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NCERT Class 11 Physics Chapter 6 Work, Energy and Power Notes - Download PDF

NCERT Class 11 Physics Chapter 6 Work, Energy and Power Notes - Download PDF

Edited By Vishal kumar | Updated on Jun 30, 2025 08:46 PM IST

Ever wondered how things move, why they stop, or how much effort it takes to get them going? That is exactly what you will explore in NCERT Notes Class 11 Physics Chapter 6: Work, Energy and Power. This chapter is very important to understand how force leads to motion and how energy flows in different forms. It forms the foundation for many real world concepts and is vital for students preparing for CBSE board exams, JEE, and NEET. These NCERT Notes are prepared by our expert faculty based on the latest CBSE syllabus.

In these Work, Energy and Power Class 11 Notes PDF, you will find clear explanations of work done by a force, kinetic and potential energy, conservation of mechanical energy, and the concept of power. These NCERT Notes for class 11 also include important formulas, and diagrams. Whether you are preparing for exams or strengthening your basics, these NCERT notes will make your learning more structured and effective.

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NCERT Notes for Class 11 Chapter 5

The Scalar Product


There are two ways of multiplying vectors (i) scalar product (ii) vector product. We shall study the latter in next chapter. The scalar product or dot product of any two vectors A and B, denoted as AB (read as AdotB ) is defined as

Background wave

AB=ABcosθ

where A and B are magnitudes of vectors A¯ and B¯ respectively and θ is the smaller angle between them. Dot product is called the scalar product as A,B and cosθ are scalars. Both vectors have a direction but their scalar product does not have a direction.

Notions of Work and Kinetic Energy: The Work-Energy Theorem

This concept explains how work done on an object results in a change in its kinetic energy. When a force causes displacement, work is said to be done. The Work-Energy Theorem states that the net work done by all forces on an object is equal to the change in its kinetic energy.

Wnet=ΔKE=12mv212mu2

Work


If a force F is applied on a body in a direction different from the direction of displacement of the body and the body undergoes a displacement d in the positive X-direction as shown in figure 2.

The work done by the force is defined as the product of component of the force in the direction of the displacement and the magnitude of this displacement.

So, W=(Fcosθ)d=Fd

(i) Absolute units: Joule [S.I.] and Erg [C.G.S.]

(ii) Gravitational units: kg-m [S.I.] and gm-cm [C.G.S.]

Nature of work

  • Positive work :W=Fd=FdcosθSo, work done is positive when θ is acute (θ<90) as cosθ is positive.
  • Negative work :W=Fd=Fdcosθ.Work done is negative when θ is obtuse (θ>90)cosθ is negative.

  • Zero Work:
    W=Fd=Fdcosθ=0 when F=0 or d=0 or cosθ=0(θ=90)

Work Done Calculation by Force Displacement Graph

The work done by a force on an object can be calculated using the area under the force-displacement graph.

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dW=FdxW=xixfdW=xixfFdxW=xixf (strip area with width dx ) W= Area under curve Between xi and xf

Energy

The energy of a body is essentially its ability or capacity to get things done, or in other words, to do work.

  • The SI unit of energy is the same as the work is Joule (J).

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion.

 Kinetic Energy (KE)=12mv2

Where: KE is the kinetic energy, m is the mass of the object and v is its velocity.

Work Done by a Variable Force

If the applied force (F) varies along the path, the work required to move a body from position A to B can be calculated by integrating the product of the force and differential displacement.

W=ABFds=AB(Fcosθ)ds

Relation of kinetic energy with linear momentum

KE(E)=12mv2=12[pv]v2{ from p=mv}E=12pvE=p22m{ from v=p/m} And Momentum (P)=2mE

The Work-Energy Theorem for a Variable Force

Confining to one dimension, rate of change of kinetic energy with time is

dKdt=ddt(12mv2)=12mddt(v2)( as m is constant) =12m(2vdvdt)=mdvdtv=Fv( as F=ma=mdvdt (from Newton's II nd  law)) dKdt=Fdxdt or dK=FdxkikfdK=xixfFdx

[Integrating from the initial position (xi) to the final position (xf) ]
Where Ki and Kf are the initial and final kinetic energies corresponding to xi and xf respectively.
or

KfKi=xixfFdx


Thus, KfKi=W

Thus, work-energy theorem is proved for a variable force.

The Concept of Potential Energy

Potential energy is the energy that an object has because of its position or state.

Types of Potential Energy

Gravitational Potential Energy (GPE): It is the energy associated with an object's height in a gravitational field.

 Potential Energy (U)=mgh

Where, U is the potential energy, m is the mass of the object, g is the acceleration due to gravity and h is the height of the object above a reference point.

Elastic Potential Energy: For objects like springs or rubber bands, the potential energy is associated with how much the material is stretched or compressed.

When an elastic spring is compressed (or strained) by a distance x from its equilibrium state, its elastic potential energy is represented by:

U=12kx2

Where, k is the force constant of a given spring.

The Conservation of Mechanical Energy

The Law of Conservation of Energy asserts that energy is neither created nor destroyed; it merely changes from one form to another.

For simplicity, we are considering one-dimensional motion. Suppose that a body undergoes displacement Δx under the action of a conservative force F. From the W-E theorem,

ΔK=F(x)Δx
If the force is conservative, the potential energy function Ux can be defined such that

ΔU=F(x)Δx

Adding the above two equations, we get
or

ΔK+ΔU=0
Δ(K+U)=0

K+U, the sum of the kinetic and potential energies of the body is a constant.
Over the whole path xi to xf

Ki+Uxi=Kf+Uxi
The quantity K+Ux is the total mechanical energy of the system. Kinetic energy K and the potential energy Ux may charge individually from point to point, but the sum is a constant.

The Potential Energy of a Spring

When a spring is either compressed or stretched from its natural (equilibrium) position, it stores potential energy due to its elasticity. This energy is known as elastic potential energy.

According to Hooke's Law, the force required to stretch or compress a spring is:

F=kx

where:
F is the restoring force,
k is the spring constant,
x is the displacement from the equilibrium position.

The potential energy stored in the spring is given by the formula:

U=12kx2


This energy increases with greater displacement and depends on the stiffness of the spring (value of k ). The spring stores energy when deformed and releases it when it returns to its natural length.

Conservative and Non-Conservative Forces

The work done against which is stored in the system as its potential energy, which can be recovered later, is called a conservative force.

The work done against which is not stored in the system as its potential energy, which can be recovered later, is called non-conservative force".

Power

The power (P) of a body is defined as the rate at which the body can do work. Mathematically, power is expressed as the amount of work done (W) divided by the time (t) taken to do that work. The formula for power is:

Average Power(Pavg )=ΔwΔt=0tpdt0tdt
Instantaneous Power(P)=dwdt=P=Fv

  • Dimension of power: [ML2T-3]
  • Unit of power: Watt or Joule/sec [S.I.], Erg/sec [C.G.S.]
  • Practical unit: Kilowatt (kW), Megawatt (MW) and Horsepower (hp)

Collision

A collision occurs when two or more objects come into contact for a short period, during which they exert forces on each other. These forces can cause changes in the motion of the objects involved. Collisions are important because they help us understand how momentum and kinetic energy are transferred and conserved.

Types of collision

On the basis of conservation of kinetic energy, there are mainly three types of collision

  • Perfectly Elastic Collision: In a perfectly elastic collision, the system's kinetic energy is conserved, which means that the total kinetic energy before and after the collision is the same. There is no net loss or gain in kinetic energy, and the objects involved bounce off each other without deforming or losing energy to other forms.
  • Inelastic collision: In an inelastic collision, the system's kinetic energy is not conserved; that is, the kinetic energy after the collision differs from the kinetic energy before the collision. Some of the initial kinetic energy is converted into different forms, such as internal energy, heat, or deformation.
  • Perfectly inelastic collision: Inelastic collisions occur when the kinetic energy after the collision is less than the kinetic energy before the collision. In inelastic collisions, some of the initial kinetic energy is converted into other forms, while the total kinetic energy is not conserved.

Types of collision based on the direction of colliding bodies

  • Head on or one dimensional collision: when the motion of colliding particles before and after the collision occurs along the same line, it is referred to as a "head-on" or "one-dimensional" collision. In such collisions, the initial and final velocities of the particles are aligned along a straight line, simplifying the analysis of the collision dynamics.

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12m1u12+12m2u22=12m1v12+12m2v22(1)m1u1+m2u2=m1v1+m2v2(2)

m1,m2 : masses
u1,v1 : initial and final velocity of the mass m1
u2,v2 : initial and final velocity of the mass m2
From equation (1) and (2) We get,

u1u2=v2v1 (3) 

From equations (1),(2), (3) We get

v1=(m1m2m1+m2)u1+2m2u2m1+m2v2=(m2m1m1+m2)u2+2m1u1m1+m2

  • Perfectly Elastic Oblique Collision:

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Let two bodies move as shown in the figure. By the law of conservation of momentum,

Along x-axis-

m1u1+m2u2=m1v1cosθ+m2v2cosϕ(1)

Along y-axis-

0=m1v1sinθm2v2sinϕ(2)

By the law of conservation of kinetic energy

12m1u12+12m2u22=12m1v12+12m2v22(3)

So along the line of impact (here along in the direction of ) We apply e =1

e=1=v2v1cos(θ+ϕ)u1cosϕu2cosϕ(4)


So we solve these equations (1),(2),(3),(4) to get unknown.

  • Perfectly Inelastic Collision
    After a collision, two bodies stick together, resulting in a final common velocity.
  • When the colliding bodies are moving in the same direction
    m1u1+m2(u2)=(m1+m2)vv=m1u1m2u2m1+m2
  • When the colliding bodies are moving in the opposite direction
    m1u1+m2u2=(m1+m2)vv=m1u1+m2u2(m1+m2)

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Frequently Asked Questions (FAQs)

1. What is the difference between work and energy?

Work is the transfer of energy when a force is applied to an object, causing displacement. Energy, on the other hand, is the capacity to do work. The unit of work and energy is the joule (J).

2. What is the work-energy theorem?

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.

3. What is the significance of the Work, Energy, and Power chapter in Physics?

This chapter introduces fundamental concepts that form the basis for understanding various physical phenomena. It helps explain how energy is transferred, how work is done, and how power is calculated in mechanical systems, making it essential for both academic exams and real-life applications.

4. Why is power important in this chapter?

Power helps quantify how quickly work is done or energy is transferred. Understanding power is crucial in practical scenarios, such as determining the efficiency of machines and engines or calculating the speed at which energy is consumed.

5. How does this chapter help in competitive exams like JEE or NEET?

The concepts of work, energy, and power are foundational for various topics in competitive exams, especially mechanics and thermodynamics. Understanding these helps in solving a wide range of problems, improving both speed and accuracy in exams.

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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

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

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

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

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

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

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

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

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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