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 θ | cos θ | Work | Example |
|---|---|---|---|
| 0° | 1 | Positive (max) | Force along displacement (pushing a box) |
| 0° to 90° | 0 to 1 | Positive | Pulling a sled at an angle |
| 90° | 0 | Zero | Centripetal force in circular motion |
| 90° to 180° | −1 to 0 | Negative | Friction opposing motion |
| 180° | −1 | Negative (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:
| Property | Conservative Force | Non-conservative Force |
|---|---|---|
| Work depends on path? | No (path-independent) | Yes (path-dependent) |
| Work in closed loop | Zero | Non-zero |
| PE can be defined? | Yes | No |
| Examples | Gravity, spring force, electrostatic | Friction, air resistance, viscous drag |
| Mechanical energy conserved? | Yes (if only conservative forces) | No (some KE converted to heat) |
| Force from PE | F = −dU/dx | Not 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)
| Quantity | Formula | Unit |
|---|---|---|
| Average power | W/t | watt (W) |
| Instantaneous power | F · v = dW/dt | watt (W) |
| Kilowatt-hour | 1 kWh = 3.6 × 10<sup>6</sup> J | commercial 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:
| Property | Elastic Collision | Inelastic Collision | Completely Inelastic |
|---|---|---|---|
| Momentum conserved? | Yes | Yes | Yes |
| KE conserved? | Yes | No (some lost) | No (maximum loss) |
| Coefficient of restitution | e = 1 | 0 < e < 1 | e = 0 |
| Bodies separate? | Yes | Yes | No (stick together) |
| Example | Billiard balls | Car crash | Lump 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.
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.
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>)
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.
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>)
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.
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.
NCERT Exercise Reference
| NCERT Exercise | Topic | Marks to Expect |
|---|---|---|
| 6.1–6.4 | Work, angle dependence | 1–2 marks (definition, formula) |
| 6.5–6.8 | KE, work-energy theorem | 3–5 marks (derivation + numerical) |
| 6.9–6.11 | PE, conservative forces | 2–3 marks (concept + derivation) |
| 6.12–6.14 | Conservation of ME | 3–5 marks (numerical) |
| 6.15–6.17 | Power | 1–2 marks (numerical) |
| 6.18–6.22 | Collisions | 3–5 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>).
