⚛️ CBSE Class 9 Science · Physics · Chapter 3 · JEE Foundation
Work, Energy & Simple Machines
In physics, "work" has one exact meaning, "energy" is the capacity to do it, and "power" is the rate of doing it. This chapter builds all three step by step — work done by a force, the work–energy theorem, kinetic & potential energy, conservation of energy, power and the kilowatt-hour, and finally simple machines and mechanical advantage. Every formula is applied in worked steps.
📌 Physics Chapter 3
W = Fs cosθ
KE = ½mv² · PE = mgh
★ Conservation of energy
Power & kWh
Simple Machines & MA
⚡ JEE Foundation set
Section 3.1 – 3.2 · The Basics
What "Work" Means in Physics
In everyday speech, studying or thinking is "work". In physics, work is done only when a force makes a body move. A man pushing hard against a wall that does not budge does no physical work at all.
Definition
Work is said to be done when a force acting on a body displaces it along the direction of the force. Work done depends on two factors: (i) the magnitude of the force applied and (ii) the distance moved by the body in the direction of the force.
🐎 Examples of work being done
A horse pulling a cart along a road.
A person climbing stairs (work against gravity).
Pulling a trolley so it moves; lifting a box to a height.
Section 3.3 · The Formula
Work Done by a Constant Force
The force and the displacement can point in the same direction, opposite directions, or at an angle. Each case gives a different result.
Work — the three forms
W = F × s (force along motion)
W = F s cos θ (force at angle θ)
W = m g h (lifting against gravity)
θ is the angle between the force and the displacement. Lifting a mass m through height h needs a minimum force equal to its weight mg.
Positive, negative & zero work
➕ Positive work (θ = 0°)
Force acts along the motion, so W = Fs (cos 0° = 1). Examples: pulling a lawn roller, gravity on a falling body, the winning team in a tug of war, a coolie lifting a box upward.
➖ Negative work (θ = 180°)
Force acts opposite to the motion, so W = −Fs (cos 180° = −1). Examples: friction on a rolling football, the losing team in a tug of war, gravity on a ball thrown upward.
⓪ Zero work (θ = 90° or s = 0)
Work is zero when the force is perpendicular to the motion (cos 90° = 0), or when there is no displacement. Examples: a coolie carrying a load on his head walking horizontally (weight is vertical, motion horizontal); a body in circular motion (centripetal force ⊥ motion — so the Earth around the Sun does zero work); pushing an immovable wall (s = 0).
Force along motion → positive; opposite → negative; perpendicular → zero work.
📊
The area under a force–displacement graph equals the work done — useful when the force is not constant.
Section 3.4 · Measuring Work
The Unit of Work — the Joule
Work is a scalar — a single number with a + or − sign, but no direction.
1 joule defined
The SI unit of work is the joule (J), named after James Prescott Joule. 1 J = 1 N·m — the work done when a force of 1 N moves a body 1 m in the direction of the force. Bigger units: 1 kJ = 1000 J, 1 MJ = 10⁶ J.
Worked Example
A man lifts a 20 kg load and places it on his head 1.4 m above the ground. Find the work done. (g = 9.8 m/s²)
Given
m = 20 kg, h = 1.4 m, g = 9.8 m/s²
Formula
W = mgh (work against gravity)
Substitute
W = 20 × 9.8 × 1.4
Answer
W = 274.4 J
Worked Example
A 60 N force displaces a body 5 m on a horizontal surface, acting at 60° to the surface. Find the work done.
Given
F = 60 N, s = 5 m, θ = 60° (cos 60° = ½)
Formula
W = Fs cos θ
Substitute
W = 60 × 5 × ½ = 300 × ½
Answer
W = 150 J
Section 3.5 · The Bridge
The Work–Energy Theorem
Work and energy are two sides of the same coin: doing work on a body changes its energy of motion.
Work–Energy Theorem
The work done by the net force on a body equals the change in its kinetic energy: W = KEfinal − KEinitial = ½mv² − ½mu². If the body speeds up, the work is positive; if it slows down, the work is negative.
Worked Example
A force changes the velocity of a 10 kg body from 5 m/s to 3 m/s. Find the work done.
Formula
W = ½m(v² − u²)
Substitute
W = ½ × 10 × (3² − 5²) = 5 × (9 − 25)
Answer
W = −80 J (the body slows, so work is negative)
Section 3.6 · The Capacity to Do Work
Energy
A body that can do work is said to possess energy. We get energy from food; machines get it from fuel or electricity.
Definition
Energy is the capacity of a body to do work. It is measured by the total work the body can do, and like work it is a scalar measured in joules (J). Whenever work is done, energy is transferred.
The many forms of energy
Kinetic
½mv²
Due to motion.
Potential
mgh
Due to position or shape.
Heat / Thermal
🔥
From burning fuels.
Light
💡
Gives the sensation of sight.
Sound
🔊
From vibrating objects.
Electrical
⚡
From moving charges.
Chemical
🍎
Stored in food and fuels.
Nuclear
⚛️
From fission and fusion.
🔗
Mechanical energy = kinetic energy + potential energy — the energy a body has due to its motion and its position together.
Section 3.7 – 3.8 · Energy of Motion
Kinetic Energy
A moving object can do work because of its motion — a moving hammer drives a nail, flowing water turns a turbine. That energy of motion is kinetic energy.
Kinetic Energy
KE = ½ m v²
Also, in terms of momentum p = mv: KE = p² / 2m. SI unit: joule. A body at rest has zero KE.
KE ∝ mass, and KE ∝ (speed)² — doubling the speed makes KE four times larger.
Deriving KE = ½mv²
A body of mass m, initially at rest (u = 0), is pushed by a constant force F and reaches velocity v after a displacement s.
From the third equation of motion: v² = u² + 2as → s = v²/2a.
Work done: W = F × s = (ma) × (v²/2a) = ½mv².
This work is stored in the body as its kinetic energy, so KE = ½mv².
Example A
Find the kinetic energy of a 2 kg body moving at 20 m/s.
Formula
KE = ½mv²
Substitute
KE = ½ × 2 × 20² = 1 × 400
Answer
KE = 400 J
Example B
How fast must a 60 kg man run to have a kinetic energy of 750 J?
Formula
KE = ½mv² → v = √(2KE/m)
Substitute
v = √(2 × 750 / 60) = √25
Answer
v = 5 m/s
Example C
An object of mass 20 kg has a momentum of 50 kg m/s. Find its kinetic energy.
Formula
KE = p² / 2m
Substitute
KE = 50² / (2 × 20) = 2500 / 40
Answer
KE = 62.5 J
⚡
Two bodies of equal mass moving at v and 3v have KE in the ratio 1 : 9 — tripling the speed multiplies KE by nine.
Section 3.9 · Stored Energy
Potential Energy
Energy can also be stored — in a raised weight, a stretched bow, a wound spring. This stored energy of position or shape is potential energy.
Gravitational Potential Energy
PE = m g h
Energy stored in a body of mass m raised to height h. It equals the work done against gravity to lift it. SI unit: joule.
Gravitational PE
Energy due to a body's position above the ground: water in a tank or dam, a lifted hammer, a raised weight. Released, it can do work as it falls.
Elastic PE
Energy due to a change of shape: a stretched bow, a wound watch spring, a compressed or stretched spring, a stretched catapult. Released, the shape springs back and does work.
Example A
A 50 kg bag of wheat is to have a potential energy of 5000 J. To what height must it be raised? (g = 10 m/s²)
Formula
PE = mgh → h = PE / (mg)
Substitute
h = 5000 / (50 × 10) = 5000 / 500
Answer
h = 10 m
Example B
A shot-put of mass 3 kg just clears a 2 m wall at a speed of 4 m/s. Find its total mechanical energy at that point. (g = 9.8 m/s²)
Formula
ME = KE + PE = ½mv² + mgh
Substitute
= ½ × 3 × 4² + 3 × 9.8 × 2 = 24 + 58.8
Answer
Total mechanical energy = 82.8 J
Section 3.10 – 3.12 · The Big Law
Transformation & Conservation of Energy
Energy is forever changing form — but the total never changes. This is one of the deepest laws in all of science.
🔄 Transformation of energy
One form of energy can change into another. A falling body turns PE into KE; a thrown ball turns KE into PE; a dam turns PE of water into electrical energy; a bulb turns electrical energy into light; a car engine turns chemical energy of fuel into kinetic energy; a microphone turns sound into electrical energy. During every change some energy may be lost to the surroundings (usually as heat), but the total stays the same.
Law of Conservation of Energy
Energy can be neither created nor destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant, so the total energy of the universe is constant.
Conservation of mechanical energy — free fall
Drop a body of mass m from height h. At every point its total mechanical energy (KE + PE) stays equal to mgh:
At the top (A): v = 0, so KE = 0 and PE = mgh. Total = mgh.
Partway down (B), after falling x: v₁² = 2gx, so KE = ½m(2gx) = mgx, and PE = mg(h − x). Total = mgx + mg(h − x) = mgh.
At the ground (C): v² = 2gh, so KE = ½m(2gh) = mgh and PE = 0. Total = mgh.
As the body falls, PE steadily becomes KE — but the sum is always mgh.
Free fall: PE converts to KE, but KE + PE = mgh at every instant.
🕰️ The simple pendulum
At the extreme positions the bob is momentarily at rest — all its energy is potential. At the mean (lowest) position it moves fastest — all its energy is kinetic. In between, energy is part PE and part KE, but the total is always the same. (The swing slowly dies down only because air resistance turns some mechanical energy into heat — energy is still conserved overall.)
Worked Example
A 5 kg ball is dropped from a height of 10 m (g = 10 m/s²). Find (a) its initial PE, (b) its KE just before landing, (c) its landing speed.
(a) PE
PE = mgh = 5 × 10 × 10 = 500 J
(b) KE
all PE → KE at the ground = 500 J
(c) Speed
½mv² = 500 → v = √(2×500/5) = √200
Answer
PE = 500 J · KE = 500 J · v ≈ 14.14 m/s
Section 3.13 · The Rate of Doing Work
Power
A machine that does the same work in less time is more powerful. Power measures how fast work is done — not how much.
Power
P = W / t = E / t = F × v
work (or energy) ÷ time, which also equals force × velocity. SI unit: watt (W), where 1 W = 1 J/s. Power is a scalar.
Average power = total work ÷ total time.
Units of power & energy
1 W = 1 J/s. Bigger: 1 kW = 1000 W, 1 MW = 10⁶ W. Practical: 1 horsepower (hp) = 746 W. The commercial unit of electrical energy is the kilowatt-hour (kWh): 1 kWh = energy used by a 1 kW appliance in 1 hour = 3.6 × 10⁶ J. One kWh is "1 unit" on an electricity bill.
Example A
A 60 kg athlete runs up a staircase of 10 steps, each 0.5 m high, in 30 s. Find the power. (g = 10 m/s²)
Height
h = 10 × 0.5 = 5 m
Power
P = mgh/t = (60 × 10 × 5)/30 = 3000/30
Answer
P = 100 W
Example B
A force of 10 N moves a body at a constant 2 m/s. Find the power.
Formula
P = F × v
Substitute
P = 10 × 2
Answer
P = 20 W
Example C
A radio of 60 W runs for 50 hours. How many "units" (kWh) of energy does it use?
Convert
P = 60 W = 0.06 kW
Energy
E = P × t = 0.06 kW × 50 h
Answer
E = 3 kWh (3 units)
Section 3.14 · Working Smarter
Simple Machines & Mechanical Advantage
A simple machine lets us overcome a large resistance by applying a smaller force at a convenient point and direction. Machines never create energy — they only help us use it more effectively.
Mechanical Advantage
MA = Load / Effort = L / E
The load (L) is the resistance to be overcome; the effort (E) is the force we apply. MA > 1 means the machine multiplies our force.
Two families of simple machines
Lever — and its modifications, the pulley and the wheel & axle.
Inclined plane — and its modifications, the wedge and the screw.
1 · The Lever & its three orders
A lever is a rigid bar that turns about a fixed point called the fulcrum (F). The distance from the fulcrum to the effort is the effort arm; to the load, the load arm.
Principle of the lever: Load × load arm = Effort × effort arm. So:
MA of a lever
MA = L / E = effort arm / load arm
▲ fulcrum · ↑red = effort · ↓green = load. The order depends on what sits in the middle.
2 · The Pulley
Single fixed pulley
A grooved wheel fixed to a support. It only changes the direction of the effort (you pull down to raise a load up), making the job convenient. It does not reduce the force, so its ideal MA = 1.
Single movable pulley
Here the pulley moves with the load. A load of, say, 100 kgf can be raised by an effort of only 50 kgf, so the ideal MA = 2. A system of pulleys (block and tackle) can give an even greater MA.
3 · The Inclined Plane
Pushing a load up a ramp needs less force than lifting it straight up, though you push it over a longer distance. Work in = work out (ideally), so:
MA of an inclined plane
MA = L / E = length of plane / height = l / h
Worked Example
A 2 kg box is pushed at steady speed up a frictionless ramp 5 m long that rises 3 m. Find the work done, the effort needed, and the MA. (g = 10 m/s²)
Work
W = mgh = 2 × 10 × 3 = 60 J
Effort
W = E × l → E = 60 / 5 = 12 N
MA
MA = l / h = 5 / 3 ≈ 1.67
Answer
W = 60 J · effort = 12 N · MA ≈ 1.67
Problem-Solving Masterclass
Every Formula, Applied Step by Step
The routine never changes — Given → Formula → Substitute → Calculate → Answer. Learn to spot which idea (work, KE, PE, power or MA) a problem is testing.
M1 · W = Fs cos θ (find angle)
A 50 N force displaces a body 5 m and does 125 J of work. Find the angle between the force and the displacement.
Formula
W = Fs cos θ → cos θ = W/(Fs)
Substitute
cos θ = 125 / (50 × 5) = 0.5
Answer
θ = 60°
M2 · negative work
A cart is pushed 50 m against a friction force of 1250 N. Find the work done by friction and state its type.
Formula
W = Fs cos 180° = −Fs (friction opposes motion)
Substitute
W = −1250 × 50
Answer
W = −62500 J (negative work)
M3 · work–energy theorem
How much work raises the speed of a 20 kg bicycle from 2 m/s to 5 m/s?
Formula
W = ½m(v² − u²)
Substitute
W = ½ × 20 × (25 − 4) = 10 × 21
Answer
W = 210 J
M4 · KE from a force over time
A 5 kg body at rest is acted on by a 20 N force for 10 s. Find the kinetic energy gained.
Acceleration
a = F/m = 20/5 = 4 m/s²
Velocity
v = u + at = 0 + 4×10 = 40 m/s
KE
KE = ½ × 5 × 40² = 4000 J
Answer
KE = 4000 J
M5 · increase in PE
A 4 kg body is taken from a height of 5 m to a height of 10 m. Find the increase in its potential energy. (g = 10 m/s²)
Formula
ΔPE = mg(Δh)
Substitute
= 4 × 10 × (10 − 5) = 40 × 5
Answer
ΔPE = 200 J
M6 · power of an engine
What power must an engine have to lift 90 metric tonnes of coal per hour from a mine 200 m deep? (g = 10 m/s²)
Given
m = 90 t = 9×10⁴ kg, h = 200 m, t = 3600 s
Work
W = mgh = 9×10⁴ × 10 × 200 = 1.8×10⁸ J
Power
P = W/t = 1.8×10⁸ / 3600 = 50000 W
Answer
P = 50 kW
M7 · time from power
An engine of 10 kW lifts a 200 kg mass to a height of 40 m. How long does it take? (g = 10 m/s²)
Formula
P = mgh/t → t = mgh/P
Substitute
t = (200 × 10 × 40) / 10000 = 80000/10000
Answer
t = 8 s
M8 · power from KE
A 1000 kg car starts from rest and reaches 72 km/h in 10 s. Find the power developed by the engine.
Convert
v = 72 × 5/18 = 20 m/s
Work
W = ½mv² = ½ × 1000 × 400 = 200000 J
Power
P = W/t = 200000/10
Answer
P = 20000 W = 20 kW
M9 · energy in joules from units
A family uses 250 units of electrical energy in a month. Express this in joules.
Given
250 units = 250 kWh; 1 kWh = 3.6×10⁶ J
Substitute
E = 250 × 3.6×10⁶
Answer
E = 9 × 10⁸ J
Textbook Exercise · Fully Solved
Exercise — Worked with Formula & Alternate Methods
Representative numericals from the chapter, each solved in full with the formula shown and, where useful, a second method to cross-check.
Q1 · 3 Marks
A boy of mass 55 kg runs up a flight of 40 stairs, each 0.15 m high. Find the work done. (g = 10 m/s²)
Height
h = 40 × 0.15 = 6 m
Work
W = mgh = 55 × 10 × 6
Answer
W = 3300 J
Q2 · 3 Marks
Calculate the work done in lifting 200 kg of water through a vertical height of 6 m. (g = 10 m/s²)
Formula
W = mgh (work against gravity)
Substitute
W = 200 × 10 × 6
Answer
W = 12000 J
Q3 · 3 Marks
A 5 N force acts on a body at 30° to the horizontal and moves it 6 m horizontally. Find the work done.
Formula
W = Fs cos θ (cos 30° = √3/2)
Substitute
W = 5 × 6 × (√3/2) = 30 × 0.866
Answer
W = 15√3 ≈ 25.98 J
Q4 · 3 Marks
A body has 5 J of kinetic energy while moving at 2 m/s. Find its mass.
Formula
KE = ½mv² → m = 2KE/v²
Substitute
m = (2 × 5)/2² = 10/4
Answer
m = 2.5 kg
Q5 · 5 Marks
What work is needed to raise a car's speed from 30 km/h to 60 km/h? Its mass is 1500 kg.
For a body moving at steady speed against a force, power = force × velocity — often faster than W/t.
9
Machines don't make energy
An MA > 1 multiplies force but over a longer distance — work in ≈ work out. Energy is conserved.
10
Cross-check by a 2nd method
JEE habit: verify a numerical with energy and with kinematics/F = ma. Two routes, one answer.
CBSE Pattern Practice
Practice Question Bank
MCQ · Assertion–Reason · VSA (1M) · SA (2M / 3M) · LA (5M) · Application (4M) — covering every question type in the chapter exercise, each with a step-by-step solution.
⚡ Beyond the Textbook
JEE Foundation Challenge Set
Original problems on the same Class 9 toolkit (work, energy, power, conservation, machines) with a little extra reasoning. Try each before opening the solution.
Bonus problems — tap a card's button to reveal the worked solution.