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

NCERT Class 11 Physics Chapter 5 Notes: Download PDF

These Work, Energy and Power Class 11 Notes provide a clear and structured explanation of the concepts of work, energy, and power, along with important formulas, derivations, and solved examples. The downloadable PDF will help students with quick revision for CBSE exams as well as competitive exams like JEE and NEET.

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Work, Energy and Power Class 11 Notes

The NCERT Notes for the chapter Work, Energy and Power provide easy-to-understand explanations and concise derivations, which help you prepare for exams. These notes are helpful in gaining better knowledge and mastering the concepts covered in the chapter.

The Scalar Product

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

$\overrightarrow{A}\cdot \overrightarrow{B}=AB\cos \theta$

where $A$ and $B$ are magnitudes of vectors $\overline{A}$ and $\stackrel{¯,}{B}$ respectively and $\theta$ is the smaller angle between them. The dot product is called the scalar product as $A,B$ and $\cos \theta$ 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.

$W_{\mathrm{net}}=\mathrm{\Delta }KE=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

Work

If a force $\overrightarrow{F}$ is applied on a body in a direction different from the direction of displacement of the body and the body undergoes a displacement $\overrightarrow{d}$ in the positive $X$-direction, as shown in Figure 2.

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

So, $W=(F\cos \theta )d=\overrightarrow{F}\cdot \overrightarrow{d}$

(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 :$W=\overrightarrow{F}\cdot \overrightarrow{d}=Fd\cos \theta$So, work done is positive when $\theta$ is acute $(\theta <{90}^{\circ })$ as $\cos \theta$ is positive.

• Negative work :$W=\overrightarrow{F}\cdot \overrightarrow{d}=Fd\cos \theta .$Work done is negative when $\theta$ is obtuse $(\theta >{90}^{\circ })\cos \theta$ is negative.

• Zero Work:
  $W=\overrightarrow{F}\cdot \overrightarrow{d}=Fd\cos \theta =0$ when $F=0$ or $d=0$ or $\cos \theta =0(\theta ={90}^{\circ })$

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.

$\begin{array}{r}dW=Fdx\\ \therefore W={\int }_{x_i}^{x_f}dW={\int }_{x_i}^{x_f}\overrightarrow{F}dx\\ \therefore W={\int }_{x_i}^{x_f}\text{(strip area with width}dx\text{)}\\ \therefore W=\text{Area under curve Between}{x}_i\text{and}{x}_f\end{array}$

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.

$\text{Kinetic Energy}(\mathrm{KE})=\frac{1}{2}m{v}^2$

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.

$W={\int }_A^B\overrightarrow{F}\cdot d\overrightarrow{s}={\int }_A^B(F\cos \theta )ds$

Relation of kinetic energy with linear momentum

$\begin{array}{r}\mathrm{KE}(\mathrm{E})=\frac{1}{2}m{v}^2=\frac{1}{2}\left[\frac{p}{v}\right]v^2\{\text{from}\mathrm{p}=\mathrm{mv}\}\\ \Rightarrow \mathrm{E}=\frac{1}{2}pv\\ \mathrm{E}=\frac{p^2}{2m}\{\text{from}\mathrm{v}=\mathrm{p}/\mathrm{m}\}\\ \text{And Momentum}(\mathrm{P})=\sqrt{2mE}\end{array}$

The Work-Energy Theorem for a Variable Force

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

$\begin{array}{rl}\frac{dK}{dt}& =\frac{d}{dt}\left(\frac{1}{2}m{v}^2\right)\\ =\frac{1}{2}m\frac{d}{dt}\left(v^2\right)(\text{as}m\text{is constant)}\\ =\frac{1}{2}m\left(2v\frac{dv}{dt}\right)\\ =m\frac{dv}{dt}v\\ =Fv(\text{as}F=ma=\frac{mdv}{dt}\text{(from Newton's II}{\text{nd}}^{}\text{law))}\\ \frac{dK}{dt}& =F\frac{dx}{dt}\text{or}dK=Fdx\\ {\int }_{k_i}^{k_f}dK& ={\int }_{x_i}^{x_f}Fdx\end{array}$

[Integrating from the initial position $\left(x_i\right)$ to the final position $\left(x_f\right)$ ]
Where $K_i$ and $K_f$ are the initial and final kinetic energies corresponding to $x_i$ and $x_{\mathrm{f,}}$ respectively.
or

$K_f-K_i={\int }_{x_i}^{x_f}Fdx$

Thus, ${\mathit{K}}_{\mathit{f}}-{\mathit{K}}_{\mathit{i}}=\mathit{W}$

Thus, the 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 = mgh

$$\text{<span style="display:none"> </span>}$$

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=\frac{1}{2} k x^2

$$$$

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 a displacement $\mathrm{\Delta }x$ under the action of a conservative force $F$. From the W-E theorem,

$\mathrm{\Delta }K=F(x)\mathrm{\Delta }x$
If the force is conservative, the potential energy function $U_x$ can be defined such that

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

Adding the above two equations, we get
or

$\mathrm{\Delta }K+\mathrm{\Delta }U=0$
$\mathrm{\Delta }(K+U)=0$

$\Rightarrow K+U$, the sum of the kinetic and potential energies of the body, is a constant.
Over the whole path $x_i$ to $x_f$

$K_i+U_{x_i}=K_f+U_{x_i}$
The quantity $K+U_x$ is the total mechanical energy of the system. Kinetic energy $K$ and the potential energy Ux may change 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:

$U=\frac{1}{2}k{x}^2$

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 $Power\left(P_{\text{avg}}\right)=\frac{\mathrm{\Delta }w}{\mathrm{\Delta }t}=\frac{{\int }_0^{t}p\cdot dt}{{\int }_0^{t}dt}$
Instantaneous $Power(P)=\frac{dw}{dt}=P=\overrightarrow{F}\cdot \overrightarrow{v}$

• 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 collisions

• 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.

$\begin{array}{r}\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2-(1)\\ m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2-(2)\end{array}$

$m_1,m_2$ : masses
$u_1,v_1$ : initial and final velocity of the mass $m_1$
$u_2,v_2$ : initial and final velocity of the mass $m_2$
From equations (1) and (2) we get,

$u_1-u_2=v_2-v_1\dots \dots \text{(3)}$

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

$\begin{array}{rl}{v}_1& =\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\frac{2m_2{u}_2}{m_1+m_2}\\ v_2& =\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2+\frac{2m_1{u}_1}{m_1+m_2}\end{array}$

Perfectly Elastic Oblique Collision:

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

Along x-axis-

$m_1{u}_1+m_2{u}_2=m_1{v}_1\cos \theta +m_2{v}_2\cos \varphi \dots -(1)$

Along y-axis-

$0=m_1{v}_1\sin \theta -m_2{v}_2\sin \varphi \dots -(2)$

By the law of conservation of kinetic energy

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\dots -(3)$

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

$e=1=\frac{v_2-v_1\cos (\theta +\varphi )}{u_1\cos \varphi -u_2\cos \varphi }\dots (4)$

So we solve these equations $(1),(2),(3),(4)$ to get the 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
$\begin{array}{r}{m}_1{u}_1+m_2(-u_2)=(m_1+m_2)v\\ v=\frac{m_1{u}_1-m_2{u}_2}{m_1+m_2}\end{array}$

When the colliding bodies are moving in the opposite direction
$\begin{array}{r}{m}_1{u}_1+m_2{u}_2=(m_1+m_2)v\\ v=\frac{m_1{u}_1+m_2{u}_2}{(m_1+m_2)}\end{array}$

Work, Energy and Power: Previous Year Questions

Two blocks $M_1$ and $M_2$ having equal mass are free to move on a horizontal frictionless surface. $M_2$ is attached to a massless spring as shown in Figure. Initially, $M_2$ is at rest and $M_1$ is moving toward $\mathrm{M}_2$ with speed v and collides head-on with $M_2$.

Then choose the correct option

(a) While the spring is fully compressed, all the KE $M_1$ is stored as PE of the spring.

(b) While spring is fully compressed, the system momentum is not conserved, though final momentum is equal to initial momentum.

(c) If spring is massless, the final state of the $M_1$ is a state of rest.

(d) None of these

Answer:

a) The kinetic energy of $M_1$ is not fully transferred to the spring as its potential energy, and hence, option a is incorrect

b) The law of conservation of mass is valid here since the surface is frictionless; hence, option b is incorrect.

c) if we consider the case where the spring is totally massless, then all the kinetic energy $M_1$ gets transferred to $M_2$. As a result, m1 comes to rest, $M_2$ acquires a velocity v and starts moving. Hence, option c is correct.

Hence, the answer is option (c).

Q2:

A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
(a) Work done by all forces on man is equal to the rise in potential energy, mgL.
(b) Work done by all forces on the man is zero.
(c) Work done by the gravitational force on man is mgL.
(d) The reaction force from a step does not work because the point of application of the force does not move while the force exists.

Answer:

The gravitational force acts in a downward direction. The displacement is labelled L, which is in the upward direction. So, the work done on a man due to the gravitational force amounts to -mgL. Also, the work done to lift the man amounts to the force in the direction of displacement. Hence, the net force amounts to -mgL+ mgL = 0. Hence, statement (b) is correct.

The displacement at the point where the force acts is zero. Hence, the amount of work done by the force is also zero. So, statement (d) is correct.

Hence, the answers are options (b) and (d).

Q3:

A body of mass 0.5 kg travels in a straight line with velocity $v=a x^{\frac{3}{2}}$ where $a=5 \mathrm{~m}^{\frac{-1}{2}} \mathrm{~s}^{-1}$. The work done by the net force during its displacement from $x=0$ to $x=2 m$ is

(a) 1.5J

(b) 50J

(c) 10J

(d) 100J

Answer:

$
\begin{aligned}
& \mathrm{m}=0.5 \mathrm{~kg} \\
& v=a x^{\frac{3}{2}} \\
& a=5 m^{\frac{-1}{2}} s^{-1} \\
& \text { now, acceleration }=a=\frac{d v}{d t}=v \frac{d v}{d x} \\
& =a x^{\frac{3}{2}} \frac{d}{d x} a x^{\frac{3}{2}}=a x^{\frac{3}{2}} \times \frac{3}{2} \times a x^{\frac{1}{2}}=\frac{3}{2} a^2 x^2 \\
& \text { Net force }=m a=m\left(\frac{3}{2} a^2 x^2\right)
\end{aligned}
$
Work done under the variable force.

$
\begin{aligned}
& =\int_{x=0}^{x=2} F . d x=\int_0^2 \frac{3}{2} m a^2 x^2 d x \\
& =\frac{1}{2} m a^2 \times 8=\frac{1}{2} \times(0.5) \times(25) \times 8=50 J
\end{aligned}
$
Hence, the answer is the option(b).

Importance of NCERT Class 11 Physics Chapter 5 Notes

• The Work, Energy and Power Class 11 Notes break down complicated learning points, such as the work done by a force, kinetic energy, potential energy, and power and enable students to find good basics to be used during exams

• A good revision of the major formulas, derivations, numbers, etc. that are recurrently tested in CBSE Class 11 exams, JEE and NEET is facilitated by exam-related notes.

• NCERT Class 11 Physics Chapter 5 Notes PDF comes in handy during last-minute revision and is the best guide with all the important definitions, laws, and problem-solving methods in a succinct form.

• Physics concepts such as work-energy theorem, conservation of energy, and collision are vital in JEE Mains, JEE Advanced, and NEET Physics, and this notes reduces the difficulty in solving problems.

• The notes are planned according to the chapters in a step-by-step manner, thus enabling students to save time and maintain focus when doing revisions.

• The notes include the solved examples and shortcuts, and help students to comprehend how to use the Physics concepts in reality and the numerical problems.

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