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

Chapter summary

Work, Energy and Power links the force you apply to the motion it produces, defining work as force times displacement and introducing kinetic and potential energy. It builds up to the work-energy theorem, the conservation of mechanical energy, power as the rate of doing work, and the analysis of elastic and inelastic collisions. The chapter is a high-yield NEET topic because its scalar shortcuts let you solve mechanics problems faster than tracking forces along the whole path.

Chapter notes

Table of Contents


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.

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.]


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.

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.

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.


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).

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

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


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.


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.

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.

TypeMomentumKinetic EnergyExample
ElasticConservedConservedCollision of two smooth billiard balls; atomic collisions
InelasticConservedNot conservedA ball hitting clay; most everyday collisions
Perfectly inelasticConservedMaximum lossBodies 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.


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

ExamTypical WeightageMost-Tested Areas
CBSE Board (Class 11)6–8 marksWork-energy theorem, conservation of energy, power, collisions
JEE Main / Advanced2–3 questionsVariable force & springs, energy on inclines, elastic/inelastic collisions
NEET2–3 questionsKE–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.]


Important Definitions

TermDefinition
WorkProduct of force and displacement in the force’s direction: W = Fs cos θ (a scalar)
Kinetic energyEnergy of motion: KE = ½mv² = p²/2m
Work-energy theoremNet work done equals change in kinetic energy: W_net = ΔKE
Potential energyEnergy of position/configuration; gravitational U = mgh, spring U = ½kx²
Conservative forcePath-independent force; work in a closed loop = 0; F = −dU/dx
Mechanical energySum of kinetic and potential energy: E = KE + PE
PowerRate of doing work: P = W/t = F·v; SI unit watt
Elastic collisionCollision in which both momentum and kinetic energy are conserved (e = 1)
Inelastic collisionMomentum conserved but kinetic energy is not (e < 1)
Coefficient of restitutionRatio of relative velocity of separation to approach: e = (v₂ − v₁)/(u₁ − u₂)

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.


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.

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)

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

🃏 Flash Cards: Work, Energy and Power

Class 11 Physics · Chapter 5 – swipe through all 9 cards to understand the whole chapter.

🏋️Start here1/9

What is Work?

A force does work only when it moves something along its line of action.

W = F⃗ · d⃗ = F d cos θ

Scalar; SI unit joule (J), 1 J = 1 N·m, dimensions [M L2 T⁻2].

  • θ < 90° → +ve work; θ = 90° → zero; θ > 90° → −ve work
  • Zero-work traps: coolie carrying load horizontally, string tension in circular motion
  • Variable force: W = ∫ F dx = area under the F–x graph
🏃Core energy2/9

Kinetic Energy

Kinetic energy is the energy a body has because it is moving.

K = ½ m v2 = p2 / 2m

Always ≥ 0; scalar; unit joule. Here p = mv is momentum.

  • Double the speed → KE becomes 4× (v2 dependence)
  • Double KE → momentum rises by √2
  • Energy carried only by motion, never negative
🔗Key theorem3/9

Work-Energy Theorem

The net work of all forces equals the change in kinetic energy.

W_net = ΔK = ½ m v2 − ½ m u2

Holds for constant & variable, conservative & non-conservative forces.

  • Constant speed (e.g. uniform circular motion) → net work = 0
  • Stopping distance from friction: d = v2 / (2 μ g), so d ∝ v2
  • A shortcut: compare end speeds instead of tracking force along the path
📦Stored energy4/9

Potential Energy

Potential energy is energy stored by position or configuration.

U_grav = m g h · U_spring = ½ k x2

Only changes in U are physical — pick any U = 0 reference.

  • Gravitational PE near Earth: U = mgh
  • Elastic (spring) PE for stretch x: U = ½ k x2
  • PE is stored, ready-to-release energy (raised brick, drawn bow)
♻️Force vs PE5/9

Conservative Forces

For conservative forces, work depends only on endpoints, not the path.

F = − dU / dx

Closed-loop work = 0. Non-conservative (friction, drag) dissipate energy.

  • Conservative: gravity, spring, electrostatic — path-independent
  • Force is the negative slope of the U–x curve
  • Equilibrium dU/dx = 0: stable if U is minimum, unstable if maximum
⚖️Conservation6/9

Conservation of Mechanical Energy

When only conservative forces act, total mechanical energy (KE + PE) stays constant.

Kᵢ + Uᵢ = K_f + U_f

With friction: ΔK + ΔU = W_friction (negative); energy turns to heat/sound.

  • Drop from height h: speed at bottom v = √(2gh), independent of mass
  • Vertical circle on a string: v_top = √(gL), v_bottom = √(5gL)
  • Energy merely trades between kinetic and potential forms
Rate of work7/9

Power

Power is how fast work is done or energy is transferred.

P_avg = W / t · P_inst = dW/dt = F⃗ · v⃗ = F v cos θ

Scalar; SI unit watt (W) = 1 J s⁻1. 1 hp = 746 W; 1 kWh = 3.6 × 106 J.

  • Same work in less time → more powerful
  • Vehicle at max speed: engine power = resistive force × speed
  • Pump lifting water: P = (m/t) g h (add ½(m/t)v2 if water leaves with speed v)
🎱Collisions8/9

Elastic Collisions

In every collision momentum is conserved; in elastic ones KE is too.

Elastic (e = 1): v2 − v1 = u1 − u2

Coefficient of restitution e = (relative speed after) / (relative speed before).

  • Equal masses (target at rest): velocities are exchanged
  • Light hits heavy: light body bounces back at nearly same speed
  • Heavy hits light: light body shoots off at ≈ 2u1
🧱Collisions9/9

Inelastic Collisions

Bodies stick or deform; momentum still conserved but kinetic energy drops.

Perfectly inelastic (e = 0): v = (m1u1 + m2u2) / (m1 + m2)

0 ≤ e ≤ 1. Ball dropped from h, bouncing to h′: e = √(h′ / h).

  • Perfectly inelastic → bodies move together, maximum KE lost
  • Lost KE becomes heat, sound, deformation — total energy conserved
  • e = 1 elastic, e = 0 perfectly inelastic, in between partially elastic
Swipe Click a card to focus 9 cards
📝 Practice Work, Energy and Power — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? (g = 10 m/s²)
Correct answer: B. Input power = (m/t)gh = 15×10×60 = 9000 W. With 10% loss, output = 0.90×9000 = 8100 W = 8.1 kW.
🔎 See the full step-by-step solution in the app →
Q2NEET 2020
The energy required to break one bond in DNA is 10⁻²⁰ J. This value in eV is nearly
Correct answer: B. E = 10⁻²⁰ J ÷ (1.6×10⁻¹⁹ J/eV) ≈ 0.06 eV.
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Q3NEET 2020
A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively
Correct answer: B. Total energy at top = mgS. When KE = 3·PE, total = 4·PE = 4mgx, so x = S/4. Then KE = 3mgS/4 = ½mv², giving v = √(3gS/2).
🔎 See the full step-by-step solution in the app →
Q4NEET 2020
A point mass m is moved in a vertical circle of radius r with the help of a string. The velocity of the mass is √(7gr) at the lowest point. The tension in the string at the lowest point is
Correct answer: C. At the lowest point T − mg = mv²/r = m(7gr)/r = 7mg, so T = 8mg.
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Q5NEET 2019
A force F = 20 + 10y acts on a particle in y-direction, where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is
Correct answer: B. W = ∫₀¹(20 + 10y)dy = [20y + 5y²]₀¹ = 20 + 5 = 25 J.
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Q6NEET 2019
An object of mass 500 g, initially at rest, is acted upon by a variable force whose X component varies with X in the manner shown. The velocities of the object at points X = 8 m and X = 12 m would be the respective values of (nearly)
Correct answer: C. Work = area under F-x graph = ΔKE. At x = 8 m, area = 130 J gives v = √(2×130/0.5) ≈ 23 m/s; at x = 12 m, area = 105 J gives v ≈ 20.6 m/s.
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Q7NEET 2019
An object flying in air with velocity (20î + 25ĵ − 12k̂) suddenly breaks in two pieces whose masses are in the ratio 1 : 5. The smaller mass flies off with a velocity (100î + 35ĵ + 8k̂). The velocity of the larger piece will be
Correct answer: A. By momentum conservation m·v = (m/6)v₁ + (5m/6)v₂, so v₂ = [6(20î+25ĵ−12k̂) − (100î+35ĵ+8k̂)]/5 = 4î + 23ĵ − 8k̂ (per the answer key).
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Q8NEET 2018
A particle moves from a point (−2î + 5ĵ) to (4ĵ + 3k̂) when a force of (4î + 3ĵ) N is applied. How much work has been done by the force?
Correct answer: C. Displacement Δs = (4ĵ+3k̂) − (−2î+5ĵ) = 2î − ĵ + 3k̂. W = F·Δs = (4î+3ĵ)·(2î−ĵ+3k̂) = 8 − 3 = 5 J.
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Q9NEET 2016
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8×10⁻⁴ J by the end of the second revolution after the beginning of the motion?
Correct answer: D. KE = ½mv² = 8×10⁻⁴ J gives v² = 0.16 m²/s². Using v² = 2aₜs with s = 2 revolutions = 2(2πr), aₜ = v²/(2s) = 0.1 m/s².
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Q10NEET 2012
The potential energy of a particle in a force field is U = A/r² − B/r, where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is
Correct answer: B. F = −dU/dr = 0 gives −2A/r³ + B/r² = 0, so 2A/r = B, hence r = 2A/B.
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Frequently Asked Questions

What is work in physics and when is it zero?

Work is the product of force and the displacement along the line of that force, given by W = F d cos theta. It is zero when the force is perpendicular to the displacement (theta = 90 degrees), such as the tension in a string during circular motion or a coolie carrying a load horizontally.

What does the work-energy theorem state?

The work-energy theorem says the net work done by all forces on a body equals the change in its kinetic energy, W_net = half m v squared minus half m u squared. It holds for both constant and variable forces and for conservative and non-conservative forces.

What is the difference between conservative and non-conservative forces?

For a conservative force like gravity or a spring force, the work done depends only on the start and end points and is zero around any closed loop, so the energy can be stored as potential energy. Non-conservative forces like friction and air drag depend on the path and dissipate energy as heat and sound.

How are elastic and inelastic collisions different?

In every collision linear momentum is conserved, but kinetic energy is conserved only in an elastic collision (coefficient of restitution e = 1). In an inelastic collision some kinetic energy is lost to heat, sound and deformation, and in a perfectly inelastic collision (e = 0) the bodies stick and move together with maximum kinetic energy loss.

Is Work, Energy and Power important for NEET?

Yes, it is a high-weightage Class 11 mechanics chapter that NEET questions regularly, often combining kinetic energy, the work-energy theorem, conservation of mechanical energy and collisions. Mastering its formulas, including P = F v for power and v = root 2gh for a free drop, helps you solve problems quickly and accurately.

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