Work, Energy & Power — Class 11 Master Guide
Class 11 · CBSE · NEET · JEE

Master Work, Energy
& Power

The chapter that toppers love — Visual explanations, deep intuition, and exam-winning strategies. From zero to hero, step by step.

15+
Core Concepts
45+
Practice Questions
10
Sections
100%
Exam Coverage
Start Learning Formula Sheet
Section 01

NCERT Exam Blueprint

Know what matters most. Focus your energy where marks are hiding.

🔥 High Weight

Work-Energy Theorem

W = ΔKE — Asked in EVERY exam. Numericals, MCQs, assertion-reason. Master this first.

🔥 High Weight

Conservation of Energy

PE + KE = constant. Free fall, projectile, pendulum. Always in NEET/JEE.

🔥 High Weight

Numerical Problems

Work done calculations, power problems, energy conversion. Guaranteed marks.

⭐ Medium Weight

Work by Variable Force

Graph area = work done. Spring force (Hooke's law). Area under F-x graph.

⭐ Medium Weight

Power

P = W/t = Fv. Often combined with kinematics. Don't ignore units!

😵 Confusing Area

Sign of Work

When is work positive? Negative? Zero? The angle θ decides everything.

Section 02

Core Concepts — Built from Zero

No jargon. Just intuition and clarity.

💪

What is Work?

Work = Force × Displacement × cos(angle). Key rule: Without displacement, NO work — no matter how hard you push!

W = F · s · cosθ
🚫 Pushing a wall → No work (no displacement)
✅ Lifting a bag → Work done!
🚫 Holding a bag still → No work
⚡

Kinetic Energy

Energy of motion. The faster you move, the more KE you have. Depends on mass AND speed.

KE = ½mv²
🚗 Car moving at 60 km/h → has KE
🏃 Running student → has KE
🪨 Stone at rest → KE = 0
🏔️

Potential Energy

Energy due to position. Stored energy waiting to be released. Height means more PE.

PE = mgh
📦 Box on a shelf → has PE
💧 Water in dam → has PE
🌊 Falls down → PE converts to KE
⚡

Power

How fast work is done. Same work done faster = more power. Your body's "engine rating".

P = W/t = Fv
🚀 Rocket engine → very high power
🐢 Same work, more time → less power
💡 1 Watt = 1 Joule per second
Section 03

Visual: The Angle That Changes Everything

Understand W = F·s·cosθ like drawing on a board. The angle θ between force and displacement decides the type of work.

F → s → θ = 0°

Same Direction

θ = 0°, cos 0° = 1
Work is maximum positive

W = +Fs (MAXIMUM) ✅
← F ← s θ = 180°

Opposite Direction

θ = 180°, cos 180° = −1
Friction doing negative work

W = −Fs (NEGATIVE) ❌
N↑ s → θ = 90°

Perpendicular

θ = 90°, cos 90° = 0
Normal force does zero work

W = 0 (ZERO) ⭕
F s → θ

At Angle θ

General case. Only cosθ component of force does work along displacement.

W = F·s·cosθ ✅
Section 04 — 🔥 Most Important

Work-Energy Theorem

The single most powerful tool in mechanics. Learn this, score big.

THE GOLDEN RULE
Wnet = ΔKE = ½mv² − ½mu²
Net work done on an object = Change in its kinetic energy
Work done → Speed increases → KE increases
Work done against → Speed decreases → KE decreases
📈

Positive Work → KE Increases

Engine pushes car forward → car speeds up → KE goes up. Force and motion are friends.

Example: Ball falling down — gravity does positive work, speed increases
📉

Negative Work → KE Decreases

Brakes applied → friction does negative work → car slows down → KE goes down.

Example: Ball thrown upward — gravity does negative work, speed decreases
💡

Exam Strategy

When you see "find velocity" after a force acts — use W-E theorem. It's faster than Newton's laws!

Shortcut: W = ΔKE → solve for v directly!
Section 05

Conservation of Energy

Energy never disappears — it just changes form. This principle solves complex problems in seconds.

The Law
Total Energy = KE + PE = Constant
mgh + ½mv² = constant (in absence of friction)
🏀 Ball Falling from Height h = 10m (mass = 1kg, g = 10 m/s²)
Top
(h=10m)
PE
100 J
+
KE
0 J
=
Total E
100 J
↓
Middle
(h=5m)
PE
50 J
+
KE
50 J
=
Total E
100 J
↓
Bottom
(h=0m)
PE
0 J
+
KE
100 J
=
Total E
100 J
✅ Total energy = 100J throughout! PE ↓ exactly as much as KE ↑
Section 06

Formula Sheet — Your Exam Weapon

Every formula you need. When to use. Units. Memory tricks.

Work (Constant Force)
W = F·s·cosθ
Unit: Joule (J) = N·m
θ = angle between F and displacement
✅ When: force and displacement given
Kinetic Energy
KE = ½mv²
Unit: Joule (J) = kg·m²/s²
Also: KE = p²/2m (p = momentum)
✅ When: mass and velocity given
Gravitational PE
PE = mgh
Unit: Joule (J)
h = height above reference point
✅ When: height and mass given
Work-Energy Theorem
W = ½mv² − ½mu²
Unit: Joule (J)
u = initial, v = final velocity
✅ When: find v after work is done
Conservation of Energy
mgh = ½mv²
Unit: Joule (J)
For free fall from rest: v = √(2gh)
✅ When: no friction, find v from height
Power
P = W/t = Fv
Unit: Watt (W) = J/s
1 HP (horsepower) = 746 W
✅ When: work done in time t
Spring PE
PE = ½kx²
k = spring constant (N/m)
x = compression/extension
✅ When: spring problems (bonus)
Variable Force (Graph)
W = Area under F-x graph
Unit: Joule (J)
Count squares or use geometry
✅ When: force changes with position
Section 07

Common Mistakes — Never Lose Marks Again

These are where students lose easy marks. Read carefully.

😵
❌ WRONG: "Work is done whenever force is applied"
Many students say a man pushing a wall does work. He doesn't — the wall doesn't move!
✅ RIGHT: Work needs BOTH force AND displacement. W = F·s·cosθ. If s = 0, W = 0.
📐
❌ WRONG: Using wrong angle in W = Fscosθ
Students use the angle with vertical when they should use angle with displacement (horizontal).
✅ RIGHT: θ is always between the Force vector and the Displacement vector. Draw both arrows first!
🔀
❌ WRONG: Confusing KE and PE
"Object at height has kinetic energy" — No! Height = PE, motion = KE.
✅ RIGHT: KE = motion energy (½mv²). PE = stored energy (mgh). At height, before falling → PE only.
➕➖
❌ WRONG: Ignoring signs of work
Treating all work as positive in energy conservation problems.
✅ RIGHT: Friction always does negative work. Check direction of force vs displacement. Net work = algebraic sum.
🪨
❌ WRONG: "Normal force does work on a box being pushed"
Normal force is vertical (upward). If box moves horizontally, θ = 90°.
✅ RIGHT: cos 90° = 0. Normal force does ZERO work when box slides horizontally.
🔢
❌ WRONG: Forgetting to square velocity in KE
Writing KE = ½mv instead of ½mv². This loses 1-2 marks.
✅ RIGHT: KE = ½mv². If speed doubles, KE becomes 4 times! Always square the velocity.
Section 08

Graph Mastery

Graphs appear every year. Learn to read them instantly.

📊 Force vs Displacement (W = Area under graph)
Displacement (m) Force (N) AREA = WORK W = F × s
🔑 Key: For CONSTANT force → rectangle → W = F × s
🔑 For variable force → any shape → count the area carefully
📊 KE vs Height for Falling Object
PE ↓ KE ↑ Total E Height fallen (m) Energy (J)
🔑 Key: PE and KE are mirror images. Where they cross → PE = KE (h = H/2)
🔑 Total (amber dashed) = always constant
Section 09

Problem Solving — Step by Step

Follow this strategy for EVERY numerical. Never get stuck again.

1

Read & Identify

What is given? Mass, velocity, height, force, angle? List them down.

2

Identify Type

Is it work? KE problem? PE problem? Energy conservation? Identify before touching formulas.

3

Choose Formula

W=Fscosθ? or W=ΔKE? or mgh=½mv²? Pick the right weapon.

4

Check Signs

Is work positive or negative? Is the object speeding up or slowing down?

5

Solve & Units

Calculate and always write units. Joules for energy, Watts for power.

6

Verify Answer

Does the answer make sense physically? Is energy conserved? Quick sanity check.

📝 Worked Example — Energy Conservation

A ball of mass 0.5 kg is dropped from a height of 20 m. Find its velocity just before hitting the ground. (g = 10 m/s²)

STEP 1: Given
m = 0.5 kg, h = 20 m, g = 10 m/s², u = 0 (dropped)
STEP 2: Choose Method
No friction → Use Energy Conservation: PE at top = KE at bottom
STEP 3: Apply Formula
mgh = ½mv²
gh = ½v²
v² = 2gh = 2 × 10 × 20 = 400
v = 20 m/s ✅
ANSWER
v = 20 m/s
Note: mass cancelled! Velocity of free fall doesn't depend on mass.
Section 10

Practice Questions — Level by Level

Start from Level 1. Master each level before moving up.

Q.01 · Level 1
A person pushes a heavy wall with force 100 N for 5 seconds but the wall doesn't move. What is the work done?
Answer: W = 0 J
Work = F × s × cosθ. Since displacement s = 0 (wall didn't move), Work = 100 × 0 × cos0° = 0 J.
Lesson: No displacement = No work, no matter how much force!
Q.02 · Level 1
A force of 50 N pushes a box 4 m in the same direction as the force. Find the work done.
Answer: W = 200 J
θ = 0° (same direction), cos 0° = 1
W = F × s × cosθ = 50 × 4 × 1 = 200 J
Q.03 · Level 1
What is the kinetic energy of a 2 kg ball moving at 5 m/s?
Answer: KE = 25 J
KE = ½mv² = ½ × 2 × 5² = ½ × 2 × 25 = 25 J
Q.04 · Level 1
A 3 kg book is kept on a shelf 2 m high. What is its potential energy? (g = 10 m/s²)
Answer: PE = 60 J
PE = mgh = 3 × 10 × 2 = 60 J
Q.05 · Level 1
A person does 500 J of work in 10 seconds. What is the power?
Answer: P = 50 W
P = W/t = 500/10 = 50 Watts
Q.06 · Level 1
A coolie carries a load on his head and walks on a horizontal road. What work does he do against gravity?
Answer: W = 0 J
Gravity acts downward (vertical). Displacement is horizontal. θ = 90°, cos 90° = 0.
Work done by gravity = 0. This is a classic NCERT example!
Q.07 · Level 1
If the speed of an object doubles, by how many times does its kinetic energy increase?
Answer: KE becomes 4 times
KE = ½mv². If v becomes 2v:
New KE = ½m(2v)² = ½m × 4v² = 4 × (½mv²)
Remember: KE ∝ v². Double v → 4× KE
Q.08 · Level 1
Work done by friction is always _____ (positive/negative/zero).
Answer: Negative
Friction always acts opposite to the direction of motion (displacement). So θ = 180°, cos 180° = -1. Work = F × s × (-1) = negative.
Friction is a "killer" of kinetic energy!
Q.09 · Level 1
State the Work-Energy Theorem in one sentence.
The net work done on a body equals the change in its kinetic energy.
Mathematically: W_net = ΔKE = ½mv² - ½mu²
Q.10 · Level 1
A satellite moves in a circular orbit. Does Earth's gravity do any work on the satellite?
Answer: Zero work
For circular orbit, gravity always points toward the center (inward). The satellite's velocity (displacement) is always tangential (perpendicular to radius). So θ = 90°, W = 0.
Q.11 · Level 2 · NCERT
A body of mass 5 kg initially at rest is subjected to a force of 20 N. What is the kinetic energy acquired by the body at the end of 10 s?
Answer: KE = 4000 J
Step 1: a = F/m = 20/5 = 4 m/s²
Step 2: v = u + at = 0 + 4×10 = 40 m/s
Step 3: KE = ½mv² = ½ × 5 × 40² = ½ × 5 × 1600 = 4000 J
Tip: Could also use W-E theorem — find displacement s=½at²=200m, then W=Fs=20×200=4000J
Q.12 · Level 2 · NCERT
A bullet of mass 0.012 kg and horizontal velocity 70 m/s strikes a wooden block of mass 0.4 kg at rest on a frictionless surface. The bullet gets embedded in the block. What is the velocity of the block after collision? (Use conservation of momentum, then find KE lost.)
v = 2.0 m/s (approx); KE lost ≈ 28.6 J
By conservation of momentum: m₁u₁ = (m₁+m₂)v
0.012 × 70 = (0.012 + 0.4) × v
0.84 = 0.412v → v ≈ 2.04 m/s
Initial KE = ½(0.012)(70²) = 29.4 J
Final KE = ½(0.412)(2.04²) ≈ 0.86 J
KE lost ≈ 28.5 J (converted to heat/sound)
Q.13 · Level 2 · NCERT
An object of mass 1 kg is raised to a height of 5 m. It is then allowed to fall. Find its KE when it is 3 m above the ground. (g = 10 m/s²)
Answer: KE = 20 J
At height 5m: Total E = mgh = 1×10×5 = 50 J (all PE, KE=0)
At height 3m: PE = mgh = 1×10×3 = 30 J
By conservation: KE = Total E - PE = 50 - 30 = 20 J
(Also: v² = 2g(5-3) = 40, KE = ½×1×40 = 20J ✓)
Q.14 · Level 2 · NCERT
A pump on the ground floor of a building can pump up water to fill a tank of 30 m³ capacity in 15 minutes. The tank is 40 m above the ground. If efficiency is 30%, find the electric power consumed by the pump. (ρ = 1000 kg/m³, g = 9.8 m/s²)
Answer: Power consumed ≈ 43.6 kW
m = ρ×V = 1000×30 = 30,000 kg
W = mgh = 30000 × 9.8 × 40 = 11,760,000 J
t = 15 × 60 = 900 s
Useful power = W/t = 11760000/900 = 13,067 W
Efficiency = 30% → Input power = 13067/0.30 ≈ 43,556 W ≈ 43.6 kW
Q.15 · Level 2 · NCERT
A 10 kg block slides 5 m down a frictionless incline of 30°. Find work done by gravity. (g = 10 m/s²)
Answer: W = 250 J
Height fallen: h = 5 × sin30° = 5 × 0.5 = 2.5 m
W by gravity = mgh = 10 × 10 × 2.5 = 250 J
(Alternative: W = F cosθ × s = mg sin30° × s = 10×10×0.5×5 = 250J ✓)
Q.16 · Level 3 · NEET Style
A particle of mass m is projected at angle θ with velocity v₀. At the highest point, its kinetic energy is:
A) ½mv₀² B) ½mv₀²cos²θ C) ½mv₀²sin²θ D) Zero
Answer: B) ½mv₀²cos²θ
At highest point, vertical velocity = 0. Only horizontal velocity remains.
vₓ = v₀cosθ (this doesn't change throughout projectile motion)
KE = ½m(v₀cosθ)² = ½mv₀²cos²θ
Q.17 · Level 3 · NEET Style
A car of mass 1000 kg moves at constant velocity of 20 m/s on a horizontal road. The engine produces a power of 5000 W. What is the force of friction?
Answer: Friction = 250 N
At constant velocity: Net force = 0 → Engine force = Friction force
P = Fv → F = P/v = 5000/20 = 250 N
Therefore, friction = 250 N
Q.18 · Level 3 · NEET Style
A body of mass 2 kg moves with velocity 10 m/s. It is brought to rest by a constant force F in distance 5 m. Find F.
Answer: F = 20 N
Use Work-Energy Theorem:
W = ΔKE = ½mv² - ½mu² = 0 - ½(2)(10²) = -100 J
W = -F × s (friction does negative work)
-100 = -F × 5
F = 20 N
Q.19 · Level 3 · NEET Style
Two bodies of masses m and 2m have equal kinetic energies. What is the ratio of their momenta (p₁:p₂)?
Answer: p₁:p₂ = 1:√2
KE = p²/2m → p = √(2m·KE)
Since KE is same: p ∝ √m
p₁/p₂ = √m/√(2m) = 1/√2
p₁:p₂ = 1:√2
Q.20 · Level 3 · NEET Style
A spring of spring constant k = 500 N/m is compressed by 0.1 m. A ball of mass 0.1 kg is placed against it. Find the velocity of the ball when the spring returns to natural length.
Answer: v = √10 ≈ 3.16 m/s
Spring PE = ½kx² = ½ × 500 × (0.1)² = 2.5 J
By energy conservation: Spring PE = KE of ball
2.5 = ½mv² = ½ × 0.1 × v²
v² = 2.5/0.05 = 50 → v = √50 = 5√2 ≈ 7.07 m/s
(Check: 2.5 = 0.05v² → v = √50 m/s)
Section 11

Quick Revision Sheet

Read this the night before the exam. 10 minutes = full chapter recall.

⚡ Key Formulas

Work: W = Fscosθ (Joule)
KE: ½mv² (Joule)
PE: mgh (Joule)
W-E Theorem: W = ½mv² - ½mu²
Conservation: KE + PE = constant
Power: P = W/t = Fv (Watt)
Spring: PE = ½kx²

🧠 Key Concepts

No displacement → No work
θ=0° → Maximum +ve work
θ=90° → Zero work
θ=180° → Maximum -ve work
Friction always does negative work
Normal force → Zero work (horizontal motion)
Energy can't be created or destroyed

🎯 Exam Tricks

Free fall: v = √(2gh) (mass cancels!)
KE ∝ v² — double speed → 4x KE
Equal KE → p ∝ √m
Equal momentum → KE ∝ 1/m
At highest point projectile: KE = ½mv₀²cos²θ
Spring: PE ↔ KE (both sum = constant)
Horsepower: 1 HP = 746 W

🏆 NEET Must-Know

Work-Energy theorem problems — very common
Projectile energy at highest point
Spring energy conservation
Power = Fv (constant velocity problems)
KE-momentum relation: KE = p²/2m
Area under F-x graph = Work
Energy lost to friction = μmg × distance
Section 12

Memory Tricks & NEET Shortcuts

Mnemonic

🎵 "FSD Cosine" Song

Force times Sdisplacement times Cosine = Work done! Say it like a rap beat when you forget.

Visual Trick

🎯 The Angle Rule

Imagine Force and Displacement as two arrows meeting at a point. The angle between them — that's your θ. Never use the angle with ground blindly.

Shortcut

⚡ Free Fall Speed

v = √(2gh). Memorize this! Works for any free fall. Drop from h=5m → v = √100 = 10 m/s. Mass doesn't matter.

Shortcut

📐 Incline Work

Work by gravity on incline = mgh (use vertical height, NOT slant length). Height = L×sinθ. Simple!

Concept

🎢 Roller Coaster Law

The "roller coaster" shows conservation perfectly: highest point → lowest speed. Lowest point → highest speed. Total energy always same.

NEET Shortcut

💡 KE-Momentum Relation

KE = p²/2m. This is gold for NEET! "Same KE → more mass → more momentum." "Same p → less mass → more KE."

⚡ Work, Energy & Power — Master Guide
Class 11 · CBSE · NEET · JEE · Complete Chapter Coverage
Study smart. Practice more. Score big. 🎯
📌 FORMULAS:
W = Fscosθ
KE = ½mv²
PE = mgh
W = ΔKE
KE+PE = const
P = W/t = Fv
v = √(2gh)
PE_spring = ½kx²
KE = p²/2m

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