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

NCERT-aligned study guide for CBSE Class 11 Physics Chapter 6 — Work, Energy and Power. Covers work-energy theorem, PE, KE, conservation laws, collisions, and coefficient of restitution with formulas, tables, and worked numericals.

Updated 11 June 202614 min read

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Chapter Introduction

Work, Energy and Power is one of the most scoring chapters in CBSE Class 11 Physics (8–10 marks). It connects force to motion through the work-energy theorem, introduces the powerful principle of conservation of mechanical energy, and covers collisions — a topic that appears in every board and competitive exam. The concepts here are used throughout Class 12 (electrostatics, current electricity) and JEE/NEET. Master the work-energy theorem and collision formulas, and you can solve many problems faster than with Newton's laws alone.

What this guide covers: work done (by constant and variable force), kinetic energy, work-energy theorem, potential energy (gravitational and elastic), conservative and non-conservative forces, conservation of mechanical energy, power, collisions (elastic, inelastic, coefficient of restitution) — all with worked numericals.

Key Definitions

  • Work (W): W = F · d = Fd cos θ. Scalar; SI unit joule (J). Positive if force has a component along displacement (0° ≤ θ < 90°), negative if against displacement (90° < θ ≤ 180°), zero if perpendicular.
  • Kinetic energy (KE): Energy possessed by a body by virtue of its motion. KE = ½mv<sup>2</sup>. Scalar; SI unit joule. Always positive or zero (never negative).
  • Potential energy (PE): Energy possessed by a body by virtue of its position or configuration. Scalar; SI unit joule.
  • Gravitational PE: PE = mgh (near Earth's surface). Reference level can be chosen arbitrarily; only changes in PE matter.
  • Elastic PE (spring): PE = ½kx<sup>2</sup>, where k = spring constant, x = extension/compression.
  • Conservative force: A force for which work done depends only on the end points, not the path. Work done in a closed loop is zero. Examples: gravity, spring force, electrostatic force.
  • Non-conservative force: A force for which work done depends on the path. Work done in a closed loop is non-zero. Examples: friction, viscous drag, air resistance.
  • Mechanical energy: KE + PE. The sum of kinetic and potential energies.
  • Power (P): Rate of doing work. P = W/t (average) or P = F · v (instantaneous). SI unit watt (W = J/s).
  • Collision: An event in which two or more bodies exert strong forces on each other for a short time. Total momentum is always conserved.
  • Coefficient of restitution (e): e = (v<sub>2</sub> − v<sub>1</sub>) / (u<sub>1</sub> − u<sub>2</sub>) = relative velocity of separation / relative velocity of approach.

Detailed Concepts

1. Work Done

General formula: W = Fd cos θ

Angle θcos θWorkExample
0°1Positive (max)Force along displacement (pushing a box)
0° to 90°0 to 1PositivePulling a sled at an angle
90°0ZeroCentripetal force in circular motion
90° to 180°−1 to 0NegativeFriction opposing motion
180°−1Negative (max)Force opposite to displacement (braking)

Work done by variable force: W = ∫F dx (area under the force-displacement graph).

Work done by spring: W = ∫(−kx)dx from 0 to x = −½kx<sup>2</sup> (work done by the spring force when stretched by x; negative because force opposes extension).

Work done by gravity: W = mgh when a body falls through height h (positive, gravity does work). W = −mgh when raised (negative, work done against gravity).

Work done is zero when:

  • Force is perpendicular to displacement (centripetal force, carrying a load horizontally)
  • Displacement is zero (pushing a wall, no motion)
  • Force is zero

2. Kinetic Energy and Work-Energy Theorem

KE = ½mv<sup>2</sup>

Work-energy theorem: The net work done on a body equals the change in its kinetic energy.

W<sub>net</sub> = ΔKE = ½mv<sup>2</sup> − ½mu<sup>2</sup>

Derivation (from Newton's second law): F = ma = m(dv/dt) Multiply both sides by dx: F dx = m(dv/dt)dx = mv dv Integrate: ∫F dx = ∫mv dv = m[v<sup>2</sup>/2] = ½mv<sup>2</sup> − ½mu<sup>2</sup> Therefore, W<sub>net</sub> = ΔKE

Steps to apply the work-energy theorem: 1. Identify all forces doing work on the body. 2. Calculate work done by each force. 3. Sum all work = change in KE. 4. Solve for the unknown.

Advantage over F = ma:

  • The work-energy theorem is a scalar equation — no vector components to resolve.
  • It directly relates force, distance, and speed without needing acceleration.
  • It works even for variable forces (just integrate F dx).
  • It gives a direct relation between the initial and final states.

3. Potential Energy and Conservative Forces

Conservative vs. Non-conservative Forces:

PropertyConservative ForceNon-conservative Force
Work depends on path?No (path-independent)Yes (path-dependent)
Work in closed loopZeroNon-zero
PE can be defined?YesNo
ExamplesGravity, spring force, electrostaticFriction, air resistance, viscous drag
Mechanical energy conserved?Yes (if only conservative forces)No (some KE converted to heat)
Force from PEF = −dU/dxNot possible

Gravitational PE: PE = mgh (near Earth's surface, h measured from chosen reference level).

Spring PE: PE = ½kx<sup>2</sup> (always positive, whether stretched or compressed).

PE and force relation: F = −dU/dx (force is the negative gradient of potential energy).

  • Gravity: U = mgh → F = −dU/dh = −mg (downward, as expected)
  • Spring: U = ½kx<sup>2</sup> → F = −dU/dx = −kx (restoring force, as expected)

PE curve and equilibrium:

  • At a minimum of the PE curve: stable equilibrium (F = 0, small displacement → restoring force)
  • At a maximum of the PE curve: unstable equilibrium (F = 0, small displacement → force away)
  • At a flat region: neutral equilibrium

4. Conservation of Mechanical Energy

Principle: If only conservative forces act on a system, the total mechanical energy (KE + PE) is conserved.

KE<sub>1</sub> + PE<sub>1</sub> = KE<sub>2</sub> + PE<sub>2</sub>

When non-conservative forces are present: KE<sub>1</sub> + PE<sub>1</sub> + W<sub>nc</sub> = KE<sub>2</sub> + PE<sub>2</sub>

where W<sub>nc</sub> is the work done by non-conservative forces (usually negative, since friction removes energy).

Applications:

  • A body dropped from height h: ½mv<sup>2</sup> = mgh → v = √(2gh)
  • A body sliding down a smooth incline: ½mv<sup>2</sup> = mgH (H = vertical height of incline)
  • A spring compressed by x and released: ½kx<sup>2</sup> = ½mv<sup>2</sup> → v = x√(k/m)
  • A pendulum at different heights: ½mv<sup>2</sup> + mgh = constant

Steps to solve problems using energy conservation: 1. Identify initial and final positions. 2. Write KE + PE at both positions. 3. If only conservative forces, set them equal. If friction, include W<sub>nc</sub>. 4. Solve for the unknown.

5. Power

P = W/t (average power over time t)

P = F · v = Fv cos θ (instantaneous power)

QuantityFormulaUnit
Average powerW/twatt (W)
Instantaneous powerF · v = dW/dtwatt (W)
Kilowatt-hour1 kWh = 3.6 × 10<sup>6</sup> Jcommercial unit of energy

Note: 1 kWh is a unit of energy (not power). It is the energy consumed by a 1 kW device in 1 hour.

Instantaneous power from force: If a force F pushes an object at velocity v, the power delivered is P = Fv. Example: A car engine providing force F at speed v delivers power P = Fv.

6. Collisions

One-dimensional (head-on) collisions:

PropertyElastic CollisionInelastic CollisionCompletely Inelastic
Momentum conserved?YesYesYes
KE conserved?YesNo (some lost)No (maximum loss)
Coefficient of restitutione = 10 < e < 1e = 0
Bodies separate?YesYesNo (stick together)
ExampleBilliard ballsCar crashLump of clay hitting wall

Elastic collision formulas (one dimension, two bodies):

  • v<sub>1</sub> = ((m<sub>1</sub> − m<sub>2</sub>)u<sub>1</sub> + 2m<sub>2</sub>u<sub>2</sub>) / (m<sub>1</sub> + m<sub>2</sub>)
  • v<sub>2</sub> = ((m<sub>2</sub> − m<sub>1</sub>)u<sub>2</sub> + 2m<sub>1</sub>u<sub>1</sub>) / (m<sub>1</sub> + m<sub>2</sub>)

Special cases:

  • Equal masses (m<sub>1</sub> = m<sub>2</sub>): v<sub>1</sub> = u<sub>2</sub>, v<sub>2</sub> = u<sub>1</sub> (velocities are exchanged)
  • m<sub>2</sub> at rest, m<sub>1</sub> = m<sub>2</sub>: v<sub>1</sub> = 0, v<sub>2</sub> = u<sub>1</sub> (first body stops, second moves with first's velocity)
  • m<sub>2</sub> >> m<sub>1</sub> (wall): v<sub>1</sub> ≈ −u<sub>1</sub> (first body rebounds with same speed)
  • m<sub>1</sub> >> m<sub>2</sub>: v<sub>1</sub> ≈ u<sub>1</sub> (first body barely affected), v<sub>2</sub> ≈ 2u<sub>1</sub>

Coefficient of restitution: e = (v<sub>2</sub> − v<sub>1</sub>) / (u<sub>1</sub> − u<sub>2</sub>)

  • e = 1: perfectly elastic; e = 0: perfectly inelastic
  • For a ball dropped from height h and bouncing to height h', e = √(h'/h)

Completely inelastic collision (bodies stick together): v = (m<sub>1</sub>u<sub>1</sub> + m<sub>2</sub>u<sub>2</sub>) / (m<sub>1</sub> + m<sub>2</sub>)

KE lost in completely inelastic collision: ΔKE = ½(m<sub>1</sub>m<sub>2</sub>)/(m<sub>1</sub> + m<sub>2</sub>) × (u<sub>1</sub> − u<sub>2</sub>)<sup>2</sup>

Collisions in two dimensions: Momentum is conserved as a vector (x and y components separately). For elastic collisions in 2D, KE is conserved as a scalar. The collision is usually resolved along the line of impact and perpendicular to it.

Important Numericals

Numerical 1: Work Done by a Force at an Angle

A force F = 3î + 4ĵ N acts on a body which moves from (0, 0) to (2, 3) m. Find the work done.

SolutionW = F · d = (3 × 2) + (4 × 3) = 6 + 12 = 18 J

Numerical 2: Work-Energy Theorem

A body of mass 2 kg moving at 10 m s<sup>-1</sup> is brought to rest by friction over 25 m. Find the friction force.

SolutionW<sub>net</sub> = ΔKE = 0 − ½ × 2 × 100 = −100 J W = −f × 25 = −100 → f = 100/25 = 4 N

Numerical 3: Conservation of Energy

A body is dropped from a height of 20 m. Find its speed when it is 5 m above the ground. (g = 10 m s<sup>-2</sup>)

SolutionUsing conservation of energy: mg(20) = mg(5) + ½mv<sup>2</sup> 200 = 50 + ½v<sup>2</sup> (dividing by m and using g = 10) ½v<sup>2</sup> = 150 → v<sup>2</sup> = 300 → v = √300 ≈ 17.3 m s<sup>-1</sup>

Numerical 4: Elastic Collision

Two balls of masses 2 kg and 3 kg move at 10 m s<sup>-1</sup> and 5 m s<sup>-1</sup> respectively in the same direction. They collide elastically. Find their velocities after collision.

Solutionv<sub>1</sub> = ((2 − 3) × 10 + 2 × 3 × 5) / (2 + 3) = (−10 + 30) / 5 = 4 m s<sup>-1</sup> v<sub>2</sub> = ((3 − 2) × 5 + 2 × 2 × 10) / (2 + 3) = (5 + 40) / 5 = 9 m s<sup>-1</sup> Verification: Initial KE = ½ × 2 × 100 + ½ × 3 × 25 = 100 + 37.5 = 137.5 J Final KE = ½ × 2 × 16 + ½ × 3 × 81 = 16 + 121.5 = 137.5 J ✓

Numerical 5: Power

An engine pumps water at 10 kg per second from a well 20 m deep. Find the power of the engine. (g = 10 m s<sup>-2</sup>)

SolutionP = mgh/t = 10 × 10 × 20 / 1 = 2000 W = 2 kW

Numerical 6: Completely Inelastic Collision

A 5 kg body moving at 8 m s<sup>-1</sup> collides with a 3 kg body at rest and they stick together. Find the final velocity and KE lost.

Solutionv = (5 × 8 + 3 × 0) / (5 + 3) = 40/8 = 5 m s<sup>-1</sup> Initial KE = ½ × 5 × 64 = 160 J Final KE = ½ × 8 × 25 = 100 J KE lost = 60 J Using formula: ΔKE = ½(5 × 3)/(5 + 3) × (8)<sup>2</sup> = ½ × 15/8 × 64 = 60 J ✓

Numerical 7: Spring Energy

A spring of constant 200 N/m is compressed by 0.1 m. A 0.5 kg block is released from it on a smooth surface. Find the velocity of the block.

Solution½kx<sup>2</sup> = ½mv<sup>2</sup> 200 × 0.01 = 0.5 × v<sup>2</sup> 2 = 0.5v<sup>2</sup> → v<sup>2</sup> = 4 → v = 2 m s<sup>-1</sup>

NCERT Exercise Reference

NCERT ExerciseTopicMarks to Expect
6.1–6.4Work, angle dependence1–2 marks (definition, formula)
6.5–6.8KE, work-energy theorem3–5 marks (derivation + numerical)
6.9–6.11PE, conservative forces2–3 marks (concept + derivation)
6.12–6.14Conservation of ME3–5 marks (numerical)
6.15–6.17Power1–2 marks (numerical)
6.18–6.22Collisions3–5 marks (derivation + numerical)

Common Mistakes

1. Forgetting cos θ in W = Fd cos θ: Work is not always Fd. If the force is at an angle to displacement, include the cosine factor. Centripetal force does zero work because θ = 90°. 2. KE is always positive: Unlike velocity, KE = ½mv<sup>2</sup> can never be negative, even if v is negative. 3. Confusing conservation of KE with conservation of momentum: Momentum is conserved in all collisions; KE is conserved only in elastic collisions. 4. Applying elastic collision formulas to inelastic collisions: The formulas for v<sub>1</sub> and v<sub>2</sub> apply only when e = 1. 5. Sign of work by friction: Friction does negative work (removes KE), not positive work. 6. PE reference level: You can choose any reference level for gravitational PE. Be consistent within a problem. 7. Spring PE is always positive: ½kx<sup>2</sup> is positive whether the spring is stretched (x > 0) or compressed (x < 0). 8. 1 kWh is energy, not power: 1 kWh = 3.6 × 10<sup>6</sup> J. It is the energy used by a 1 kW appliance in 1 hour.

Exam Tips

  • Work-energy theorem is the most frequently asked derivation (3–5 marks). State it clearly, derive it from F = ma, and apply it to a numerical.
  • Collision problems (3–5 marks): Know the elastic collision formulas and the coefficient of restitution definition. Practise at least 5 problems.
  • Conservation of energy is the simplest way to solve problems involving height and speed — use it instead of kinematic equations when possible.
  • Power appears as 1–2 mark questions: know P = Fv and P = W/t.
  • For derivations, start from the definition, show each step, and write the final formula clearly.
  • In collision numericals, always verify momentum conservation after finding final velocities.
  • Spring problems: Remember that spring PE = ½kx<sup>2</sup> and the maximum velocity occurs when all spring PE converts to KE (at equilibrium position).
  • Variable force work: Use the area under the F-x graph. This is a common 2-mark question.

Quick Revision Notes

  • Work: W = Fd cos θ. Positive, negative, or zero. Area under F-x graph = work.
  • KE: ½mv<sup>2</sup>. Always positive. Scalar. Proportional to v<sup>2</sup>.
  • Work-energy theorem: W<sub>net</sub> = ΔKE. Works for all forces, scalar equation. Derived from F = ma.
  • Gravitational PE: mgh; Spring PE: ½kx<sup>2</sup>. Both are scalars.
  • Conservative force: path-independent, closed-loop work = 0, PE definable. F = −dU/dx.
  • Non-conservative force: path-dependent. Friction removes mechanical energy (converts to heat).
  • Conservation of ME: KE + PE = constant (only conservative forces). With friction: W<sub>friction</sub> = Δ(ME).
  • Power: P = W/t = F · v. Unit: watt. 1 kWh = 3.6 × 10<sup>6</sup> J (energy, not power).
  • Elastic collision: e = 1, momentum + KE conserved. Equal masses → velocities exchange.
  • Inelastic collision: e < 1, only momentum conserved. Completely inelastic (e = 0) → bodies stick together.
  • Coefficient of restitution: e = (v<sub>2</sub> − v<sub>1</sub>) / (u<sub>1</sub> − u<sub>2</sub>). For bouncing ball: e = √(h'/h).
  • KE lost in inelastic: ΔKE = ½m<sub>1</sub>m<sub>2</sub>(u<sub>1</sub>−u<sub>2</sub>)<sup>2</sup>/(m<sub>1</sub>+m<sub>2</sub>).

Frequently Asked Questions

Work is zero when the force is perpendicular to displacement (θ = 90°, W = Fd cos 90° = 0), when displacement is zero, or when force is zero. Example: centripetal force does zero work on a body in circular motion.