Work, Energy and Power Class 11: Notes, Collisions and Solved Numericals

What Work, Energy and Power is about

Weightage: This chapter is part of Unit IV, Work, Energy and Power. The syllabus groups Unit IV with Unit V, Motion of System of Particles and Rigid Body and Unit VI, Gravitation: together they carry 17 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit IV Work, Energy and Power

Key concepts

Work done by a constant force

NCERT section 5.1.1, 5.2 and 5.3

Work done by a constant force $\vec{F}$ during a displacement $\vec{d}$ is the scalar product:
$W=\vec{F}\cdot \vec{d}=Fd\cos \theta$
where θ is the angle between force and displacement. Work is a scalar; SI unit the joule (1 J = 1 N m). Its dimensions are $[M{L}^2{T}^{-2}]$.

• θ < 90°: positive work (gravity on a falling ball).
• θ = 90°: zero work (a porter carrying a load on level ground does no work against gravity on the load; centripetal force does no work).
• θ > 90°: negative work (friction on a sliding block, gravity on a rising ball).
• No displacement means no work, however hard you push a wall.

In components, $W=F_x{d}_x+F_y{d}_y+F_z{d}_z$.

Worked example

A trunk is pulled 10 m along a level floor by a rope with a tension of 50 N at 60° to the horizontal. Find the work done by the tension. [1 mark]

Try it first, then show the solutionSolution

1. $W=Fd\cos \theta =50\times 10\times \cos 60°$
2. $W=500\times 0.5=250J$

Answer: 250 J

Kinetic energy and the work-energy theorem

NCERT section 5.2, 5.4 and 5.6

Kinetic energy $K=\frac{1}{2}m{v}^2$ is the energy a body has because of its motion. It is a scalar and never negative. In terms of momentum, $K=\frac{p^2}{2m}$.

Work-energy theorem: the work done by the net force on a body equals the change in its kinetic energy:
$W_{net}=K_f-K_i$

For a constant force (proof): $v^2-u^2=2as$; multiply by $\frac{m}{2}$: $\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=mas=Fs$.

The theorem holds for variable forces too and is often the quickest way to find a stopping distance or an average resistive force, because it does not need the time.

Worked example

A 20 g bullet moving at 300 m s${}^{-1}$ enters a wooden block and stops after going 6 cm. Find the average resistive force of the wood. [2 marks]

Try it first, then show the solutionSolution

1. $K_i=\frac{1}{2}\times 0.02\times 300^2=900J$, $K_f=0$
2. Work-energy theorem: $-F\times 0.06=0-900\Rightarrow F=15000N$

Answer: $1.5\times 10^4$ N

Work done by a variable force

NCERT section 5.5

If the force changes with position, split the displacement into small steps Δx over which F is nearly constant and add up: $W=\sum F(x)\Delta x$. In the limit,
$W={\int }_{x_i}^{x_f}F(x)dx$

Graphically, work = area under the F-x graph (area below the x-axis counts as negative).

Example: a woman pushing a trunk with a force that falls as friction changes; a spring, where F = kx grows with stretch.

Worked example

A force $F=3x^2$ N acts along the x-axis. Find the work it does as a body moves from x = 1 m to x = 2 m. [2 marks]

Try it first, then show the solutionSolution

1. $W={\int }_1^{2}3x^{2}dx=[x^3{]}_1^2$
2. $W=8-1=7J$

Answer: 7 J

Potential energy; conservative and non-conservative forces

NCERT section 5.7

Potential energy is energy stored because of position or configuration. For a force F(x) we define the change in potential energy by $\Delta U=-W$, so $F(x)=-\frac{dU}{dx}$.

Gravitational PE near the Earth: $U=mgh$ (height measured from any chosen zero level; only differences matter).

A force is conservative if:
1. the work it does depends only on the end points, not the path;
2. the work round any closed path is zero;
3. it can be written as $F=-\frac{dU}{dx}$.

Gravity, the spring force and the electrostatic force are conservative. Friction and air resistance are non-conservative: their work depends on the path and turns mechanical energy into heat. Potential energy can be defined only for conservative forces.

Worked example

The potential energy of a particle is $U=4x^2-2x$ joules. Find the force on it at x = 1 m. [1 mark]

Try it first, then show the solutionSolution

1. $F=-\frac{dU}{dx}=-(8x-2)$
2. At x = 1 m: F = −6 N (towards negative x)

Answer: −6 N

Conservation of mechanical energy

NCERT section 5.8

If only conservative forces do work, the total mechanical energy stays constant:
$K+U=\text{constant}$, i.e. $K_i+U_i=K_f+U_f$

Proof idea: for a conservative force, $\Delta K=W=-\Delta U$, so $\Delta (K+U)=0$.

For a body falling freely from height H: at the top E = mgH (all potential); at height h, $\frac{1}{2}m{v}^2+mgh=mgH$; at the ground E = $\frac{1}{2}m{v}^2$ (all kinetic). So $v=\sqrt{2g(H-h)}$.

If non-conservative forces such as friction act, $W_{nc}=\Delta K+\Delta U$, and the 'lost' mechanical energy appears as heat.

Worked example

A ball is dropped from a height of 20 m. Find its speed when it is 5 m above the ground. Ignore air resistance; g = 10 m s${}^{-2}$. [2 marks]

Try it first, then show the solutionSolution

1. Loss in PE = gain in KE: $mg(20-5)=\frac{1}{2}m{v}^2$
2. $v=\sqrt{2\times 10\times 15}=\sqrt{300}\approx 17.3m{s}^{-1}$

Answer: About 17.3 m s${}^{-1}$

Potential energy of a spring

NCERT section 5.9

An ideal spring obeys Hooke's law: the restoring force is $F_s=-kx$, where x is the extension or compression and k is the spring constant (N m${}^{-1}$); a stiff spring has a large k.

Work done in stretching it from 0 to x (by the pulling force):
$W={\int }_0^{x}kxdx=\frac{1}{2}k{x}^2$
This is stored as elastic potential energy $U=\frac{1}{2}k{x}^2$ (area of the triangle under the F-x line). The spring force is conservative.

A block of mass m moving at v that compresses a spring on a smooth floor stops when $\frac{1}{2}m{v}^2=\frac{1}{2}k{x}_m^2$, so $x_m=v\sqrt{\frac{m}{k}}$.

Worked example

A 1 kg block moving at 2 m s${}^{-1}$ on a smooth floor hits a spring of constant 100 N m${}^{-1}$. Find the maximum compression. [2 marks]

Try it first, then show the solutionSolution

1. $\frac{1}{2}\times 1\times 2^2=\frac{1}{2}\times 100\times x_m^2$
2. $x_m^2=0.04\Rightarrow x_m=0.2m$

Answer: 0.2 m (20 cm)

Motion in a vertical circle

NCERT section 5.8 (Example 5.7)

A bob of mass m on a light string of length L is whirled in a vertical circle. Its speed is not constant: it is fastest at the bottom and slowest at the top. Tension does no work (it is perpendicular to the motion), so mechanical energy is conserved.

• At the top: $T_t+mg=\frac{m{v}_t^2}{L}$. The string stays taut only if $T_t\ge 0$, so $v_t\ge \sqrt{gL}$.
• Energy from bottom to top: $\frac{1}{2}m{v}_b^2=\frac{1}{2}m{v}_t^2+2mgL$, so the minimum speed at the bottom is $v_b=\sqrt{5gL}$.
• At the bottom: $T_b-mg=\frac{m{v}_b^2}{L}$; with the minimum speeds, $T_b-T_t=6mg$.

If the speed at the bottom is less than $\sqrt{5gL}$, the string goes slack before the top and the bob leaves the circle.

Worked example

A small ball is whirled in a vertical circle of radius 1 m on a string. Find the least speed it must have at the bottom and at the top to complete the circle. (g = 10 m s${}^{-2}$) [2 marks]

Try it first, then show the solutionSolution

1. Top: $v_t=\sqrt{gL}=\sqrt{10}\approx 3.16m{s}^{-1}$
2. Bottom: $v_b=\sqrt{5gL}=\sqrt{50}\approx 7.07m{s}^{-1}$

Answer: About 7.07 m s${}^{-1}$ at the bottom and 3.16 m s${}^{-1}$ at the top

Power

NCERT section 5.10

Power is the rate of doing work (or of transferring energy).

• Average power $P_{av}=\frac{W}{t}$; instantaneous power $P=\frac{dW}{dt}$.
• Since $dW=\vec{F}\cdot d\vec{r}$, $P=\vec{F}\cdot \vec{v}$.

SI unit: the watt (1 W = 1 J s${}^{-1}$). 1 horsepower (hp) = 746 W. The kilowatt hour is a unit of energy, not power: 1 kWh = $3.6\times 10^6$ J (the 'unit' on an electricity bill).

For a lift or a vehicle moving at constant speed, the engine must supply power $P=Fv$ where F balances weight, friction and other resistances.

Worked example

A lift of total mass 1200 kg moves up at a steady 3 m s${}^{-1}$ against a frictional force of 3000 N. Find the power the motor delivers, in watts and horsepower. (g = 10 m s${}^{-2}$) [2 marks]

Try it first, then show the solutionSolution

1. Force needed $=mg+f=12000+3000=15000N$
2. $P=Fv=15000\times 3=45000W=\frac{45000}{746}\approx 60hp$

Answer: 45 kW, about 60 hp

Collisions: elastic and inelastic, in one and two dimensions

NCERT section 5.11

In every collision of an isolated system, total linear momentum is conserved. Total energy is also conserved, but kinetic energy may change into heat, sound or deformation.

• Elastic collision: kinetic energy is also conserved.
• Inelastic collision: some kinetic energy is lost. Perfectly (completely) inelastic: the bodies stick together.

Perfectly inelastic, 1-D ($m_2$ at rest): $v=\frac{m_1{u}_1}{m_1+m_2}$; KE lost $=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{u}_1^2$.

Elastic, 1-D ($m_2$ at rest), from momentum and KE conservation:
$v_1=\frac{(m_1-m_2)}{m_1+m_2}{u}_1$, $v_2=\frac{2m_1}{m_1+m_2}{u}_1$

• Equal masses: they exchange velocities (the moving one stops).
• Heavy hitting light: light one moves off at nearly $2u_1$. Light hitting heavy: light one bounces back.

Two dimensions: conserve the x and y components of momentum separately; if elastic, also KE. For an elastic glancing collision of equal masses with one at rest, the two move off at 90° to each other.

Worked example

A 2 kg ball moving at 3 m s${}^{-1}$ makes a head-on elastic collision with a 1 kg ball at rest. Find their velocities after the collision. [2 marks]

Try it first, then show the solutionSolution

1. $v_1=\frac{2-1}{3}\times 3=1m{s}^{-1}$; $v_2=\frac{2\times 2}{3}\times 3=4m{s}^{-1}$
2. Check KE: before $\frac{1}{2}(2)(9)=9J$; after $\frac{1}{2}(2)(1)+\frac{1}{2}(1)(16)=1+8=9J$ ✓

Answer: 1 m s${}^{-1}$ and 4 m s${}^{-1}$, both forwards

Physics in real life: Work, Energy and Power

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Elastic and Inelastic Collisions in One Dimension
• Work-Energy Theorem, Variable Force and Power

Work, Energy and Power formula sheet

Work by a constant force

$W=\vec{F}\cdot \vec{d}=Fd\cos \theta$

Work by a variable force

$W={\int }_{x_i}^{x_f}F(x)dx$

Kinetic energy

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

Work-energy theorem

$W_{net}=K_f-K_i$

Force from potential energy

$F=-\frac{dU}{dx}$

Potential energies

$U_g=mgh,U_s=\frac{1}{2}k{x}^2$

Vertical circle (minimum speeds)

$v_{top}=\sqrt{gL},v_{bottom}=\sqrt{5gL},T_b-T_t=6mg$

Power

$P=\frac{dW}{dt}=\vec{F}\cdot \vec{v},1hp=746W,1kWh=3.6\times 10^{6}J$

Elastic 1-D collision (m₂ at rest)

$v_1=\frac{m_1-m_2}{m_1+m_2}{u}_1,v_2=\frac{2m_1}{m_1+m_2}{u}_1$

Perfectly inelastic 1-D collision (m₂ at rest)

$v=\frac{m_1{u}_1}{m_1+m_2},\Delta K=\frac{m_1{m}_2{u}_1^2}{2(m_1+m_2)}$

How the school exam asks Work, Energy and Power

The question patterns to expect, and the method that works for each.

Tricks to remember Work, Energy and Power

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Work needs F, d and cos

  No displacement, no work. Perpendicular, no work. Opposite, negative work.

  Use it when: Sign-of-work questions.

• Net work = change in KE

  Add up every force's work with its sign; the total is $\Delta K$. It does not care about time.

  Use it when: Stopping distances and resistive forces.

• Vertical circle: 1, 5, 6

  $\sqrt{1gL}$ at the top, $\sqrt{5gL}$ at the bottom, tension difference 6mg.

  Use it when: Looping questions.

• Equal masses swap

  Elastic head-on collision of equal masses: they exchange velocities. Glancing: they fly apart at 90°.

  Use it when: Collision MCQs.

• kWh is energy, kW is power

  kilo-watt × hour = 1000 × 3600 J = 3.6 MJ.

  Use it when: Electricity bill and unit conversion items.

Work, Energy and Power: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Work, Energy and Power

The slips examiners see most often, and what to write instead.

• ✗Mistake: Forgetting that friction's work is negative

  Adding all works as positive.

  ✓Do this: Friction acts opposite to displacement: $W_f=-fd$. Then $W_{applied}+W_f=\Delta K$.

  Usually costs: 1-2 marks.

• ✗Mistake: Assuming KE is conserved in every collision

  Mixing up the two conservation laws.

  Wrong working

  Bullet embeds in a block: $\frac{1}{2}m{u}^2=\frac{1}{2}(m+M)v^2$

  Right working

  $mu=(m+M)v$ (momentum only)

  ✓Do this: Momentum is always conserved in a collision; KE only in elastic collisions.

  Usually costs: Whole numerical.

• ✗Mistake: Using $\sqrt{gL}$ as the minimum speed at the bottom

  Mixing the top and bottom conditions.

  ✓Do this: $\sqrt{gL}$ is at the top; energy conservation gives $\sqrt{5gL}$ at the bottom.

  Usually costs: 1-2 marks.

• ✗Mistake: Treating kWh as a unit of power

  Seeing 'watt' in it.

  ✓Do this: kWh = power × time = energy; 1 kWh = $3.6\times 10^6$ J.

  Usually costs: 1 mark.

• ✗Mistake: Missing the minus sign in $F=-dU/dx$

  Differentiating without thinking about direction.

  ✓Do this: The force points towards decreasing potential energy.

  Usually costs: 1 mark.

NCERT exercise guide: Work, Energy and Power

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Work, Energy and Power

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   One kilowatt hour is equal to:
   (a) $3.6\times 10^3$ J
   (b) $3.6\times 10^5$ J
   (c) $3.6\times 10^6$ J
   (d) $1000$ J

   Hint

   1000 W × 3600 s.

   Answer

   (c) $3.6\times 10^6$ J

2. Q2 · 1 mark · MCQExam level

   The extension of a spring is doubled. Its potential energy becomes:
   (a) double
   (b) half
   (c) four times
   (d) unchanged

   Hint

   $U\propto x^2$.

   Answer

   (c) four times

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): The work done by friction on a body sliding over a fixed rough surface is always negative.
   Reason (R): Kinetic friction on the body acts opposite to its displacement relative to the surface.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   Fixed surface: relative displacement = actual displacement.

   Answer

   (a)

4. Q4 · 1 mark · Very short answerExam level

   A satellite moves in a circular orbit round the Earth. How much work does gravity do on it in one orbit? Explain.

   Hint

   Angle between force and velocity.

   Answer

   Zero: the gravitational force is always along the radius, perpendicular to the velocity, so no work is done (and the speed stays constant).

5. Q5 · 2 marks · Very short answerHarder exam level

   A force acts on a body along x. It is 10 N from x = 0 to x = 2 m, and then falls uniformly to zero at x = 4 m. Find the total work done.

   Hint

   Rectangle + triangle.

   Answer

   $W=10\times 2+\frac{1}{2}\times 2\times 10=20+10=30$ J.

6. Q6 · 2 marks · Very short answerHarder exam level

   A 1000 kg car accelerates uniformly from rest to 20 m s${}^{-1}$ in 10 s. Find the work done by the engine (ignore friction) and the average power.

   Hint

   Work-energy theorem.

   Answer

   $W=\frac{1}{2}\times 1000\times 400=2\times 10^5$ J; average power $=\frac{2\times 10^5}{10}=2\times 10^4$ W = 20 kW.

7. Q7 · 2 marks · Very short answerHarder exam level

   A 0.5 kg ball moving at 6 m s${}^{-1}$ makes a head-on elastic collision with a 1.5 kg ball at rest. Find their velocities after the collision.

   Hint

   Use the 1-D elastic formulae.

   Answer

   $v_1=\frac{0.5-1.5}{2}\times 6=-3$ m s${}^{-1}$ (bounces back); $v_2=\frac{2\times 0.5}{2}\times 6=3$ m s${}^{-1}$ forwards.

8. Q8 · 3 marks · Short answerHarder exam level

   The potential energy of a particle is $U=\frac{1}{2}k{x}^2$ with k = 200 N m${}^{-1}$, and its total mechanical energy is 4 J. Find the range of x over which it moves, and its speed at x = 0 if its mass is 0.5 kg.

   Hint

   K ≥ 0 so U ≤ E.

   Answer

   $\frac{1}{2}\times 200x^2\le 4\Rightarrow x^2\le 0.04$, so −0.2 m ≤ x ≤ 0.2 m. At x = 0, K = 4 J: $\frac{1}{2}(0.5)v^2=4\Rightarrow v=4$ m s${}^{-1}$.

9. Q9 · 2 marks · Very short answerHarder exam level

   A body of mass 2 kg is lifted through 5 m at constant speed and then lowered back to its starting point at constant speed. Find the work done by gravity in each part and in the round trip. What does this show about gravity?

   Hint

   g = 10 m s${}^{-2}$.

   Answer

   Up: −100 J; down: +100 J; round trip: 0. Zero work round a closed path shows gravity is a conservative force.

10. Q10 · 3 marks · Short answerHOTS / competency

    A car of mass m is driven from rest by an engine that delivers constant power P. Ignoring friction, show that its speed grows as $\sqrt{t}$ and its distance as $t^{3/2}$.

    Hint

    P = Fv = mv dv/dt.

    Answer

    $P=mv\frac{dv}{dt}\Rightarrow {\int }_0^{v}vdv=\frac{P}{m}{\int }_0^{t}dt\Rightarrow v=\sqrt{\frac{2Pt}{m}}\propto t^{1/2}$. Then $x={\int }_0^t\sqrt{\frac{2P}{m}}{t}^{1/2}dt=\frac{2}{3}\sqrt{\frac{2P}{m}}{t}^{3/2}\propto t^{3/2}$.

11. Q11 · 2 marks · Very short answerHOTS / competency

    In a perfectly inelastic collision, a moving body of mass m hits an identical body at rest. What fraction of the original kinetic energy is lost? What if the target has mass 4m?

    Hint

    Fraction lost = $\frac{m_2}{m_1+m_2}$.

    Answer

    Equal masses: v = u/2, KE after = $\frac{1}{2}(2m)(u/2{)}^2=\frac{1}{4}m{u}^2$, half of $\frac{1}{2}m{u}^2$: half is lost. Target 4m: fraction lost $=\frac{4m}{5m}=80\%$.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.When is work done by a force zero?Show answer

   When the displacement is zero or perpendicular to the force.

2. 2.State the work-energy theorem.Show answer

   The work done by the net force on a body equals the change in its kinetic energy.

3. 3.Give two properties of a conservative force.Show answer

   Its work is independent of path and is zero round a closed path.

4. 4.Write the potential energy of a spring.Show answer

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

5. 5.What is the minimum speed at the top of a vertical circle of radius L?Show answer

   $\sqrt{gL}$.

6. 6.Define power and give its SI unit.Show answer

   Rate of doing work; watt (J s${}^{-1}$).

7. 7.In which collisions is kinetic energy conserved?Show answer

   Elastic collisions.

8. 8.What happens when two equal masses collide head-on elastically, one at rest?Show answer

   They exchange velocities: the moving one stops and the other moves off with its speed.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you give an example each of positive, negative and zero work?
• ☐Can you prove the work-energy theorem for a constant force?
• ☐Can you find work from an F-x graph?
• ☐Can you find the force from a given U(x)?
• ☐Can you show that the spring potential energy is $\frac{1}{2}k{x}^2$?
• ☐Can you derive the minimum speeds at the top and bottom of a vertical circle?
• ☐Can you find the power of a pump or a lift?
• ☐Can you convert kWh to joules?
• ☐Can you derive $v_1$ and $v_2$ for a 1-D elastic collision?
• ☐Can you find the KE lost in a perfectly inelastic collision?

Work, Energy and Power: frequently asked questions