What is Work-Energy Theorem?

The Work-Energy Theorem states that the net work done on an object equals the change in its kinetic energy. Mathematically, W_net = ΔKE = ½mv² - ½mu². This is one of the most important theorems for solving JEE problems.

What is the difference between elastic and inelastic collision?

In elastic collision, both momentum and kinetic energy are conserved, and coefficient of restitution e=1. In inelastic collision, only momentum is conserved while kinetic energy is lost, with 0<e<1. In perfectly inelastic collision, bodies stick together after collision with e=0.

What is the formula for kinetic energy?

Kinetic energy formula is KE = ½mv² where m is mass and v is velocity. Alternative forms include KE = p²/2m (in terms of momentum) and KE = ½Iω² for rotational motion.

How many questions come from Work, Energy and Power in JEE?

Work, Energy and Power typically has 12-16% weightage in JEE. In JEE Main, expect 3-4 questions (12-16 marks), and in JEE Advanced, expect 4-5 questions (15-20 marks). Collisions and work-energy theorem are most frequently asked.

What is the coefficient of restitution?

Coefficient of restitution (e) is the ratio of relative velocity of separation to relative velocity of approach: e = (v₂-v₁)/(u₁-u₂). For perfectly elastic collision e=1, for inelastic 0<e<1, and for perfectly inelastic e=0.

What is the SI unit of Power?

The SI unit of power is Watt (W), where 1 W = 1 J/s = 1 kg·m²/s³. Other commonly used units include: 1 HP (Horse Power) = 746 W, 1 kW = 1000 W. Note that kWh is a unit of energy, not power.

Work Energy and Power Class 11 Physics Notes for JEE 2025

Work

In physics, work is done when a force causes displacement of an object. Unlike the everyday meaning, work in physics has a precise mathematical definition involving force, displacement, and the angle between them.

1.1 Definition of Work

Work Done by Constant Force

W = \vec{F} \cdot \vec{s} = Fs\cos\theta

Force applied

Displacement

Angle between F and s

SI Unit

Joule (J)

1 J = 1 N·m = 1 kg·m²/s²

Dimensional Formula

[ML²T⁻²]

Same as Energy

1.2 Nature of Work (Sign Convention)

Condition	Angle θ	cos θ	Work	Example
Positive Work	0° ≤ θ < 90°	Positive	W > 0	Pushing a cart forward
Negative Work	90° < θ ≤ 180°	Negative	W < 0	Friction on moving block
Zero Work	θ = 90°	Zero	W = 0	Carrying bag horizontally

💡 Quick Memory Trick

Positive Work: Force helps motion (same direction)
Negative Work: Force opposes motion (opposite direction)
Zero Work: Force ⊥ displacement OR no displacement

1.3 Work Done by Various Forces

Work by Gravity

\[W_g = mgh\]

Positive when falling down, negative when rising up

Work by Friction

W_f = -f \cdot s = -\mu N s

Always negative (opposes motion)

Work by Normal Force

W_N = 0 \text{ (usually)}

Zero when surface doesn't move in normal direction

Work by Spring Force

W_s = \frac{1}{2}kx_1^2 - \frac{1}{2}kx_2^2

From position x₁ to x₂

1.4 Work Done by Variable Force

W = \int_{x_1}^{x_2} F(x) \, dx = \text{Area under F-x curve}

⚠️ Important for JEE

Work is a scalar quantity (not vector). Even though it involves vectors (F and s), the dot product gives a scalar result. Work can be positive, negative, or zero.

📝 Solved Example 1 (JEE Main Pattern)

Question: A force F = (3î + 4ĵ) N acts on a body producing displacement s = (3î + 4ĵ) m. Find the work done.

Solution:

W = \vec{F} \cdot \vec{s} = (3\hat{i} + 4\hat{j}) \cdot (3\hat{i} + 4\hat{j})

W = 3 \times 3 + 4 \times 4 = 9 + 16 = 25 \text{ J}

\boxed{W = 25 \text{ J}}

Kinetic Energy

Kinetic Energy is the energy possessed by an object due to its motion. Any object that is moving has kinetic energy, and the faster it moves, the more kinetic energy it has.

2.1 Formula for Kinetic Energy

KE = \frac{1}{2}mv^2

where m = mass, v = velocity

Alternative Forms:

In terms of momentum

KE = \frac{p^2}{2m}

Relation with p

p = \sqrt{2mKE}

For rotation

KE = \frac{1}{2}I\omega^2

2.2 Properties of Kinetic Energy

✓ Always Positive or Zero

KE ≥ 0 (mass and v² are always positive)

✓ Scalar Quantity

No direction, only magnitude

✓ Frame Dependent

Different observers measure different KE

✓ Proportional to v²

Double speed → 4× kinetic energy

💡 JEE Trick: KE-Momentum Relations

If p is constant:

KE ∝ 1/m

Lighter object has more KE

If KE is constant:

p ∝ √m

Heavier object has more momentum

Potential Energy

Potential Energy is the energy stored in an object due to its position or configuration. It represents the capacity to do work when the object's position changes.

3.1 Gravitational Potential Energy

Near Earth's Surface

\[U = mgh\]

h = height above reference point

General Form (Large distances)

U = -\frac{GMm}{r}

r = distance from Earth's center

3.2 Elastic (Spring) Potential Energy

U = \frac{1}{2}kx^2

where k = spring constant, x = displacement from natural length

3.3 Conservative vs Non-Conservative Forces

Aspect	Conservative Forces	Non-Conservative Forces
Definition	Work done is path independent	Work done depends on path
Work in closed loop	Zero	Non-zero
PE exists?	Yes, PE can be defined	No PE can be defined
Examples	Gravity, Spring, Electrostatic	Friction, Air resistance, Viscous
Relation with PE	W = -ΔU	No relation

⚠️ Key Relation (Very Important!)

\vec{F} = -\frac{dU}{dx} \quad \text{or} \quad \vec{F} = -\nabla U

Force = negative gradient of potential energy. Force points from high PE to low PE.

📝 Solved Example 2

Question: If potential energy U = 3x² + 4, find the force at x = 2.

Solution:

F = -\frac{dU}{dx} = -\frac{d(3x^2 + 4)}{dx} = -6x

At x = 2:

F = -6(2) = -12 \text{ N}

Work-Energy Theorem

📜 Statement

"The net work done on an object equals the change in its kinetic energy."

W_{net} = \Delta KE = KE_f - KE_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

4.1 Important Points

W_net > 0

KE increases (object speeds up)

W_net < 0

KE decreases (object slows down)

W_net = 0

KE constant (constant speed)

Valid for ALL forces

Conservative + Non-conservative

4.2 Extended Form

W_{gravity} + W_{friction} + W_{applied} + W_{other} = \Delta KE

Sum of work by all individual forces = Change in KE

📝 Solved Example 3 (JEE Main 2023 Type)

Question: A 2 kg block is pushed by a 20 N force on a rough horizontal surface (μ = 0.3) for 5 m. If initial velocity is 4 m/s, find final velocity. (g = 10 m/s²)

Given: m = 2 kg, F = 20 N, μ = 0.3, s = 5 m, u = 4 m/s

Step 1: Calculate work by applied force

W_F = F \times s = 20 \times 5 = 100 \text{ J}

Step 2: Calculate work by friction

f = \mu mg = 0.3 \times 2 \times 10 = 6 \text{ N}

W_f = -f \times s = -6 \times 5 = -30 \text{ J}

Step 3: Apply Work-Energy Theorem

W_{net} = \Delta KE

100 - 30 = \frac{1}{2}(2)v^2 - \frac{1}{2}(2)(4)^2

\[70 = v^2 - 16\]

\[v^2 = 86\]

v = \sqrt{86} \approx 9.27 \text{ m/s}

💡 When to Use Work-Energy Theorem?

When asked to find velocity after displacement
When force varies with position
When path is complex but start/end points known
Problems involving friction and other forces together

Conservation of Mechanical Energy

📜 Law of Conservation of Energy

"In an isolated system where only conservative forces act, the total mechanical energy (KE + PE) remains constant."

Mathematical Form

\[KE_i + PE_i = KE_f + PE_f\]

E_{total} = KE + PE = \text{constant}

5.1 Conditions for Conservation

✓ Energy IS Conserved When:

• Only conservative forces act
• System is isolated
• No friction, air resistance

✗ Energy NOT Conserved When:

• Friction is present
• External forces do work
• Inelastic collisions occur

5.2 Modified Energy Conservation (With Non-Conservative Forces)

KE_i + PE_i + W_{nc} = KE_f + PE_f

W_nc = Work done by non-conservative forces (usually negative for friction)

📝 Solved Example 4 (JEE Advanced Pattern)

Question: A ball is dropped from height 10 m. It bounces back to 6 m. Find energy lost and coefficient of restitution. (g = 10 m/s²)

Solution:

Initial energy:

E_i = mgh_1 = m \times 10 \times 10 = 100m \text{ J}

Final energy:

E_f = mgh_2 = m \times 10 \times 6 = 60m \text{ J}

Energy lost:

\Delta E = 100m - 60m = 40m \text{ J}

\text{Fraction lost} = \frac{40m}{100m} = 40\%

Coefficient of restitution:

e = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{6}{10}} = \sqrt{0.6}

e \approx 0.775

5.3 Energy Conservation in Springs

Spring-Mass System (Horizontal)

\frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2 = \frac{1}{2}mv_2^2 + \frac{1}{2}kx_2^2

At extreme position (x = A):

v = 0, PE = max, KE = 0

At mean position (x = 0):

v = max, PE = 0, KE = max

Power

Power is the rate at which work is done or energy is transferred. It tells us how quickly energy is being used or transferred.

6.1 Definition and Formulas

Average Power

P_{avg} = \frac{W}{t} = \frac{\Delta E}{t}

Instantaneous Power

P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}

SI Unit

Watt (W)

1 W = 1 J/s = 1 kg·m²/s³

Other Units

1 HP (Horse Power) = 746 W

1 kW = 1000 W

1 MW = 10⁶ W

6.2 Power in Different Scenarios

Scenario	Power Formula
Constant force, constant velocity	P = Fv
Force at angle θ to velocity	P = Fv cos θ
Lifting an object at constant speed	P = mgv
Vehicle on incline at speed v	P = (mg sin θ + f)v
Rotational motion	P = τω

6.3 Vehicle Power Problems (JEE Favorite)

Vehicle Moving on Road

Maximum speed (constant P):

v_{max} = \frac{P}{F_{resistance}}

Acceleration at speed v:

a = \frac{P - fv}{mv}

💡 Important: Power & Energy Units

1 kWh (kilowatt-hour) is a unit of ENERGY, not power!

1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}

📝 Solved Example 5

Question: A car of mass 1000 kg can develop maximum power of 50 kW. Find maximum speed on a road with friction coefficient 0.2. (g = 10 m/s²)

Solution:

At maximum speed, all power overcomes friction:

f = \mu mg = 0.2 \times 1000 \times 10 = 2000 \text{ N}

P = f \times v_{max}

50000 = 2000 \times v_{max}

v_{max} = 25 \text{ m/s} = 90 \text{ km/h}

Collisions

A collision is a short-duration interaction between two or more bodies where they exert forces on each other. Understanding collisions is crucial for JEE as 2-3 questions are asked every year.

7.1 Types of Collisions

Property	Elastic Collision	Inelastic Collision	Perfectly Inelastic
Momentum	Conserved ✓	Conserved ✓	Conserved ✓
Kinetic Energy	Conserved ✓	NOT Conserved ✗	Maximum Loss ✗
Coefficient of Restitution (e)	e = 1	0 < e < 1	e = 0
Bodies after collision	Separate	Separate	Stick together
Example	Ideal gas molecules	Most real collisions	Bullet in block

7.2 Coefficient of Restitution (e)

e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}

e = 1

Perfectly Elastic

0 < e < 1

Inelastic

e = 0

Perfectly Inelastic

7.3 Elastic Collision Formulas (1D)

For m₁ hitting stationary m₂ (u₂ = 0)

Velocity of m₁ after collision:

v_1 = \frac{(m_1 - m_2)}{m_1 + m_2}u_1

Velocity of m₂ after collision:

v_2 = \frac{2m_1}{m_1 + m_2}u_1

Special Cases (Elastic, u₂ = 0):

m₁ = m₂

v₁ = 0, v₂ = u₁

Complete transfer!

m₁ >> m₂

v₁ ≈ u₁, v₂ ≈ 2u₁

Heavy hits light

m₁ << m₂

v₁ ≈ -u₁, v₂ ≈ 0

Light bounces back

7.4 Perfectly Inelastic Collision

Bodies stick together (v₁ = v₂ = v)

\[m_1u_1 + m_2u_2 = (m_1 + m_2)v\]

v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}

Energy Lost:

\Delta KE = \frac{1}{2}\frac{m_1m_2}{m_1 + m_2}(u_1 - u_2)^2

7.5 2D Collisions (Oblique)

Conservation in 2D

x-direction:

m_1u_{1x} + m_2u_{2x} = m_1v_{1x} + m_2v_{2x}

y-direction:

m_1u_{1y} + m_2u_{2y} = m_1v_{1y} + m_2v_{2y}

📝 Solved Example 6 (JEE Main Pattern)

Question: A 2 kg ball moving at 4 m/s collides head-on with a 3 kg ball at rest. If collision is perfectly elastic, find velocities after collision.

Given: m₁ = 2 kg, m₂ = 3 kg, u₁ = 4 m/s, u₂ = 0

Using elastic collision formulas:

v_1 = \frac{(m_1 - m_2)u_1}{m_1 + m_2} = \frac{(2-3)(4)}{2+3} = \frac{-4}{5} = -0.8 \text{ m/s}

v_2 = \frac{2m_1u_1}{m_1 + m_2} = \frac{2(2)(4)}{5} = \frac{16}{5} = 3.2 \text{ m/s}

v_1 = -0.8 \text{ m/s (bounces back)}, \quad v_2 = 3.2 \text{ m/s}

⚠️ Height After Bouncing (Ball drop problems)

When a ball is dropped from height h and bounces:

h_n = e^{2n} \cdot h_0

where n = number of bounces, e = coefficient of restitution

Springs & Variable Force Problems

8.1 Spring Force (Hooke's Law)

\[F = -kx\]

k = spring constant (N/m), x = displacement from natural length

Negative sign indicates restoring nature

8.2 Work Done on/by Spring

Work done BY spring

W_s = \frac{1}{2}kx_1^2 - \frac{1}{2}kx_2^2

From position x₁ to x₂

Work done ON spring

W_{on} = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2

Stored as PE

8.3 Springs in Series and Parallel

Configuration	Equivalent Spring Constant
Series	\(\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + ...\)
Parallel	\(k_{eq} = k_1 + k_2 + ...\)

💡 Memory Trick

Springs are opposite to resistors: Series springs add like parallel resistors, and parallel springs add like series resistors!

8.4 Block-Spring Problems (Energy Method)

📝 Solved Example 7

Question: A 2 kg block moving at 3 m/s hits a spring (k = 200 N/m). Find maximum compression.

Solution:

Using energy conservation (KE → Spring PE):

\frac{1}{2}mv^2 = \frac{1}{2}kx_{max}^2

\frac{1}{2}(2)(3)^2 = \frac{1}{2}(200)(x_{max})^2

9 = 100 \cdot x_{max}^2

x_{max}^2 = 0.09

x_{max} = 0.3 \text{ m} = 30 \text{ cm}

📋 Complete Formula Sheet - Work, Energy & Power

Work

W = F·s = Fs cos θ

W = ∫F dx (variable force)

W_gravity = mgh

W_friction = -μNs

W_spring = ½kx₁² - ½kx₂²

Energy

KE = ½mv² = p²/2m

PE (gravity) = mgh

PE (spring) = ½kx²

PE (general) = -GMm/r

F = -dU/dx

Theorems

W_net = ΔKE (Work-Energy)

KE_i + PE_i = KE_f + PE_f

E_i + W_nc = E_f

Power

P = W/t = dW/dt

P = F·v = Fv cos θ

P = τω (rotational)

1 HP = 746 W

1 kWh = 3.6 × 10⁶ J

Elastic Collision (u₂=0)

v₁ = (m₁-m₂)u₁/(m₁+m₂)

v₂ = 2m₁u₁/(m₁+m₂)

e = (v₂-v₁)/(u₁-u₂) = 1

Inelastic Collision

m₁u₁ + m₂u₂ = (m₁+m₂)v

ΔKE = ½[m₁m₂/(m₁+m₂)](u₁-u₂)²

e = 0 (perfectly inelastic)

h_n = e²ⁿh₀ (bouncing)

Download Formula PDF

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Work-Energy Problems: 30%
✓ Collisions: 25%
✓ Power & Vehicle Problems: 20%
✓ Spring-Block Systems: 15%
✓ Conservation of Energy: 10%

JEE Advanced (Last 5 Years)

✓ Variable Force Integration: 25%
✓ Multi-body Collisions: 25%
✓ Energy + Momentum Combined: 20%
✓ Chain/Rope Problems: 15%
✓ Coefficient of Restitution: 15%

Weightage Analysis

JEE Main: 12-16 marks (3-4 questions)

JEE Advanced: 15-20 marks (4-5 questions)

Difficulty Level: Medium to Hard

Study Time Required: 15-20 hours

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

A force of 10 N acts on a 2 kg body at 60° to displacement of 4 m. Find work done.
Find kinetic energy of a 5 kg body moving at 6 m/s.
A spring of k = 100 N/m is compressed by 10 cm. Find stored potential energy.
A 1000 W pump lifts water to height 10 m. How much water per second? (g = 10 m/s²)
A 2 kg ball moving at 5 m/s stops after traveling 10 m. Find work by friction.
A ball is dropped from 20 m. Find velocity just before hitting ground.
Two balls of mass 2 kg and 4 kg undergo elastic collision. If 2 kg ball has 6 m/s and 4 kg is at rest, find final velocities.
A 50 kg person climbs stairs of height 10 m in 20 s. Find power.

Level 2: Intermediate (JEE Main/Advanced)

A force F = 2x + 3 acts on body. Find work done in moving from x = 0 to x = 5 m.
A 2 kg block slides down a rough incline (30°, μ = 0.2) of length 5 m. Find velocity at bottom using energy method.
A spring gun shoots a 50 g ball. If spring (k = 500 N/m) is compressed 10 cm, find muzzle velocity.
A car of mass 2000 kg with engine power 100 kW starts from rest. Find time to reach 30 m/s (ignore resistance).
A bullet of mass 20 g moving at 400 m/s embeds in a 2 kg wooden block. Find velocity of system and energy lost.
A ball dropped from height h bounces back to h/2. Find coefficient of restitution.
If U = 3x² - 6x + 5, find equilibrium position and nature of equilibrium.
A 10 kg block on smooth surface is pushed by force F = 20 + 2t for 5 seconds. Find velocity.

Level 3: Advanced (JEE Advanced/Olympiad)

A chain of mass m and length L on table starts sliding when L/4 hangs over. Find velocity when completely off.
A ball moving with velocity v collides obliquely with identical ball at rest. Show that after elastic collision, velocities are perpendicular.
Three identical balls are placed in a line. First ball hits second with velocity u. Find velocities after all collisions (elastic).
A particle moves from origin under force F = (3x² + 2y)î + 2xĵ. Is force conservative? If yes, find PE function.
Two blocks m₁ and m₂ connected by spring are placed on smooth surface. Find velocity of center of mass and oscillation amplitude when m₁ is pushed with velocity v₀.
A ball falls from height H on inclined plane at angle θ. It bounces elastically. Find distance where it lands next.
A particle slides from top of smooth hemisphere of radius R. At what angle does it leave surface?
A sand car of mass M moving at v₀ starts collecting sand at rate dm/dt. Find velocity as function of time.

Get Solutions PDF with Step-by-Step Working

📑 Quick Navigation - Work, Energy and Power

Work, Energy & Power JEE Main & Advanced 2025-26

Work

1.1 Definition of Work

Work Done by Constant Force

SI Unit

Dimensional Formula

1.2 Nature of Work (Sign Convention)

💡 Quick Memory Trick

1.3 Work Done by Various Forces

Work by Gravity

Work by Friction

Work by Normal Force

Work by Spring Force

1.4 Work Done by Variable Force

⚠️ Important for JEE

📝 Solved Example 1 (JEE Main Pattern)

Kinetic Energy

2.1 Formula for Kinetic Energy

Alternative Forms:

2.2 Properties of Kinetic Energy

💡 JEE Trick: KE-Momentum Relations

Potential Energy

3.1 Gravitational Potential Energy

Near Earth's Surface

General Form (Large distances)

3.2 Elastic (Spring) Potential Energy

3.3 Conservative vs Non-Conservative Forces

⚠️ Key Relation (Very Important!)

📝 Solved Example 2

Work-Energy Theorem

📜 Statement

4.1 Important Points

4.2 Extended Form

📝 Solved Example 3 (JEE Main 2023 Type)

💡 When to Use Work-Energy Theorem?

Conservation of Mechanical Energy

📜 Law of Conservation of Energy

Mathematical Form

5.1 Conditions for Conservation

5.2 Modified Energy Conservation (With Non-Conservative Forces)

📝 Solved Example 4 (JEE Advanced Pattern)

5.3 Energy Conservation in Springs

Spring-Mass System (Horizontal)

Power

6.1 Definition and Formulas

Average Power

Instantaneous Power

SI Unit

Other Units

6.2 Power in Different Scenarios

6.3 Vehicle Power Problems (JEE Favorite)

Vehicle Moving on Road

💡 Important: Power & Energy Units

📝 Solved Example 5

Collisions

7.1 Types of Collisions

7.2 Coefficient of Restitution (e)

7.3 Elastic Collision Formulas (1D)

For m₁ hitting stationary m₂ (u₂ = 0)

Special Cases (Elastic, u₂ = 0):

7.4 Perfectly Inelastic Collision

Bodies stick together (v₁ = v₂ = v)

7.5 2D Collisions (Oblique)

Conservation in 2D

📝 Solved Example 6 (JEE Main Pattern)

⚠️ Height After Bouncing (Ball drop problems)

Springs & Variable Force Problems

8.1 Spring Force (Hooke's Law)

8.2 Work Done on/by Spring

Work done BY spring

Work done ON spring

8.3 Springs in Series and Parallel

💡 Memory Trick

8.4 Block-Spring Problems (Energy Method)

📝 Solved Example 7

📋 Complete Formula Sheet - Work, Energy & Power

Work

Energy

Theorems