Work Energy and Power Class 11 Notes | CBSE Physics Chapter 5

Work, Energy and Power is Chapter 5 of CBSE Class 11 Physics — and one of the most rewarding chapters you will study all year. It takes the forces you learned in Laws of Motion and asks a deeper question: how much work does a force do, and where does that energy go? Master this chapter and a huge slice of mechanics in JEE, NEET, and your board exam suddenly becomes a one-line energy equation.

By the end of these notes you will be able to calculate work done by constant and variable forces, apply the work-energy theorem, switch fluently between kinetic and potential energy, use conservation of mechanical energy to shortcut tough problems, and solve every type of collision. This is a high-weightage chapter carrying roughly 6–8 marks in boards, and the foundation for Rotational Motion, Gravitation, and most of Class 12 Physics.

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Table of Contents

• Key Concepts — Work, energy, the work-energy theorem, potential energy, conservation, power, collisions
• Weightage in Board & Entrance Exams
• Important Definitions
• Solved Examples
• Important Questions for Board Exams
• Quick Revision Points

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Key Concepts

1. Work Done by a Constant Force

Work is done when a force acting on a body produces a displacement in its direction. It is a scalar quantity equal to the dot product of force and displacement.

W = F·s = Fs cos θ, where θ is the angle between the force and the displacement.

• SI unit: joule (J); 1 J = 1 N·m. Dimensions: [ML²T⁻²].
• Positive work (θ < 90°): force aids motion, e.g. a falling body’s weight.
• Negative work (θ > 90°): force opposes motion, e.g. friction.
• Zero work (θ = 90°): force is perpendicular to displacement, e.g. centripetal force, or carrying a bag horizontally.

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2. Work Done by a Variable Force

When the force changes with position, we add up the work over tiny displacements where the force is nearly constant. This sum becomes an integral.

W = ∫ F dx (from x₁ to x₂)

Graphically, the work done by a variable force equals the area under the force–displacement (F–x) graph.

[DIAGRAM: A force–displacement graph; the shaded area between the curve and the x-axis from x₁ to x₂ represents the work done.]

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3. Kinetic Energy

Kinetic energy (KE) is the energy a body possesses by virtue of its motion. It equals the work needed to bring the body from rest to its current speed.

KE = ½mv²

• SI unit: joule (J); it is a scalar and always positive.
• Relation with momentum: KE = p²/2m and p = √(2m·KE).
• If momentum is constant, a lighter body has more kinetic energy.

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4. Work-Energy Theorem

The work-energy theorem states that the net work done by all forces on a body equals the change in its kinetic energy.

W_net = ΔKE = ½mv² − ½mu²

• It holds for constant and variable forces alike.
• Positive net work speeds the body up; negative net work slows it down.
• This single equation replaces messy force-and-acceleration calculations in many problems.

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5. Potential Energy

Potential energy (PE) is the energy stored in a body by virtue of its position or configuration. It is defined only for conservative forces.

Gravitational Potential Energy

Near the Earth’s surface, lifting a mass m through a height h stores energy:

U = mgh

The reference level (where U = 0) is chosen for convenience, usually the ground.

Elastic (Spring) Potential Energy

A spring obeying Hooke’s law (restoring force F = −kx) stores energy when stretched or compressed by x:

U = ½kx²

This U is the area under the F = kx line — a triangle of area ½kx².

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6. Conservative and Non-Conservative Forces

A conservative force does work that depends only on the start and end positions, not the path taken; the work done in a closed loop is zero. Potential energy can be defined for it.

A non-conservative force does path-dependent work, so no potential energy can be defined; mechanical energy is lost (usually as heat).

Conservative	Non-Conservative
Gravity, spring force, electrostatic force	Friction, air resistance, viscous drag
Work in a closed loop = 0	Work in a closed loop ≠ 0
PE can be defined; F = −dU/dx	PE cannot be defined

Force from potential energy: for a conservative force, F = −dU/dx — the force points towards decreasing potential energy.

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7. Conservation of Mechanical Energy

If only conservative forces act on a body, its total mechanical energy (KE + PE) stays constant throughout the motion.

E = KE + PE = ½mv² + mgh = constant

Freely Falling Body

As a body of mass m falls from height H, PE converts to KE while the total stays mgH. At the bottom, all of it is kinetic: ½mv² = mgH, giving the familiar v = √(2gH).

[DIAGRAM: A ball falling from height H — at the top all energy is PE (mgH), midway it is half PE half KE, at the ground all KE (½mv²); the total bar stays the same height.]

When friction acts: mechanical energy is not conserved; the lost energy equals the work done against friction: W_friction = E_initial − E_final.

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8. Power

Power is the rate at which work is done, or equivalently the rate at which energy is transferred.

P = W/t (average power) and P = dW/dt = F·v (instantaneous power)

• SI unit: watt (W); 1 W = 1 J/s. It is a scalar.
• Commercial unit of energy: 1 kWh = 3.6 × 10⁶ J (one “unit” of electricity).
• 1 horsepower (hp) = 746 W.

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9. Collisions

A collision is a brief, strong interaction between bodies in which momentum is exchanged. Linear momentum is conserved in every collision (no net external force during the impact). Kinetic energy may or may not be conserved.

Type	Momentum	Kinetic Energy	Example
Elastic	Conserved	Conserved	Collision of two smooth billiard balls; atomic collisions
Inelastic	Conserved	Not conserved	A ball hitting clay; most everyday collisions
Perfectly inelastic	Conserved	Maximum loss	Bodies stick and move together; bullet embeds in block

Elastic Collision in One Dimension

For two masses m₁ (velocity u₁) and m₂ (velocity u₂), solving conservation of momentum and kinetic energy together gives:

v₁ = [(m₁ − m₂)u₁ + 2m₂u₂] / (m₁ + m₂)

v₂ = [(m₂ − m₁)u₂ + 2m₁u₁] / (m₁ + m₂)

• Equal masses (m₁ = m₂): the bodies simply exchange velocities.
• Heavy hits light at rest: heavy body continues, light body shoots off at ≈ 2u₁.
• Light hits heavy at rest: light body rebounds with nearly the same speed.

Perfectly Inelastic Collision in One Dimension

The bodies stick and move with a common velocity v:

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

Kinetic energy lost = ½·[m₁m₂/(m₁ + m₂)]·(u₁ − u₂)² — this is the maximum possible loss.

Collision in Two Dimensions (Oblique)

When the bodies do not move along the same line after impact, momentum is conserved separately along two perpendicular axes:

• Along x: m₁u₁ = m₁v₁ cos θ₁ + m₂v₂ cos θ₂
• Along y: 0 = m₁v₁ sin θ₁ − m₂v₂ sin θ₂

For an elastic 2-D collision, kinetic energy conservation gives a third equation.

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10. Coefficient of Restitution

The coefficient of restitution (e) measures how elastic a collision is. It is the ratio of the relative velocity of separation to the relative velocity of approach.

e = (v₂ − v₁) / (u₁ − u₂)

• e = 1: perfectly elastic collision (KE conserved).
• e = 0: perfectly inelastic collision (bodies stick).
• 0 < e < 1: real, partially inelastic collision.

For a ball dropped from height h₁ and rebounding to height h₂: e = √(h₂/h₁).

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Weightage in Board & Entrance Exams

Exam	Typical Weightage	Most-Tested Areas
CBSE Board (Class 11)	6–8 marks	Work-energy theorem, conservation of energy, power, collisions
JEE Main / Advanced	2–3 questions	Variable force & springs, energy on inclines, elastic/inelastic collisions
NEET	2–3 questions	KE–momentum relation, power, coefficient of restitution

[TABLE: Question-type split — VSA (1 mark): definitions, zero/negative work, units; SA (2–3 marks): work-energy theorem, spring PE, power numericals; LA (5 marks): conservation of mechanical energy derivations, elastic-collision velocity formulae.]

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Important Definitions

Term	Definition
Work	Product of force and displacement in the force’s direction: W = Fs cos θ (a scalar)
Kinetic energy	Energy of motion: KE = ½mv² = p²/2m
Work-energy theorem	Net work done equals change in kinetic energy: W_net = ΔKE
Potential energy	Energy of position/configuration; gravitational U = mgh, spring U = ½kx²
Conservative force	Path-independent force; work in a closed loop = 0; F = −dU/dx
Mechanical energy	Sum of kinetic and potential energy: E = KE + PE
Power	Rate of doing work: P = W/t = F·v; SI unit watt
Elastic collision	Collision in which both momentum and kinetic energy are conserved (e = 1)
Inelastic collision	Momentum conserved but kinetic energy is not (e < 1)
Coefficient of restitution	Ratio of relative velocity of separation to approach: e = (v₂ − v₁)/(u₁ − u₂)

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Solved Examples

Example 1

A force of 50 N pulls a box through 8 m at 60° to the horizontal. Find the work done.

Answer: W = Fs cos θ = 50 × 8 × cos 60° = 50 × 8 × 0.5 = 200 J.

Example 2

A 2 kg body moving at 4 m/s is brought to rest by a constant force. Find the work done by the force.

Answer: By the work-energy theorem, W = ΔKE = 0 − ½ × 2 × 4² = −16 J (negative — the force opposes motion).

Example 3

A spring of force constant 200 N/m is compressed by 0.1 m. Find the elastic potential energy stored.

Answer: U = ½kx² = ½ × 200 × (0.1)² = ½ × 200 × 0.01 = 1 J.

Example 4

A pump lifts 600 kg of water to a height of 20 m in 1 minute. Find its power. (g = 10 m/s²)

Answer: W = mgh = 600 × 10 × 20 = 120000 J. P = W/t = 120000/60 = 2000 W (2 kW).

Example 5

A bullet of mass 20 g moving at 300 m/s embeds into a 4.98 kg block at rest. Find the common velocity. (perfectly inelastic)

Answer: v = m₁u₁/(m₁ + m₂) = (0.02 × 300)/(0.02 + 4.98) = 6/5 = 1.2 m/s.

Example 6

A ball is dropped from a height of 2 m and rebounds to 1.28 m. Find the coefficient of restitution.

Answer: e = √(h₂/h₁) = √(1.28/2) = √0.64 = 0.8.

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Important Questions for Board Exams

1-Mark Questions (VSA)

1. Define one joule of work.
2. When is the work done by a force zero even though the force and displacement are non-zero?
3. What is the relation between kinetic energy and linear momentum?
4. State the SI unit of power and its relation to horsepower.
5. What is the value of the coefficient of restitution for a perfectly elastic collision?

2–3-Mark Questions (SA)

1. State and prove the work-energy theorem for a constant force.
2. Derive the expression for the elastic potential energy of a stretched spring.
3. Show that for a perfectly inelastic collision in one dimension the common velocity is (m₁u₁ + m₂u₂)/(m₁ + m₂).
4. Distinguish between conservative and non-conservative forces with one example each.

5-Mark Questions (LA)

1. State the law of conservation of mechanical energy and verify it for a freely falling body at the top, midway, and the ground.
2. For an elastic collision in one dimension, derive the expressions for the final velocities of the two bodies. Discuss the case of equal masses.
3. Define power and obtain the relation P = F·v. A car of mass 1000 kg moves up an incline against friction — explain how power is used.

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Quick Revision Points

• Work W = Fs cos θ (scalar, joule); positive, negative, or zero depending on θ
• Variable force: W = ∫F dx = area under the F–x graph
• Kinetic energy KE = ½mv² = p²/2m; always positive
• Work-energy theorem: W_net = ΔKE — works for any force
• PE: gravitational U = mgh, spring U = ½kx²; F = −dU/dx
• Conservative force: path-independent, closed-loop work = 0 (gravity, springs)
• Conservation of mechanical energy: KE + PE = constant when only conservative forces act
• Power P = W/t = F·v (watt); 1 kWh = 3.6 × 10⁶ J; 1 hp = 746 W
• All collisions conserve momentum; only elastic ones conserve KE
• Elastic, equal masses: velocities are exchanged; perfectly inelastic: bodies stick
• Coefficient of restitution e = (separation speed)/(approach speed); e = √(h₂/h₁)

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Next Chapter: Chapter 6 — Systems of Particles and Rotational Motion