energy possessed by a body due to its position. It is given by P.E. = mgh 

Mechanical energy 

It is the sum of kinetic energy (K.E.) and potential energy (P.E.)

Power: 

The rate of doing work is called power. It is given by

P = $\frac{w}{t}$ = $\frac{\overrightarrow{F}.\overrightarrow{S}}{t}$ = $\overrightarrow{F}$. $\frac{\overrightarrow{S}}{t}$ =$\overrightarrow{F}$ . $\overrightarrow{V}$ = FVcos$\theta $ 

If the displacement is produced is the direction of applied force, i.e. $\theta $ = 0º,

then P = $\frac{w}{t}$ = FV 

Its unit is Js –1 or watt (w)

1 K.W = 10 3 w

1MW = 10 6 w

1 H.P. = 746 w

1mW = 10 –3 w

1$\mu $W = 10 –6 w

1nw = 10 –9 w

Relation between K.E. and linear momentum (P)

Let us consider a body of mass ‘m’ moving with velocity ‘v’. Then we have 

Linear momentum (P) = mv 

and K.E. = $\frac{1}{2}$ mv 2  

= $\frac{1}{2}$ mv 2$\times $$\frac{m}{m}$

= $\frac{1}{2}$ $\frac{{{m}^{2}}{{v}^{2}}}{m}$

=$\frac{1}{2}$$\frac{{{(mv)}^{2}}}{m}$

$\therefore $ K.E. = $\frac{{{P}^{2}}}{2m}$

Work – energy Theorem: 

Work-energy theorem states that the change in kinetic energy of a body is equal to work done. 

Let us consider a body of mass ‘m’ moving with initial velocity ‘u’ at a point. If force ‘F’ is applied on it, it moves with constant acceleration ‘a’. After time ‘t’, its velocity becomes ‘v’ at point B. Let ‘S’ be the displacement covered by the body in time ‘t’. 

Then, the amount of work done on moving from A to B is 

W = Fs 

Or, W = mas …………… (i)   [F = ma]

We have, 

v 2 = u 2 + 2as   

Or, s = $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$ …… (ii) 

From (i) and (ii), 

W = ma $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$

Or, W = $\frac{1}{2}$ mv 2 –$\frac{1}{2}$ mu 2  

Or, W = (K.E) final – (K.E) initial

$\therefore $W = $\Delta $K.E

which is the work energy theorem. 

If the body is moving in vertical direction under the action of gravity, then 

$\Delta $K.E. = $\Delta $P.E.

$\therefore $W = $\Delta $K.E. = $\Delta $P.E.

Formulae 

1. W = FScos$\theta $ 

2. P = $\frac{W}{t}$ = FVcos$\theta $ 

For $\theta $= 0 O , P = Force× Velocity

3. K.E = $\frac{{{P}^{2}}}{2m}$

4. W = $\Delta $K.E = $\Delta $P.E

Work done by Variable force 

A force is said to be variable if it changes with respect to position.

Let us consider a variable force F(x) is applied on a body and the body displaces from A to B in a fixed direction (X-axis). We can consider the entire displacement (AB) as a sum of a number of infinitesimal displacements. Let PQ = dx is one of the infinitesimal displacements. Then the small amount of work done in moving the body from P to Q is given by

dw = F × dx = PS × PQ 

or, dw = Area of rectangle PQTS 

For dx$\to $ 0, ST $\approx $ $\overset\frown{SR}$

∴ dw = area of strip PQRS ………… (i) 

Now, the total work done in moving the object from A to B can be obtained by integrating equation (i) from x = x 1 to x = x 2 , we get, 

w = $\int\limits_{x1}^{x2}{dw}$ = $\int\limits_{x1}^{x2}{area\text{ }of\text{ }strip\text{ }PQRS}$ 

$\therefore $ w = area of ABCDA 

This shows that the work done by the variable force is numerically equal to the area between force curve and displacement axis. 

Work done by constant force 

Force is said to be constant if it does not change with respect to position.

Or, W = ma s …………… (i)   [$\because $F = ma]

Or, s = $\frac{{{v}^{2}}-{{u}^{2}}}{2a}$ ………… (ii) 

Hence, Work done by constant force is equal to change in kinetic energy of the body.

Principle of Conservation of energy (Mechanical):

Principle of Conservation of energy states that energy can neither be created nor destroyed, but can be changed from one form to another form.

For a freely falling body, the total mechanical energy of the body remains constant throughout the motion.

Let us consider an object of mass ‘m’ is initially at rest at point A, which is at a height of ‘h’ from the ground level. The object is allowed to fall freely under the action of gravity. After covering a distance ‘x’, it reaches point B and finally reaches the ground level at point C.

At position A

Height = h and velocity = V A = 0

$\therefore $ K.E A = $\frac{1}{2}$ mV A 2 = 0

And P.E A = mgh

Now, mechanical energy at point A = (M.E.) A = K.E A + P.E A  

= 0 + mgh 

= mgh 

$\therefore $ M.E A = mgh ……….. (i) 

At position B

Height = h – x and velocity = V B  

We know, V B 2 = V A 2 + 2aS 

or, V B 2 = V A 2 + 2ghx 

= 0 + 2gx 

Now, K.E B = $\frac{1}{2}$m (V B ) 2  

= $\frac{1}{2}$m × 2gx = mgx 

P.E B = mg(h – x)

Now, mechanical energy at B, (M.E) B = K.E B + P.E B

= mgx + mg(h – x)

= mgx + mgh – mgx 

$\therefore $ M.E B = mgh ………… (ii)

At position C 

Height = 0 and velocity = V C  

We know, 

V C 2 = V A 2 + 2gS

or, V C 2 = 0 + 2gh 

Now, K.E C = $\frac{1}{2}$ m(V C ) 2 = $\frac{1}{2}$m × 2gh = mgh 

P.E C = mgh × 0 = 0 

Now, mechanical energy at C, M.E C = K.E C + P.E C  

$\therefore $ M.E C = mgh ……… (iii)

From (i), (ii) and (iii)

M.E A = M.E B = M.E C  

This shows that for a freely falling body, the total energy (mechanical) remains constant throughout the motion.

Fig: Variation of K.E. and P.E. with height

One dimensional collision 

A. elastic collision .

A collision is said to be elastic if both linear momentum (P) and kinetic energy (K.E.) remain conserved. There is no loss of energy in elastic collisions. 

Let us consider two objects of masses ‘m 1 ‘ and ‘m 2 ‘ moving with velocities ‘u 1 ‘ and ‘u 2 ‘ respectively. Let ‘u 1 ‘ is greater than ‘u 2 ‘, then the two objects collide and more separately with final velocities ‘v 1 ‘ and ‘v 2 ‘ respectively. Since the collision in elastic, from conservation of linear momentum.

m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2  

or, m 1 (u 1 – v 1 ) = m 2 (v 2 – u 2 ) ……….. (i) 

And from conservation of kinetic energy, 

$\frac{1}{2}$m 1 u 1 2 +$\frac{1}{2}$ m 2 u 2 2   =$\frac{1}{2}$ m 1 v 1 2 +$\frac{1}{2}$ m 2 v 2 2  

or, m 1 (u 1 2 – v 1 2 ) = m 2 (u 2 2 – v 2 2 ) ………. (ii) 

Dividing (ii) by (i),

u 1 + v 1 = u 2 + v 2  

or, u 1 – u 2 = v 2 – v 1  

$\therefore $ u 1 – u 2 = v 2 – v 1

Above result shows that velocity of approach is equal to the velocity of separation in elastic collision, i.e. relative velocity of before collision is equal to that after collision. 

v 1 = v 2 + u 2 – u 1 ………. (iii)

And, v 2 = v 1 + u 1 – u 2 ……….. (iv)

From equation (i) and (iii),

m 1 [u 1 – (v 2 + u 2 – u 1 )] = m 2 (v 2 – u 2 )

or, m 1 [u 1 – v 2 – u 2 + u 1 ) = m 2 v 2 – m 2 u 2  

or, 2m 1 u 1 – m 1 u 2 – m 1 v 2 = m 2 v 2 – m 2 u 2 

or, 2m 1 u 1 – m 1 u 2 + m 2 u 2 = m 1 v 2 – m 2 v 2 

or, 2m 1 u 1 – u 2 (m 1 – m 2 ) = v 2 (m 1 + m 2 )

or, v 2 = $\frac{2{{m}_{1}}u{}_{1}+{{u}_{2}}({{m}_{2}}-{{m}_{1}})}{{{m}_{1}}+{{m}_{2}}}$……….. (v)

From (i) and (iv),

m 1 (u 1 – v 1 ) = m 2 [(v 1 + u 1 – u 2 ) – u 2 ]

or, m 1 u 1 – m 1 v 1 = m 2 v 1 + m 2 u 1 – 2m 2 u 2  

or, m 21 – m 2 u 1 + 2m 2 u 2 = m 2 v 1 + m 1 v 1  

or, 2m 2 u 2 + u 1 (m 1 – m 2 ) = v 1 (m 1 + m 2 )

or, v 1 =$\frac{2{{m}_{2}}u{}_{2}+{{u}_{1}}({{m}_{1}}-{{m}_{2}})}{{{m}_{1}}+{{m}_{2}}}$ ……….. (vi)

From (v) and (vi) give the final velocities of m 2 and m 1 respectively.

Special cases 

1. Case I 

When m 1 = m 2 = m (Say).

Then from equations (v) and (vi), we get,

v 1 = $\frac{2m{{u}_{2}}}{2m}$

$\therefore $ v 1 = u 2 ……….. (vii)

And, v 2 = $\frac{2m{{u}_{1}}}{2m}$

$\therefore $ v 2 = u 1 ……….. (viii)

Equations (vii) and (viii) show that in elastic collisions the colliding bodies exchange their velocities if they have the same masses.

2. Case II 

When m 1 >>>> m 2 and u 2 = 0

v 1 $\approx $ $\frac{0+{{u}_{1}}{{m}_{1}}}{{{m}_{1}}}$

  $\therefore $v 1$\approx $ u 1

And, v 2 $\approx $ 2u 1

3. Case III 

When m 2 >>>> m 1 and u 2 = 0

Then from equations (v) and (vi), we get, 

v 1 $\approx $ – u 1

And v 2 $\approx $ 0  

Q.1. Show that in one dimensional elastic collision, velocity of approach is equal to velocity of separation. 

i.e. [u 1 – u 2 = v 2 – v 1 ]

Q.2. Show that in one dimensional elastic collision, the colliding bodies exchange their velocities (if they have the same masses). i.e. [u 1 = v 2 and u 2 = v 1 ]

b Inelastic collision 

The collision is said to be inelastic if linear momentum is conserved but not the kinetic energy. In inelastic collision, there is loss of kinetic energy. 

Let us consider an object of mass ‘m 1 ‘ moving with initial velocity ‘u 1 ‘ collides with another object of mass ‘m 2 ‘, which is initially at rest (i.e. u 2 = 0). After collision, the colliding bodies combine and move together with common velocity ‘v’. Since the collision is inelastic, only linear momentum is conserved, not the kinetic energy.

From the conservation of linear momentum,

m 1 u 1 + m 2 u 2 = m 1 v + m 2 v

Or, m 1 u 1 + 0 = (m 1 + m 2 )v [$\because $ m 2 is initially at rest]

$\therefore $ v = $\frac{{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}$……… (i) 

The sum of K.E. before collision is 

K.E. 1 = $\frac{1}{2}$m 1 u 1 2 + $\frac{1}{2}$m 1 u 2 2  

$\therefore $ K.E. 1 = $\frac{1}{2}$m 1 u 1 2  …. (ii) 

Now, the sum of K.E. after collision is 

K.E. 2 =  $\frac{1}{2}$m 1 v 2 + $\frac{1}{2}$m 2 v 2  

$\therefore $ K.E. 2 =  $\frac{1}{2}$(m 1 + m 2 )v 2 …… (iii)

Dividing (ii) by (iii)

$\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{({{m}_{1}}+{{m}_{2}}){{v}^{2}}}{{{m}_{1}}{{u}_{1}}^{2}}$

Using equation (i),

$\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{({{m}_{1}}+{{m}_{2}})}{{{m}_{1}}{{u}_{1}}^{2}}{{(\frac{{{m}_{1}}{{u}_{1}}}{{{m}_{1}}+{{m}_{2}}})}^{2}}$

Or, $\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ = $\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}$

Or, $\frac{K.E{{.}_{2}}}{K.E{{.}_{1}}}$ < 1 

$\therefore $K.E. 2 < K.E. 1

Above relation shows that the kinetic energy after collision is less than that before energy, i.e. there is loss in kinetic energy in inelastic collision. 

Conservative Force 

A force is said to be conservative if work done by it in a closed path (loop) is zero. The work done by conservative force is path independent.

e.g. gravitational force, electrostatic force, magnetic force. 

Non–conservative Force 

A force is said to be non – conservative if work done by it in a closed path (loop) is not zero. The work done by non – conservative force is path dependent.

e.g. viscous force, frictional force, etc. 

Coefficient of Restitution (e):

It is defined as the ratio of velocity of separation to the velocity of approaches. It is denoted by ‘e’ and given by

e = $\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}$

(i) For perfectly elastic collision, e = 1 

(ii) For perfectly inelastic collision, e = 0

(iii) Practice in collision, 0 < e < 1 

Work, Energy & Power

Numerical Problems

Q. No: 1, 2, 3, 4, 5, 6 are similar

Q.1. A stationary mass explodes into two parts of mass 4 units and 40 units respectively. If the larger mass has an initial K.E. 10 J, what is the initial kinetic energy of the smaller mass?        Ans: 100 J

Q.2. A stationary mass explodes into two parts of mass 4 kg and 40 kg. The initial kinetic energy of larger mass is 10 J. Find the velocity of the smaller mass.                  Ans: 7.07 m/sec

Q.3. A stationary mass explodes into two parts of mass 4 units and 40 units respectively, If the larger mass has an initial K.E. 100J, what is the initial K.E. of the smaller mass?                Ans: 1000 J

Q.4. An explosive of mass M placed at a point explodes into one-third and two-third parts. If the initial kinetic energy of the smaller part is 1000 J. What will be the initial K.E. of the bigger part?      Ans: 500 J

Q.5. A ball of mass 4 kg moving with a velocity 10 ms -1 collides with another body of mass 16 kg moving with 4 ms -1 from the opposite direction and then coalesces into a single body. Compute the loss of energy on impact.        Ans: 313.6J

Q.6. A bullet of mass 10 g is fired from a gun of mass 1 kg with a velocity of 100 ms -1 . Calculate the ratio of the kinetic energy of the bullet and the gun.              Ans: 100 : 1

Q.7, 8 are similar

Q.7. A 0.15 kg glider is moving to the right a frictionless horizontal air track with a speed of 0.80 ms -1 .It has a head on collision with a 0.300 kg glider that is moving to the left with a speed of 2.2 ms –1 . Find the final velocity (magnitude and direction) of each glider if the collision is elastic.                      Ans:  3.2ms -1 and 0.2ms -1 (both towards left)

Q.8. A ball A of mass 0.1 kg moving with a velocity of 6ms -1 collides directly with a ball B of mass 0.2 kg at rest. Calculate their common velocity if both balls move off together. If ball A had rebounded with a velocity of 2ms -1 in the opposite direction after collision, what would be the new velocity of B? Ans: 2 m/s, 4 m/s

Q.9, 10, 11, 12 are similar

Q. 9 A typical car weighs about 1200 N. If the coefficient of rolling friction is m r = 0.015. What horizontal force is needed to make the car move with constant speed of 72 km/hr on a level road? Also calculate the power developed by the engine to maintain this speed.                        Ans. 18 N, 360 W

Q.10. A car of mass 1000 kg moves at a constant speed of 20 m/s along a horizontal road. If a constant frictional force of 200 N. is acting between the car and the road. Calculate the power developed by the engine. Ans: 4 KW

Q. 11 A car of mass 1000 kg moves at a constant speed of 25 m/s along a horizontal road where frictional force is 200 N. Calculate the power developed by the engine.                                         Ans: 5 kw

Q.12. A car of mass 1000 kg. moves at a constant speed of 20 ms -1 along a horizontal road where the friction force is 200 N. Calculate the power developed by the engine.                         Ans: 4 KW

Q.13, 14, 15 are similar

Q.13. A 650 KW power engine of a vehicle of mass 1.5 x 10 5 Kg is rising on an inclined plane of inclination 1 in 100 with a constant speed of 60 km/hr. Find the frictional force between the wheels of the vehicle and the plane.                   Ans: 24000 N

Q.14. A train of mass 2 x 10 5 kg moves at a constant speed of 72 kmh -1 up a straight inclined against a frictional force of 1.28 x 10 4 N. The incline is such that the train rises vertically 1.0 m for every 100 m travelled along the incline. Calculate the necessary power developed by the train.    Ans: 656 KW

Q.15. Find the power of an engine in kilowatts, which pulls a train of mass 600 tonnes up an incline of 1 in 100 at the rate of 60 km/hr. The weight of the engine is 200 tonnes and the resistance due to friction is 50 Newton’s per tonne.        Ans: 2000 KW

Hint: 1 tonne = 1000 kg

Q.16. A water reservoir tank of capacity 250 m 3 is situated at a height of 20 m from the water level. What will be the power of an electric motor to be used to fill the tank in 3 hours? Efficiency of motor is 70%. Ans: 6613.76 Watt

Q.17. A block of weight 150N is pulled 20 m along a horizontal surface at constant velocity. Calculate the work done by the pulling force if the coefficient of Kinetic friction is 0.20 and the pulling force makes an angle of 60° with the vertical.                              Ans: 537.93 J

Q.18. A bullet of mass 20 g. travelling horizontally at 100 ms -1 embeds itself in the centre of a block of wood mass 1 kg, which is suspended by light vertical string 1 m. in length. Calculate the maximum inclination of the string to the vertical.                                     Ans: 35.9°

Q.19. You throw a 20 N rock vertically into the air from ground level. You observe that when it is 15 m above the ground, it is travelling at 25 m/s upward. Use the work-energy theorem to find (i) its speed as it left the ground and (ii) its maximum height.                       Ans: 30.41 m/sec, 46.24 m

Q.20. The constant force resisting the motion of a car of mass 1500 kg is equal to one fifteenth of its weight if, when travelling at 48 km/h, the car is brought to rest in a distance of 50 m by applying the brakes, find the additional retarding force due to the brakes (assumed constant) and heat developed in the brakes.                                               Ans: 1666.67 N, 83333.5 J

Also Read: Laws of Motion Class 11

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Conceptual Questions Based on Class 11 Physics Work Energy and Power

• Last modified on: 1 year ago
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Here we are providing conceptual questions for Class 11 Physics  Work Energy and Power . Answers are also provided along with questions. So that students can read and understand the question. These questions are prepared by the subject experts. Students can read these conceptual questions based on  Work Energy and Power  for better topic clarity and therefore will be able to score better marks in their exam.

Conceptual questions based on Class 11 Physics Chapter 6 Work Energy and Power   is given below. Go through each and every question to understand the topic clearly.

Q.1 . A body is moving along a circular path. How much work is done by the centripetal force?

Solution. For a body moving along a circular path, the centripetal force acts along the radius while the displacement is tangential i.e., 0 = 90°, therefore W = F.S. cos 90° = 0.

Q.2. How much work is done by a coolie walking on a horizontal platform with a load on his head?

Solution. Zero. In order to balance the load on his head, the coolie applies a force on it in the upward direction equal to its weight. His displacement is along the horizontal direction. Thus, the angle between force and displacement is 90°. Therefore, work done, W = Fs cos θ=Fs cos 90°=0.

Q.3. The earth moving around the sun in a circular orbit is acted upon by a force and hence work must be done on the earth by the force. Do you agree by this statement?

Solution. The statement is wrong. The gravitational force is a conservative force. So, the work done by the gravitational force over every complete orbit of the earth is zero.

Q.4. Is it possible that a body be in accelerated motion under a force acting on the body, yet no work is being done by the force? Explain your answer giving a suitable example.

Solution. Yes, it is possible, when the force is perpendicular to the direction ot motion. The moon revolves round the earth under the centripetal force of attraction of the earth, but earth does no work on the moon.

Q.5. A porter moving vertically up the stairs with a suitcase on his head does work. Why?

Solution. The porter lifts the suitcase vertically to the upstairs. Force has to be applied on the suitcase against the force of gravitation. Hence the porter does work.

Q.6. A man rowing boat upstream is at rest with respect to the shore. Is he doing work?

Solution. The boat is at rest with respect to the shore, but it is moving upstream with respect to water. The man is doing work relative to the stream because he is applying force to produce relative motion between the boat and the stream. But he does no work relative to the shore as displacement relative to the shore is zero.

Q.7. Does the work done in raising a suitcase on to a platform depend upon how fast it is raised up?

Solution. No. The work done depends on the force of gravity and the height through which the suitcase is raised. It does not depend on the time rate with which the suitcase is raised.

Q.8. Can a body have momentum without energy?

Solution. No, if a body has momentum, it must be in motion and consequently possess kinetic energy.

Q.9. In a tug of war one team is slowly giving way to the other. What work is being done and by whom?

Solution. The work is done by the winning team and is equal to the product of resultant force applied by the two teams and displacement that the losing team suffers.

Q.10. Explain, throwing mud on a wall is an example of perfectly inelastic collision.

Solution. When mud is thrown on a wall, it sticks to the wall. The kinetic energy of the mud is reduced to zero and there is non-conservation of kinetic energy. Hence it is a case of perfectly inelastic collision.

Q.11. Mountain roads rarely go straight up but wind up gradually. Why?

Solution. If the roads go straight up, the angle of slope θ would be large. In that case frictional force (f = μ my cos θ) would be less and the vehicles may slip. Moreover, while going up a large slope, a greater power would be required (as it will take less time).

Q.12. Can a body have energy without momentum?

Solution. Yes, there is an internal energy in a body due to the thermal agitation of the particles of the body, while the vector sum of the momenta of the moving particles may be zero.

Q.13. Can the overall energy of a body be negative?

Solution. Yes. As E = K + U, when U is negative and has magnitude greater than that of K, E is negative. For example, the energy of an electron bound to the nucleus of an atom is negative.

Q.14. Two bodies moving towards each other collide and move away in opposite directions. There is some rise of temperature of the bodies in the process. Explain the reason for the rise of temperature and state what type of collision is it.

Solution. Involved collision of the process is inelastic because the bodies suffer loss of kinetic energy which appears in the form of heat energy raising the temperature of the bodies.

Q.15. Does the potential energy of a spring decrease or increase when it is compressed or stretched?

Solution. The potential of the spring increases because work is done on it when it is compressed or stretched.

Q.16. What happens to the potential energy when (a) two protons are brought close together (b) one proton and one electron are brought close together?

(a) The potential energy increases when two protons are brought closer because work has to be done against the force of repulsion between them.

(b) The potential energy decreases when a proton and an electron are brought closer because work is done by the force of attraction between them (origin of the force being the system itself).

Q.17. “Chemical, gravitational and nuclear energies are nothing but potential energies for different types of forces in nature.” Explain the statement clearly with examples.

Solution. A system has potential energy, when various objects are held at certain distance against some force, by virtue of their position or configuration. Chemical potential energy results from the chemical bonding of atoms, gravitational potential energy results when objects are held at some distance against gravitational force while nuclear energy arises from the nuclear force between the nucleons.

Q.18. In a thermal power station, coal is used for the generation of electricity. Mention how energy changes from one form to another before it is transformed into electrical energy?

Solution. The heat energy produced due to combustion of coal converts water into steam. The heat energy of steam is converted into mechanical energy when it turns blades of a turbine. The mechanical energy so obtained is converted into electrical energy by the generators.

Conceptual Questions Based on Class 11 Physics

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• Class 11 Physics MCQs
• Chapter 6 Work Energy And Power

Class 11 Physics Chapter 6 Work Energy and Power MCQs

Class 11 Physics Chapter 6 Work, Energy and Power MCQs are provided here with answers. These questions are designed as per the latest CBSE syllabus and NCERT curriculum. Solving these chapter-wise MCQs will help students to score good marks in the final exam. Work, Energy and Power Class 11 physics MCQs are prepared for a better understanding of the concept. It allows students to test their knowledge and answering skills in the given time frame.

MCQs on Class 11 Chapter 6 Work, Energy and Power

Check the multiple-choice questions for the 11th Class Physics Work, Energy and Power chapter. Each MCQ will have four options here, out of which only one is correct. Students have to pick the correct option and check the answer provided here.

Download Chapter 6 Work, Energy and Power MCQs PDF by clicking on the button below. Download PDF

1. The rate of doing work is called ——————-

• Acceleration
• Displacement

Answer: (c) Power

Explanation: Power is defined as the rate of doing work.

2. Which is the type of collision in which both the linear momentum and the kinetic energy of the system remain conserved?

• Inelastic Collision
• Elastic Collision
• Destructive collision
• None of the options

Answer: (b) Elastic Collision

Explanation: In an elastic collision, the linear momentum and the kinetic energy of the system remain conserved.

3. Collision between marble balls is which type of collision?

Explanation: Collision between marble balls is an example of elastic collision.

4. The energy possessed by the body by virtue of its motion is known as?

• Chemical energy
• Thermal energy
• Potential energy
• Kinetic energy

Answer: (d) Kinetic energy

5. Find the potential energy stored in a ball of mass 5 kg placed at a height of 3 m above the ground.

Answer: (b) 147.15 J

Explanation: m = 5 kg, h = 3 m, g = 9.81 m/s -2

We know that, Potential energy = mgh

= 5 * 9.81 * 3 = 147.15 J

6. What is the power utilised when work of 1000 J is done in 2 seconds?

Answer: (d) 500 W

Explanation: W=1000J, t=2 seconds

Power=work/time = 1000/2 = 500 W

7. State true or false: According to Equivalence of Mass and Energy, it states that mass and energy are NOT interconvertible.

Answer: (b) False

8. Which one has higher kinetic energy? Both light and heavy bodies have equal momenta.

Answer: (b) Light body

Explanation: Since, the momenta of the heavy body and light body are same, thus the velocity of light body is greater than that of heavy body. Also, kinetic energy is proportional to the square of velocity. Hence, the lighter body has greater kinetic energy.

9. An electric heater of rating 1000 W is used for 5 hrs per day for 20 days. What is the electrical energy utilized?

Answer: (a) 100 kWh

Explanation: The power of the electric heater is 1000 W, and the time period is 20×5 = 100 hr.

Electrical energy = Power × Time

Electrical energy = 1000 × 100 = 100000 Wh

Electrical energy = 100 kWh

10. A ball moves in a frictionless inclined table without slipping. The work done by the table surface on the ball is

Answer: (b) Zero

Explanation: The work done by a ball when it moves on a frictionless inclined table without slipping is zero.

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Work, Energy and Power Class 11 Notes CBSE Physics Chapter 6 (Free PDF Download)

Revision Notes for CBSE Class 11 Physics Chapter 6 (Work, Energy and Power) - Free PDF Download

Work, Energy, and Power – a captivating chapter in CBSE Class 11 Physics! This chapter unravels the intricacies of some of the most fundamental concepts in physics that govern the motion and transformations of energy in our universe. From understanding the concept of work done by forces to exploring the different forms of energy and their interconversion, this chapter offers a comprehensive insight into the dynamics of energy in various physical systems. 

To aid your learning journey, we present you with these Class 11 Notes on Work, Energy, and Power – a free PDF download that condenses the key points, equations, and practical applications of the topic. With these notes at your disposal, you can embark on a fascinating voyage to comprehend the core principles that shape the behavior of physical systems and their energetic interplay. Let's dive in and unlock the secrets of this captivating realm!

Revision Notes for Class 11 Physics Chapter 6 - Work, Energy, and Power are available in Vedantu. These Revision Notes are written as per the latest Syllabus of NCERT. We hear the words 'work,' 'energy,' and 'power' all the time. A person carrying materials, a farmer cultivating, and a student studying for exams are all said to be performing their work. Work has a specific and definite meaning in Physics.

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

Download CBSE Class 11 Physics Revision Notes 2024-25 PDF

Also, check CBSE Class 11 Physics revision notes for other chapters:

Work, Energy and Power Chapter-Related Important Study Materials It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Work, Energy and Power Class 11 Notes Physics - Basic Subjective Questions

Section – a (1 mark questions).

1. What is the unit of Power?

Ans. Watt (W)

2. Name the factors on which work is done depends.

Ans. Force, displacement and angle between force and displacement.

3. Work done by external forces is always equal to the gain in kinetic energy. Is it always true ? 

Ans. Yes. This is the universal work-energy theorem.

4. If energy is neither created nor destroyed, then from where do we get energy?

Ans. By transforming energy from one form to another.

5. A force $F=5\hat{i}+6\hat{j}-4\hat{k}$ acting on a body, produce a displacement $s=6\hat{i}+5\hat{k}$ Work done by the force.

Ans. $W=F\cdot s=\left ( 5\hat{j}+6\hat{j}-4\hat{k} \right )\left ( 6\hat{i}+5\hat{k} \right )$

=30-20=10 units

Section – B (2 Marks Questions)

6. List two conditions which need to be satisfied for the work to be done on an object?

(i) Force should be applied on the body.

(ii) Body should move in the direction of force.

7. An archer stretches a bow to release an arrow to hit the target at a distance of 10 m. Explain who does the work, in which form is the energy possessed by the bow and the arrow.

Ans. The archer does the work in pulling the bow string taut. The muscular energy of the archer arm a → potential energy of the taut string → kinetic energy of the arrow.

8. A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.

Ans. Work done = Force × displacement

= 15 × 10 × 1.5

9. A man of mass 60 kg runs up a flight of 30 steps in 40 s. If each step is 20 cm high, calculate his power. (take g = 10 m/s 2 )

Ans. Total height reached by $man=30\times \dfrac{20}{100}m=6m$

Power = work done/time = mgh / time 

$=\dfrac{60\times 10\times 6}{40}=90V$

10. An electric bulb of 100 W works for 4 hours a day. Calculate the units of energy consumed in 15 days.

Ans. Energy consumed = Power × Time

= 100 × 4 × 15

= 6 KWh = 6 units

PDF Summary - Class 11 Physics Work, Energy and Power Notes (Chapter 6)  

In Physics, work refers to ‘mechanical work’. Work is said to be done by a force on a body when the body is actually displaced through some distance in the direction of the applied force.

However, when there is no displacement in the direction of the applied force, there is no work done, i.e., work done is zero, when displacement of the body in the direction of the force is zero.

Consider a constant force ‘F’ acting on a body to produce a displacement ‘s’ in the body along the positive x-direction as shown in the following figure:

(Image will be updated soon)

If $\theta $ is the angle which F makes with the positive x-direction of the displacement, then the component of F in the direction of displacement is given by $F\cos \theta $. Since the work done by the force is the product of component of force in the direction of the displacement and the magnitude of the displacement, we can write:

$W=(F\cos \theta )s$

Now, when the displacement is in the direction of force applied, i.e., when $\theta ={{0}^{0}}$;

\[\Rightarrow W=\left( F\cos 0{}^\circ  \right)s=\vec{F}.\vec{s}\]

Clearly, work done by a force is the dot product of force and displacement.

In terms of rectangular components, $\vec{F}$ and $\vec{s}$ may be written as

\[\vec{F}=\hat{i}{{F}_{x}}+\hat{j}{{F}_{Y}}+\hat{k}{{F}_{Z}}\] and $\vec{s}=\hat{i}x+\hat{j}y+\hat{k}z$

$\Rightarrow W=\left( \hat{i}{{F}_{x}}+\hat{j}{{F}_{Y}}+\hat{k}{{F}_{Z}} \right).\left( \hat{i}x+\hat{j}y+\hat{k}z \right)$

$\Rightarrow W=x{{F}_{x}}+y{{F}_{y}}+z{{F}_{z}}$

Work is a scalar quantity, i.e., it has magnitude only and no direction. However, work done by a force can be positive, negative or zero.

2. DIMENSIONS AND UNITS OF WORK

As work $=$ force × distance;

$\Rightarrow W=({{M}^{1}}{{L}^{2}}{{T}^{-2}})\times L$

$\Rightarrow W=[{{M}^{1}}{{L}^{2}}{{T}^{-2}}]$

This is the dimensional formula of work.

The units of work are of two kinds: a) Absolute units and b) Gravitational units

a) Absolute Units

Joule: It is the absolute unit of work in the SI system of units. Work done is said to be one joule, when a force of one newton actually moves a body through a distance of one meter in the direction of applied force.

$\Rightarrow 1joule=1newton\times 1metre\times \cos {{0}^{0}}=1N.m$

Erg: It is the absolute unit of work in the CGS system of units. Work done is said to be one erg, when a force of one dyne actually moves a body through a distance of one cm in the direction of applied force.

$\Rightarrow 1erg=1dyne\times 1cm\times \cos {{0}^{0}}=1dyne.cm$

b) Gravitational Units

These are also known as practical units of work.

Kilogram-meter (kg-m): It is the gravitational unit of work in the SI system of units. Work done is said to be one kg-m, when a force of 1kgf moves a body through a distance of 1m in the direction of the applied force.

$\Rightarrow 1kg-m=1kgf\times 1m\times \cos {{0}^{0}}=9.8N\times 1m=9.8joules$, i.e.,

$\Rightarrow 1kg-m=9.8J$

Gram-centimeter (g-cm): It is the gravitational unit of work in the CGS system of units. Work done is said to be one g-cm, when a force of 1gf

moves a body through a distance of 1cm in the direction of the applied force.

$\Rightarrow 1g-cm=1gf\times 1cm\times \cos {{0}^{0}}$

$\Rightarrow 1g-cm=980dyne\times 1cm\times 1$

$\Rightarrow 1g-m=980ergs$

3. NATURE OF WORK DONE

Although work done $\left( W=(F\cos \theta )s \right)$ is a scalar quantity, its value may be positive, negative, negative or even zero, as detailed below:

Positive Work is said to be done on a body when $\theta $ is acute ($<{{90}^{0}}$). Clearly, $\text{cos}\theta $ turns out to be positive and hence, the work done is positive.

For example, when a body falls freely under the action of gravity,$\theta ={{0}^{0}};\cos \theta =\cos {{0}^{0}}=+1$. Clearly, work done by gravity on a body falling freely is positive.

Negative Work is said to be done on a body when $\theta $ is obtuse ($>{{90}^{0}}$). Clearly, $\text{cos}\theta $  is negative and hence, the work done is negative.

For example, when a body is thrown up, its motion is opposed by gravity. The angle $\theta $ between gravitational force and the displacement is ${{180}^{0}}$. Since $\cos \theta =\cos {{180}^{0}}=-1$; work done by gravity on a body moving upwards is negative.

(image will be updated soon)

Zero Work is said to be done on a body when force applied on it or the displacement caused or both of them are zero. Here, when angle $\theta $ between force and displacement is ${{90}^{0}}$; $\cos \theta =\cos {{90}^{0}}=0$ and hence, the work done is zero.

For example, when we push hard against a wall, the force we exert on the wall does no work because displacement is zero in this case. However, in this process, our muscles are contracting and relaxing alternately and internal energy is being used up. This is why we do get tired.

4. WORK DONE BY A VARIABLE FORCE

Graphical Method:

A constant force is rare. It is the variable force which is encountered more commonly. 

To evaluate the work done by a variable force, let us consider a force acting along a fixed direction, say x–axis, but having a variable magnitude.

We have to compute work done in moving the body from A to B under the action of this variable force. 

To facilitate this, we assume that the entire displacement from A to B is made up of a large number of infinitesimal displacements.

One such displacement shown in the following figure from P to Q.

Since the displacement $PQ=dx$ is infinitesimally small, we consider that all along this displacement, force is constant in magnitude as well in the same direction.

Now, a small amount of work done in moving the body from P to Q is given by,

$dW=F\times dx=(PS)(PQ)=area\text{ }of\text{ }strip\text{ }PQRS$

Therefore, the total work done in moving the body from A to B is given by

$\Rightarrow W=\sum{dW}$

$\Rightarrow W=\sum{F\times dx}$

Here, when the displacement is allowed to approach zero, then the number of terms in the sum increases without a limit. And the sum approaches a definite value equal to the area under the curve CD.

Thus, we may rewrite that 

$\Rightarrow W=\underset{dx\to x}{\mathop{\lim }}\,\sum{F(dx)}$

Using integral calculus, we may write it as

$\Rightarrow W={}_{{{X}_{A}}}^{{{X}_{B}}}\int{A(dx)}$ 

${{x}_{A}}={{O}_{A}}$ and ${{x}_{B}}=OB$

$\Rightarrow W={}_{{{X}_{A}}}^{{{X}_{B}}}\int{area\text{ }of\text{ }strip\text{ }PQRS}$

Which is nothing but the total area under the curve between F and x-axis from $x={{x}_{A\text{ }}}\text{ }to\text{  }x={{x}_{B}}$.

$\Rightarrow W=Area\text{ }of\text{ }ABCDA$

Clearly, the work done by a variable force is numerically equal to the area under the force curve and the displacement axis.

Mathematical Treatment (of work done by a variable force)

Suppose we have to evaluate the work done in moving a body from a point A (${{S}_{A}}$) to point B (${{S}_{B}}$) under the action of a varying force as shown in the following figure. Here, ${{S}_{A}}$ and ${{S}_{B}}$ are the distance of the points A and B with respect to some reference point.

At any stage, let the body be at P, where force on the body is $\vec{F}$. 

Under the action of this force, let the body undergo an infinitesimally small displacement $d\vec{s}$

During such a small displacement, if we assume that the force remains constant, then small amount of work done in moving the body from P to Q is

$dW=\vec{F}.d\vec{s}$

Now, when the displacement is zero, the total work done in moving the body from A to B can be obtained by integrating the above expression between ${{S}_{A}}$and ${{S}_{B}}$ as follows:

\[\Rightarrow W={}_{{{S}_{A}}}^{{{S}_{B}}}\int{\vec{F}.d\vec{s}}\] 

5. CONSERVATIVE & NON­CONSERVATIVE FORCES

Conservative Force

A force is said to be conservative when the work done by or against the force in moving a body is dependent only on the initial and final positions of the body, and not on the nature of path followed between the initial and the final positions.

This suggests that the work done by or against a conservative force in moving a body over any path between fixed initial and final positions would be the same.

For instance, gravitational force is a conservative force.

Properties of Conservative Forces :

Work done by or against a conservative force in moving a body from one position to the other depends only on the initial position and final position of the body.

Work done by or against a conservative force does not depend upon the nature of the path followed by the body in going from initial position to the final position.

Work done by or against a conservative force in moving a body through any round trip (i.e., closed path, where final position coincides with the initial position of the body) is always zero.

Non-conservative Forces

A force is said to be non-conservative when the work done by or against the force in moving a body from one position to another, is dependent on the path followed between these two positions.

For instance, frictional forces are non-conservative forces.

Power of a person or machine refers to the time rate at which work is done by it.

Mathematically,

Power $=$ Rate of doing work $=$ $\frac{work\text{ }done}{time\text{ }taken}$

Thus, power of a body measures how fast it can do the work.

$\Rightarrow P=\frac{dW}{dt}$

Now, it is known that $\text{dW=\vec{F}}\text{.d\vec{s}}$;

$\Rightarrow P=\frac{\vec{F}.d\vec{s}}{dt}$

But $\frac{d\vec{s}}{dt}=\vec{v}$, which is the instantaneous velocity.

$\Rightarrow P=\vec{F}.\vec{v}$

Dimensions of power is given by

$P=\frac{W}{t}=\frac{{{M}^{1}}{{L}^{2}}{{T}^{-}}}{{{T}^{1}}}=[{{M}^{1}}{{L}^{2}}{{T}^{-3}}]$

Units of power

The absolute unit of power in SI system of units is watt, which is denoted by $W$.

$\Rightarrow P=\frac{W}{t}$

$\Rightarrow 1watt=\frac{1joule}{1\text{ }\sec }\Rightarrow 1W=1J{{s}^{-1}}$

Clearly, power of a body is said to be one watt, when it can do one joule of work in one second. A bigger unit if power is horsepower (hp), given by

Energy of a body refers to the capacity or ability of the body to do work.

8. KINETIC ENERGY

The kinetic energy of a body refers to the energy possessed by the body by virtue of its motion.

Here are some examples:

A bullet fired from a gun can pierce through a target on account of kinetic energy of the bullet.

Wind mills work on the kinetic energy of air. For instance, sailing ships use the kinetic energy of wind.

Water mills work on the kinetic energy of water. For instance, fast flowing streams are utilized to grind corn.

A nail is driven into a wooden block on account of kinetic energy of the hammer striking the nail.

Formula for Kinetic Energy

Kinetic Energy of a body can be obtained either from

the amount of work done in stopping the moving body, or from

the amount of work done in giving the present velocity to the body from the state of rest.

Let us first consider the second method:

Suppose that, 

m$=$ mass of a body at rest (i.e., u$=$0).

F$=$Force applied on the body

a$=$acceleration produced in the body in the direction of force applied.

v$=$velocity acquired by the body in moving through a distance ‘s’, as shown in the following diagram.

Now, consider the equation of motion: $v-u=2as$;

$\Rightarrow {{v}^{2}}-0=2as$

$\Rightarrow a=\frac{{{v}^{2}}}{2s}$

It is known that 

$\Rightarrow F=m\left( \frac{{{v}^{2}}}{2s} \right)$

Clearly, work done on the body, (W $=$ Force × distance)

$\Rightarrow W=m\frac{{{v}^{2}}}{2s}\times s$

$\Rightarrow W=\frac{1}{2}m{{v}^{2}}$

This work done on the body is a measure of kinetic energy (K.E.) acquired by the body,

$\Rightarrow K.E\text{ }Of\text{ }the\text{ }body=W=\frac{1}{2}m{{v}^{2}}$

Alternative Method

The formula for kinetic energy of a body can also be obtained by the method of calculus as follows:

Let m$=$mass of a body, which is initially at rest (i.e., u$=$0)

$\vec{F}$$=$Force applied on the body,

$d\vec{s}$$=$ small displacement produced in the body in the direction of the force applied.

A small amount of work done by the force is given by,

$dW=\vec{F}.d\vec{s}=Fds\cos {{0}^{0}}=Fds$

If ‘a’ is the acceleration produced by the force, then from

$F=ma=m\frac{dv}{dt}$ and from 

$dW=\left( m\frac{dv}{dt} \right)ds=m\left( \frac{ds}{dt} \right)dv$;

$\Rightarrow dW=mvdv\text{     }\left( \because \frac{ds}{dt}=v \right)$

Thus, the total work done by the force in increasing the velocity of the body from zero to v is given by

$W=\int\limits_{0}^{v}{mvdv}=\frac{1}{2}m{{v}^{2}}$

Thus, kinetic energy of a body is half the product of mass of the body and square of velocity of the body.

9. RELATION BETWEEN KINETIC ENERGY AND LINEAR MOMENTUM

If m is the mass of a body and v is the velocity of the body;

Linear momentum of the body is given by \[p=mv\] and K.E. of the body is given by $KE=\frac{1}{2}m{{v}^{2}}=\frac{1}{2m}({{m}^{2}}v)$

$\Rightarrow KE=\frac{p}{2m}$

This is an important relation. It shows that a body cannot have kinetic energy without having linear momentum. The reverse is also true.

Further, if linear momentum (p) is constant, then,

$KE \propto \frac{1}{m}$

This is shown in figure (a).

On the other hand, if kinetic energy (KE) is constant, then,

${{p}^{2}}\propto m\text{ }or\text{ }p\propto \sqrt{m}$

This is shown in figure (b).

Also, if mass (m) is constant, the,

${{p}^{2}}\propto KE\text{ }or\text{ }p\propto \sqrt{KE}$

This is shown in figure (c).

10. WORK ENERGY THEOREM OR WORK ENERGY PRINCIPLE

According to this principle, work done by net force in displacing a body is the same as the change in kinetic energy of the body.

Thus, when a force does some work on a body, the kinetic energy of the body increases by the same amount. Conversely, when an opposing (retarding) force is applied on a body, its kinetic energy decreases. The decrease in kinetic energy of the body is equal to the work done by the body against the retarding force. Thus, according to work energy principle, work and kinetic energy are equivalent quantities.

Proof: To prove the work-energy theorem, we confine ourselves to motion in one dimension.

Suppose that m$=$mass of a body, u$=$initial velocity of the body, F$=$force applied on the body along it direction of motion, a$=$acceleration produced in the body, v$=$final velocity of the body after t second.

Small amount of work done by the applied force on the body is given by $dW=F(ds)$, when ds is the small distance moved by the body in the direction of the force applied.

$F=ma=m\left( \frac{dv}{dt} \right)ds=m\left( \frac{ds}{dt} \right)dv=mvdv\left( \therefore \frac{\text{ds}}{\text{dt}}=v \right)$

Total work done by the applied force on the body in increasing its velocity from u to v is given by

\[W=\int\limits_{u}^{v}{mvdv}=m\left[ \frac{{{v}^{2}}}{2} \right]_{u}^{v}\]

$\Rightarrow W=\frac{1}{2}m(v-u)=\frac{1}{2}mv-\frac{1}{2}mu$

But $\frac{1}{2}m{{v}^{2}}={{K}_{f}}=final\text{ }KE\text{ }of\text{ }the\text{ }body\text{ }and\text{ }\frac{1}{2}m{{u}^{2}}={{K}_{i}}=initial\text{ }KE\text{ }of\text{ }the\text{ }body$

$\Rightarrow W={{K}_{f}}-{{K}_{i}}=increases\text{ }in\text{ }KE\text{ }of\text{ }body$

i.e., work done on the body is equal to the increase in KE of the body.

11. POTENTIAL ENERGY

The potential energy of a body refers to the energy possessed by the body by virtue of its position or configuration in some field.

Thus, potential energy is the energy that can be associated with the configuration (or arrangement) of a system of objects that exert forces on one another. Obviously, if configuration of the system changes, then its potential energy changes.

Two important types of potential energy are:

Gravitational potential energy

Elastic potential energy.

11.1 Gravitational Potential Energy

Gravitational potential energy of a body refers to the energy possessed by the body by virtue of its position above the surface of the earth.

To calculate gravitational potential energy, suppose

m$=$mass of a body

g$=$acceleration due to gravity on the surface of earth.

h$=$height through which the body is raised, as shown in the following figure.

If we assume that height ‘h’ is not too large and the value of ‘g’ is practically constant over this height, then the force applied just to overcome gravitational attraction is given by,

As the distance moved is in the direction of the force applied, work can be expressed as:

Work done$=$force × distance

$\Rightarrow W=F\times h=mgh$

Notice that we have taken the upward direction to be positive. Therefore, work done by applied force$=+mgh$. However, work done by gravitational force$=-mgh$.

This work gets stored as potential energy. The gravitational potential energy of a body, as a function of height (h) is denoted by V(h), and it is negative of work done by the gravitational force in raising the body to that height.$\Rightarrow Gravitational\text{ }PE=V(h)=mgh$

11.2 Potential Energy of a Spring

Potential energy of a spring refers to the energy associated with the state of compression or expansion of an elastic spring.

To compute it, consider an elastic spring OA of negligible mass. The end O of the spring is fixed to a rigid support and a body of mass ‘m’ is attached to the free end A. Let the spring be oriented along the x–axis and the body of mass ‘m’ lie on a perfectly frictionless horizontal table.

The position of the body A, when spring is unstretched, is chosen as the origin. Now, when the spring is compressed or elongated, it tends to recover to its original length, on account of elasticity. The force trying to bring the spring back to its original configuration is termed restoring force or spring force.

For a small stretch or compression, spring obeys Hooke's law, i.e., for a spring,

Restoring Force $\propto $ stretch or compression

$\Rightarrow -F\propto x\text{ }or\text{ }-F=kx$

where k is a constant of the spring called the spring constant.

It is established that for a spring, $k\propto \frac{1}{l}$. i.e., smaller the length of the spring, greater would be the force constant and vice-versa.

The negative sign in the equation indicates that the restoring force is always directed towards the equilibrium position.

Now, consider that the body be displaced further through an infinitesimally small distance dx, against the restoring force.

A small amount of work done in increasing the length of the spring by dx is given by,

$dW=-Fdx=kxdx$

Thus, the total work done in giving displacement x to the body can be obtained by integrating from $x=0$ to $x=x$, i.e.,

$W=\int\limits_{x=0}^{x=x}{kxdx}=\frac{1}{2}k{{x}^{2}}$

This work done is stored in the spring at the point B.

$\Rightarrow PE\text{ }at\text{ }B=W=\frac{1}{2}k{{x}^{2}}$

The variation of potential energy with distance x is as shown in the following figure.

12. MECHANICAL ENERGY AND ITS CONSERVATION

The mechanical energy (E) of a body refers to the sum of kinetic energy (K) and potential energy (V) of the body

i.e., $E=K+V$

Obviously, mechanical energy of a body is a scalar quantity measured in joules.

We can show that the total mechanical energy of a system is conserved if the force doing work on the system is conservative.

This is known as the principle of conservation of total mechanical energy.

For simplicity, we assume the motion to be one dimensional only. Suppose a body undergoes a small displacement ‘x’ under the action of a conservative force F. According to the work energy theorem, change in kinetic energy is equal to the work done.

$\Rightarrow \Delta K=F(x)\Delta x$

Now, as the force is conservative, the potential energy function V(x) is defined as

\[-\Delta V=F(x)\Delta x\text{ }or\text{ }\Delta V=-F(x)\Delta x\]

Adding both the above expressions, we get,

$\Rightarrow \Delta K+\Delta V=0\text{ }or\text{ }\Delta (K+V)=0$,

which means that

$(K+V)=E=constant$

12.1 Illustration of the Law of Conservation of Mechanical Energy

To illustrate the law further, let us evaluate kinetic energy, potential energy, and total energy of a body falling freely under gravity.

Let ‘m’ be the mass of the body held at A, at a height h above the ground, as shown in the following figure.

As the body is at rest at A, therefore,

KE of the body is zero.

PE of the body is equal to mg, where g is acceleration due to gravity at A.

$\Rightarrow TE\text{ }of\text{ }the\text{ }body=KE+PE=0+mgh$

$\Rightarrow {{E}_{A}}=mgh$….(1)

Now, let the body be allowed to fall freely under gravity, when it strikes the ground at C with a velocity ‘v’.

From ${{v}^{2}}-{{u}^{2}}=2as$;

$\Rightarrow {{v}^{2}}-0=2(g)h$

$\Rightarrow {{v}^{2}}=2gh$….(2)

Therefore, at C;

KE of the body$=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m(2gh)=mgh$

PE of the body\[=mgh=mg\left( 0 \right)=0\]

Total energy of the body is given by,

$TE={{E}_{C}}=mgh+0=mgh$.... (3)

Now, in free fall, let the body cross any point B with a velocity ${{v}_{1}}$, where AB is equal to ‘x’. Thus, from ${{v}^{2}}-{{u}^{2}}=2as$;

$\Rightarrow v_{1}^{2}-0=2(g)x$…. (4)

$\Rightarrow v_{1}^{2}=2gx$

Clearly, at B;

KE of the body$=\frac{1}{2}mv_{1}^{2}=\frac{1}{2}m(2gx)=mgx$

Height of the body at B above the ground\[=CB=\left( hx \right)\]

PE of the body at B\[=mg\left( hx \right)\]

Total energy of the body at B\[=KE+PE\]

$\Rightarrow {{E}_{B}}=mgx+mg(h-x)=mgx+mgh-mgx$

$\Rightarrow {{E}_{B}}=mgh$.... (5)

Clearly, from (1), (3), and (5); we find that

${{E}_{A}}={{E}_{C}}={{E}_{B}}=mgh$

13. DIFFERENT FORMS OF ENERGY

We have learnt some details of potential energy and kinetic energy. It should be understood that these are not the only two forms of energy. Energy may manifest itself in several other forms. Some of the examples are:

Heat Energy

It is the energy possessed by a body by virtue of random motion of the molecules of the body.

Heat is also associated with the force of friction. When a block of mass ‘m’ sliding on a rough horizontal surface with speed ‘v’, stops over a distance ‘x’, work done by the force of kinetic friction ‘f’ over a distance ‘x’ is given by $-f(x)$. By the work energy theorem, $\frac{1}{2}m{{v}^{2}}=f(x)$ .We often say that kinetic energy of the block is lost due to frictional force. However, when we examine the block and the horizontal surface carefully, we detect a slight increase in their temperatures. Thus, work done by friction is not lost, but is transferred as heat energy of the system.

Internal Energy

It is the total energy possessed by the body by virtue of particular configuration of its molecules and also their random motion. Thus, internal energy of a body is the sum of potential energy and kinetic energy of the molecules of the body.

Electrical Energy

The flow of electric current causes bulbs to glow, fans to rotate and bells to ring. A definite amount of work has to be done in moving the free charge carriers in a particular direction through all the electrical appliances.

Chemical Energy

Chemical energy arises from the fact that the molecules participating in the chemical reaction have different binding energies. A chemical reaction is basically a rearrangement of atoms. For example, coal consists of carbon and a kilogram of it. When burnt, it releases $3\times {{10}^{7}}J$ of energy.

Nuclear Energy

It is the energy obtainable from an atomic nucleus. Two distinct modes of obtaining nuclear energy are nuclear fission nuclear fusion.

Nuclear fission involves splitting of a heavy nucleus into two or more lighter nuclei, whereas nuclear fusion involves fusing of two or more lighter nuclei to form a heavy nucleus.

14. MASS ENERGY EQUIVALENCE

Einstein made an incredible discovery that energy can be transformed into mass and mass can be transformed into energy. To put it precisely, one energy can be obtained at the cost of the other energy.

The mass energy equivalence relation as put forth by Einstein is

$E=m{{c}^{2}}$

m is the mass that disappears; 

E is the energy that appears;

C is the velocity of light in vacuum.

Mass and energy are not conserved separately, but are conserved as a single entity called ‘mass-energy’.

15. THE PRINCIPLE OF CONSERVATION OF ENERGY

If we account for all forms of energy, the total energy of an isolated system does not change.

The principle of conservation of energy cannot be proved as such. However, no violation of this principle has ever been observed.

16. WORK DONE BY A VARIABLE FORCE

When the force is an arbitrary function of position, we require the techniques of calculus to determine the work done by it. The figure shows F(x) as some function of the position x. To calculate work done by F from A to B, we find the area under the graph from ${{X}_{A}}\text{ }to\text{ }{{X}_{B}}$.

Thus, the work done by a force F(x) from an initial point A to final point B is given by,

${{W}_{A\to B}}=\int\limits_{{{X}_{A}}}^{{{X}_{B}}}{{{F}_{X}}dx}$

17. CONSERVATIVE & NON­CONSERVATIVE FORCES

17.1 Conservative Forces

A force is conservative if the net work done against the force in moving a mass between two points depends only on the location of two points and not on the path followed.

17.2 Non-Conservative Forces

Those forces which do not satisfy the above-mentioned criteria are termed non-conservative forces. Friction and viscous forces are the most common examples of non-conservative forces.

17.3 Conservative Forces and Potential Energy

For every conservative force, there is a corresponding potential energy function. In each case, the potential energy expression is dependent only on position. For every conservative force ${{F}_{X}}$, that depends only on the position ‘x’, there is an associated potential energy function U(x). When conservative force does positive work, the potential energy of the system decreases. Work done by conservative force is given as:

$F(x)\Delta x=-\Delta U$

$\Rightarrow F(x)=\frac{-\Delta U}{\Delta x}$

which, in the limit, becomes, 

$\Rightarrow F(x)=-\frac{dU}{dx}$

Integrating both sides for a displacement from\[x=a\text{ }to\text{ }x=b\], we have,

\[\Rightarrow {{U}_{b}}-{{U}_{a}}=-\int\limits_{b}^{a}{F(x)dx}\]

18. DYNAMICS OF CIRCULAR MOTION

18.1 Force on the Particle

In uniform circular motion, acceleration is of magnitude $\frac{{{v}^{2}}}{r}$ and is directed towards the center. Thus, a force of magnitude $\frac{m{{v}^{2}}}{r}$ and directed towards the center is needed to keep a particle in circular motion. 

This force (acting toward center) is called the centripetal force. Centripetal force is not an extra force on a body. Whatever force is responsible for circular motion becomes the centripetal force.

Examples  

When a satellite revolves around the earth, the gravitational attraction of earth becomes the centripetal force for the circular motion of that satellite; 

When an electron revolves around the nucleus in an atom, the electrostatic attraction of the nucleus becomes the centripetal force for the electron’s circular motion. 

In case of a conical pendulum, \[Tsin\theta \](component of tension) becomes the centripetal force.

18.2 Main steps for analyzing forces

Consider an axis along the radius of circle (i.e., in the direction of acceleration) and another axis perpendicular to the radius. Resolve all the forces into components.

Net force along perpendicular axis is equal to zero.

Net force along radial axis (towards center)$=\frac{m{{v}^{2}}}{r}=m{{\omega }^{2}}r$.

18.3 Main Steps for Analyzing Forces in Non–Uniform Circular Motion

Once we resolve all the forces along tangential and radial axes;

Net tangential force $={{F}_{t}}=m{{a}_{t}}$

Net radial force = $={{F}_{r}}=m{{a}_{r}}=\frac{m{{v}^{2}}}{r}$

Example of non-uniform circular motion  

The motion of particles in a vertical circle. If a particle is revolved in a vertical circle with the help of a string, the forces are: tension (T) towards center and weight (mg). 

In case of a particle moving along the outside surface of a circular track (or sphere), the forces are: normal reaction (N) away from the center and weight (mg).

18.4 Conical Pendulum

A small block of mass ‘m’ is rotated in a horizontal circle with the help of a string of length ‘l’ connected to ‘m’. The other end of the string is fixed to a point O vertically above the center of the circle so that the string is always inclined with the vertical at an angle. Such an arrangement is referred to as a conical pendulum as shown in the following diagram.

With respect to the force diagram of the block;

Along the vertical:

$T\cos \theta =mg$…(1)

Net force towards center,

\[Tsin~\theta =ma\]

$\Rightarrow \text{Tsin}\theta \text{=m}{{\omega }^{2}}r$…(2)

From (1) and (2), we have,

\[{{\omega }^{2}}=\frac{g\tan \theta }{r}=\frac{g\tan \theta }{l\sin \theta }=\frac{g}{l\cos \theta }\]

$\Rightarrow Time\text{ }period=T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{l\cos \theta }{g}}$

If ‘h’ is the height of point O above the center of the circle, then time period is equal to $2\pi \sqrt{\frac{h}{g}}$.

For a conical pendulum, 

${{\omega }^{2}}l\cos \theta =g$

$\Rightarrow \omega >\sqrt{\frac{g}{l}}$ (Because \[cos~\theta <l\])

18.5 Motion in a Vertical Circle

For a mass ‘m’ tied to a string of length ‘l’ and rotated in a vertical circle with center at the other end of the string, let is determine:

(a) the minimum velocity of the mass at the top of the circle so that it is able to complete the circle.

(b) the minimum velocity at the bottom of the circle.

At all positions, there are two forces acting on the mass: its own weight and the tension in the string.

Let the radius of the circle be equal to one unit.

(a) At the Top 

Let ${{v}_{t}}=$velocity at the top;

Net force towards center$=\frac{mv_{t}^{2}}{l}$

$T+mg=\frac{mv_{t}^{2}}{l}\Rightarrow T=\frac{mv_{t}^{2}}{l}-mg$

For the movement in the circle, the string must remain tight i.e., the tension should be positive at all positions.

As the tension is minimum at the top ${{\text{T}}_{top}}\ge 0$;

$\Rightarrow \frac{mv_{t}^{2}}{l}-mg\ge 0\Rightarrow {{v}_{t}}\ge \sqrt{{{l}_{g}}}$

$\Rightarrow $ minimum or critical velocity at the top = $=\sqrt{{{l}_{g}}}$

(b) At the Bottom

Let ${{v}_{b}}$ be the velocity at the bottom. As the particle goes up, its kinetic energy decreases and gravitational potential energy increases.

$\Rightarrow $loss in KE is equal to gain in GPE

$\Rightarrow \frac{1}{2}mv_{b}^{2}-\frac{1}{2}mv_{1}^{2}=mg(2l)$

$\Rightarrow v_{b}^{2}=v_{t}^{2}+4gl$

$\Rightarrow {{({{v}_{b}})}_{\min }}=\sqrt{{{(v_{t}^{2})}_{\min }}+4gl}=\sqrt{5gl}$

When a particle moves in a vertical circle, its speed reduces as it goes up and its speed rises as it comes down. Clearly, it is an example of non-uniform circular motion.

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CBSE Class 11 Physics Notes Chapter 6 Work, Power and Energy

In this context, we are going to discuss the list of content on what is work, power and energy. When we talk about Physics, the chapter of work, power, and energy is one of the most important chapters which involves concepts of mechanics.

The utmost collective illustrations that aid a concept of work, power and energy are a car in motion, bracing heavy objects, walking upstairs, an aeroplane flying and so on.

Some concepts regarding Chapter Work, Power, and Energy are precisely described in physics. So, these concepts can help you to do some measurements. Also, these perceptions can be utilized to describe and calculate the motion and its behaviour among multiple figures.

Chapter 6 Physics Class 11 Notes

What do you mean by Work?

Work can be defined as the amount of energy that transfers when a body is moved by an outside (external) force propagated in the displacement’s direction.

What do you mean by Power?

The definition of power can be explained as the rate at which the work is accomplished. Mathematically, Power = Work/time

What do you mean by Energy?

The definition of energy can be explained as the capability of doing work.

Energy has many forms. The most popular forms of energy are kinetic, thermal, potential, electrical, chemical, nuclear, etc.

The SI unit of work and energy is the same.

There is a list of physics of class 11 chapter 6 work, energy, and power.

Work Energy and Power Class 11 Notes (Table will be updated soon)

Notes of work energy and power class 11.

Here are some important questions under Class 11 Chapter 6 Physics Notes.

Q1. The aircraft casing burns up by friction. Find the energy obtained required for the burning of the casing.

Ans: The mass of the rocket reduces at the time of the burning of the casing due to the friction.

No reaction is beyond the law of conservation of energy. 

So, E Total = Potential energy + Kinetic energy = mgh + ½ mv 2

It is noticed that a drop in total energy happens due to the reduction in the mass of the aircraft. That is the reason for which energy is required for the burning of the casing which is obtained from the rocket.

Q2. If a ball of mass 5 kg is placed on a higher ground of 3 meters, find the potential energy stored in that body.

Ans: Given, m = 5 kg

We know that g = 9.81 m/s -2

So, Potential energy = m * g * h = 5 * 3 * 9.81 = 147.15 J

Work Done by a Variable Force and Conservative and Non-Conservative Forces

These two sections in Physics Class 11 Chapter 6 Notes explain about the work done by variable forces and details regarding conservative and non-conservative forces. 

Power And Energy

In this section of the Notes of Physics Class 11 Chapter 6 the concepts of power and energy along with their measurable units, expressions, etc.

Kinetic Energy, Relation Between Kinetic Energy and Linear Momentum

In this segment, students will get to know about different real-life applications of kinetic energy, expression to determine K.E of a body and its relation with linear momentum.

Work Energy Theorem or Work Energy Principle

As per the work energy principle, the amount of work done to move a body is equivalent to kinetic energy change. Precise analysis of the theory is laid out in this section.

Potential Energy

Another crucial topic is potential energy. Go through Physics Chapter 6 Class 11 Notes and be familiar with the different types of potential energy – gravitational and elastic potential energy.

Mechanical Energy and its Conservation

What is mechanical energy? It is nothing but the sum of the kinetic and potential energy of a body. Comprehensive elucidation of the same is provided in class 11 Chapter 6 Physics Notes to clear your concepts.

Different Energy Forms, Mass Energy Equivalence and Principle of Conservation of Energy

This section discusses different energy forms available like heat energy, internal energy, etc. furthermore, you will also get to know the mass energy equivalence expression stated by Einstein and conservation of energy principle.

Work Done by a Variable Force and Dynamics of Circular Motion

The last portion of the notes of Physics Class 11 Chapter 6 explain topics like work done by a force from point A to B, and forces related to uniform and non-uniform circular motion.

Proper understanding of the concepts is necessary to write precise answers and solve numerical problems. Hence, along with the text, refer to Class 11 Chapter 6 Physics notes offered by Vedantu.

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?

Why Choose Vedantu for CBSE Class 11 Physics Notes?

Vedantu provides a definitive study tool for students in the form of concepts of chapters along with the arranged question sets as per CBSE guidelines.

The chapters of Physics and its respective questions are based on the CBSE board syllabus. The notes also come with numerous concepts and practice papers.

Scholars can have the best education along with practice sets and mock tests for fluency in the specific chapter. They can learn time-management skills by practising Vedantu’s curated questions.

Key Features of Revision Notes for Class 11 Physics Chapter 6 - Work, Energy, and Power

Revision Notes are curated in order to help students in quickly finding important concepts from Work, Energy, and Power.

All concepts are explained in a detailed manner.

Revision Notes are clear and easy to understand as they are prepared by subject experts to match the syllabus.

These Revision Notes 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 Important Questions are available in PDF format and can be downloaded for free.

The CBSE Class 11 Notes on Work, Energy, and Power offer a comprehensive and concise resource for understanding the fundamental concepts that govern the dynamics of energy in the physical world. The chapter explores the concept of work done, the various forms of energy, and their interconversion, providing students with a solid foundation in this essential aspect of physics. With these free PDF notes at their disposal, students can reinforce their understanding, revise key points, and practice important equations. By mastering the principles of work, energy, and power, students can confidently approach complex physics problems and delve deeper into the fascinating world of energy transformations and their practical applications.

FAQs on Work, Energy and Power Class 11 Notes CBSE Physics Chapter 6 (Free PDF Download)

1. Is it Possible to Create Infinite Energy?

No, it is not possible. The universe does not have any infinities. In this universe, everything is finite in mass, size, time and energy. Infinity means something big which can’t be counted, which is impossible.

2. Calculate the Kinetic Energy Attained by a Football of Mass 0.46 kg Travelling at a Speed of 60 m/s.

We know that m = 0.46 kg

Velocity, v = 60 m/s

So, Kinetic energy = ½ mv 2 = ½ * 0.46 * 60 = 13.8 J

3. Mention Some Important Features of Vedantu for CBSE Board Exams.

Vedantu has taken part in so many educational activities. Vedantu has always focused on quality studies for Physics also.

The responsiveness, along with thorough question and answer practice papers, is very helpful to the students as per the CBSE board exam point of view.

4. What do you Mean by Internal Forces? Give Examples.

Internal forces are the types of forces that are present inside the body and act upon it internally. Examples of internal forces are the gravity forces, magnetic force, electrical force and spring force. We can’t see the gravitational force, but it acts upon every object that stays on earth. The concept is the same for all the examples given above.

5. Can you please provide a detailed stepwise study plan to ace Class 11 Physics, Chapter 6 - Work, Energy and Power?

The first step to ace Class 11 Physics, Chapter 6 - Work, Energy and Power is to thoroughly read the chapter from the NCERT textbook. Try to clear all doubts as soon as possible and aim for a crystal-clear understanding of the concepts rather than mugging up the NCERT text. Refer to Vedantu's Class 11 Physics Chapter 6 Revision Notes for this chapter to understand the chapter well. Practice all the NCERT questions and solve previous year questions from this chapter to perform well in the exam.

6. What are the best Revision Notes for NCERT Class 11 Physics, Chapter 6 - Work, Energy and Power?

The best revision notes for Class 11 Physics , Chapter 6 - “Work, Energy and Power” are Vedantu's Revision Notes. These are the best quality notes for this chapter as they are error-free, credible, and compiled by a team of expert Physics teachers based on the latest syllabus, pattern and marking scheme. These notes are easy to understand and very efficient for revising the maximum syllabus in less time. Revise this chapter from these notes to perform well in the Physics exam. You can access the study material on Vedantu’s App. All the resources are available free of cost.

7. What are the basics of Work, Energy and Power?

Class 11 Physics, Chapter 6 - “Work, Energy and Power” deals with the foundational Physics concepts. The chapter begins with the concept of work in Physics. Then, the chapter discusses energy. Under this, the concepts covered are kinetic energy and potential energy. The law of conservation of energy is discussed next. The other basic concepts discussed in the chapter are power, collision and its types. The application-based numerical problems on the concepts of work, energy and power, conclude the chapter. 

8. What are the real-time applications of Work, Energy and Power?

Work, Energy and Power have several real-time applications. All these three terms are interrelated. We perform a lot of work daily - pushing a car at rest horizontally, driving a truck uphill, a horse pulling the plough across the field etc. Similarly, some real-life instances which require the use of some form of energy are watching television, washing clothes in a washing machine and lighting the home with the help of electricity.

9. What is the law of conservation of energy in reference to Class 11 Chapter 6?

In physics, the law of conservation of energy states two things. First, energy is neither created nor destroyed. Second, energy can be converted from one form to another. Thus, the total energy of any object is never lost, and hence, the name of this law is the law of conservation of energy. Therefore, a system has a fixed amount of energy when it is in isolation i.e., no energy is added from an external source of energy.

CBSE Study Materials for Class 11