Work, Energy, and Simple Machines

CLASS 9 SCIENCE · CHAPTER 7

Work, Energy, and Simple Machines

Study notes to score 100% — concepts, definitions, examples & self-check.
🧠 What This Chapter Covers Work, Energy & Machines
Work
  • W = F × s
  • Unit: joule (J)
  • Positive / negative / zero
Energy
  • Capacity to do work
  • Kinetic = ½mv²
  • Potential = mgh
Conservation & Power
  • KE + PE = constant
  • Power = W/t
  • Unit: watt (W)
Simple Machines
  • Pulley, inclined plane, lever
  • Mechanical advantage
  • MA = load/effort
7.1 Work Done by a Constant Force
  • In science, work has a precise meaning — it is done only when a force moves an object.
  • Lifting a bag to a height needs an upward force = mg; the force acts through the displacement → work is done.
  • A larger force over the same distance → proportionally more work (3 bags = 3× work).
  • The same force over a larger distance → proportionally more work (1 bag to 3 m = 3× work).
Work Done EXAM · 2 marks
Work done by a constant force = force applied × displacement in the direction of the force.   W = F × s
Work Done by a Constant Force W = F × s Works for horizontal, vertical or any direction of force
SI Unit of Work
  • SI unit of work = joule (J).
  • 1 J = 1 N × 1 m — work done when 1 N moves an object 1 m in the direction of the force.
  • 1 J = 1 kg m² s⁻².
Force-displacement graph, area equals work done
👀 LOOK HERE: On a force–displacement graph, the area under the line = work done. Here 10 N × 1 m = 10 J.
📌 Note Work has no direction — it is a number with a positive or negative sign.
7.1.1 When is Work Done Equal to Zero?
  • Work is zero if force = 0.
  • Work is zero if displacement = 0 — e.g. pushing a rigid wall.
  • Work is zero if force is perpendicular to displacement — e.g. carrying a box while walking.
  • You feel tired pushing a wall, but scientifically no work is done — muscles use internal energy.
7.1.2 Positive and Negative Work
  • Positive work: displacement in the same direction as force — pushing a wheelchair.
  • Negative work: displacement opposite to force — a goalkeeper stopping a ball.
Positive and negative work
👀 LOOK HERE: Force & displacement same way → positive work. Opposite → negative work.
✍️ Example 7.1 (method — study this)

A girl lifts a dumbbell and slowly lowers it. When is the work positive/negative?

Lifting up: force (up) same as displacement → positive work.
Lowering down: force (up) opposite to displacement → negative work.
✍️ Example 7.2 (method — study this)

A goalkeeper hand moves back 15 cm stopping a ball with 200 N. Work done on the ball?

Force opposes motion → displacement negative. W = 200 × (−0.15) = −30 J.
📝 Check Your Concepts
QDefine work and give its SI unit.
Work = force × displacement in the direction of the force. SI unit = joule (J); 1 J = 1 N × 1 m.
QGive three cases when work done is zero.
(1) Force = 0. (2) Displacement = 0 (pushing a wall). (3) Force perpendicular to displacement (carrying a box while walking).
QWhen is work positive and when negative?
Positive: displacement same direction as force. Negative: displacement opposite to force.
🧮 Numerical Practice — try, then tap to check
NA force of 15 N moves a box 4 m in its direction. Find the work done.
W = 15 × 4 = 60 J.
NA player stops a ball applying 250 N; the ball moves 0.2 m against the force. Work done on the ball?
W = 250 × (−0.2) = −50 J.
PART 2
Work-Energy Theorem & Kinetic Energy
🧠 Section at a Glance Work → Energy → Motion
Energy
  • Capacity to do work
  • Unit: joule (J)
  • Many forms
Work-Energy Theorem
  • Work = change in energy
  • Holds for systems too
Forms of Energy
  • Mechanical, thermal
  • Light, sound, electrical
  • Nuclear, chemical
Kinetic Energy
  • Energy of motion
  • K = ½mv²
  • Doubles v → 4× KE
7.2 The Work-Energy Theorem
  • When positive work is done on an object, it gains energy.
  • An object with the capacity to do work is said to possess energy.
  • A thrown ball or a raised flowerpot gains energy from the work done on it, and can then do work on something else.
Work-Energy Theorem EXAM · 2 marks
Work done on an object = change in its energy. It holds for a system of objects and even when forces are not constant.
  • SI unit of energy = joule (J) — same as work.
✍️ Example 7.3 (method — study this)

In carrom, a striker hits the white coin, which hits the black coin. Identify the work and energy changes.

Striker does positive work on the white coin (energy increases); the white coin does negative work on the striker (its energy decreases). Similarly, the white coin does positive work on the black coin, and the black coin does negative work on the white coin.
7.3 Forms of Energy
  • Energy exists in many forms and can be converted from one to another.
  • Examples: electrical → light (bulb); chemical (food) → mechanical (muscles); mechanical → sound (bell).
Different forms of energy chart
👀 LOOK HERE: Mechanical · Thermal · Light · Sound · Electrical · Nuclear · Chemical — energy takes many forms.
7.4 Mechanical Energy
  • Mechanical energy = energy due to an object’s motion or position.
  • It has two types: kinetic energy (motion) and potential energy (position).
7.4.1 Kinetic Energy
  • Kinetic energy = energy possessed by an object due to its motion.
  • All moving objects have kinetic energy (a rolling ball, a moving bicycle).
Kinetic energy using work-energy theorem
👀 LOOK HERE: Work done by force F over displacement s gives the object its kinetic energy.
Kinetic Energy EXAM · formula
The energy an object has due to its motion. For mass m and velocity v:
K = ½ m v²  (SI unit: joule)
Kinetic Energy K = ½ m v² Derived from W = ½m(v² − u²), with u = 0
  • KE has no direction. More positive work → more speed → more KE.
  • If v doubles, KE becomes 4 times (since KE ∝ v²).
✍️ Example 7.4 (method — study this)

If a vehicle’s velocity doubles, what happens to its kinetic energy?

KE ∝ v². Doubling v → KE = ½m(2v)² = 4 × ½mv². So KE becomes 4 times the original.
✍️ Example 7.5 (method — study this)

A cricket ball of mass 0.2 kg is bowled at 154.8 km/h. Find its kinetic energy.

v = 154.8 km/h = 43 m/s.
K = ½mv² = ½ × 0.2 × 43² = 184.9 J.
✍️ Example 7.6 (method — study this)

A 15000 kg jet lands and is stopped in 100 m by a wire exerting 367500 N backward. Find its landing velocity.

Work by wire = F × s = 367500 × (−100) = −36750000 J.
By work-energy theorem this equals the change in KE (0 − ½mv²).
½ × 15000 × v² = 367500 × 100 → v² = 4900 → v = 70 m/s (252 km/h).
📝 Check Your Concepts
QState the work-energy theorem.
The work done on an object equals the change in its energy.
QDefine kinetic energy and give its formula.
The energy an object possesses due to its motion; K = ½mv².
QIf the speed of a body doubles, how does its KE change?
KE ∝ v², so it becomes four times.
🧮 Numerical Practice — try, then tap to check
NFind the KE of a 2 kg ball moving at 10 m/s.
K = ½ × 2 × 10² = 100 J.
NTwo objects of mass m and 4m have the same KE. Find the ratio of their speeds.
½mv₁² = ½(4m)v₂² → v₁² = 4v₂² → v₁/v₂ = 2 : 1.
PART 3
Potential Energy & Conservation of Energy
🧠 Section at a Glance Stored Energy & Conservation
Potential Energy
  • Energy of position/shape
  • U = mgh
  • Unit: joule (J)
Sources of PE
  • Stretched band, spring
  • Separated magnets/charges
  • Raised object (gravity)
Mechanical Energy
  • KE + PE
  • Stays constant
  • If no external force
Conservation
  • Falling body: PE→KE
  • Pendulum swing
  • ME = mgh throughout
7.4.2 Potential Energy
  • A stretched rubber band or bent bow stores energy — released, it does work on an object (gives it KE).
  • A stretched/compressed spring stores energy in its deformed shape.
  • Separated magnets or charges store energy due to their relative positions.
  • A raised object + Earth store energy due to their separation.
Potential Energy EXAM · 1 mark
The energy stored in an object due to its deformation, or in a system due to the relative positions of objects.
Gravitational Potential Energy
  • The simplest case: the Earth–ball system. Since Earth barely moves, we call it the PE of the ball.
  • Raising a ball higher needs more work → it stores more PE (drops deeper into sand).
  • Work to raise mass m to height h = mg × h → this becomes the PE.
Raising an object to height h, U = mgh
👀 LOOK HERE: Lifting mass m through height h stores potential energy U = mgh.
Gravitational Potential Energy U = m g h Unit: joule (J) — same as work and kinetic energy
✍️ Example 7.7 (method — study this)

A fielder throws a 200 g ball 10 m high. Find its PE at the top (g = 10 m/s²).

U = mgh = 0.2 × 10 × 10 = 20 J.
7.4.3 Conservation of Mechanical Energy
  • Mechanical energy = KE + PE.
  • For a freely falling body, as it drops: PE decreases, KE increases — but their sum stays constant.
  • At the top: all PE (= mgh), KE = 0. As it falls, PE converts to KE. Total ME = mgh throughout.
Conservation of Mechanical Energy EXAM · 3 marks
When only the gravitational force acts (no friction/external force), the total mechanical energy (KE + PE) stays constant.
Pendulum showing PE and KE conversion
👀 LOOK HERE: At the ends → only PE. At the bottom → only KE. Total energy stays the same (ignoring friction).
  • Pendulum: at the extreme points, KE = 0, PE = max; at the bottom, PE = 0, KE = max.
  • In real life a pendulum slows down — energy is lost to friction and air resistance.
✍️ Example 7.8 (method — study this)

Find the speed of a child reaching the bottom of a slide of height h.

PE at top = mgh converts fully to KE at the bottom (ignoring friction):
½mv² = mgh → v = √(2gh).
Speed depends only on height h, not on the slide’s shape or the child’s mass.
✍️ Example 7.9 (method — study this)

A 10000 kg truck at 72 km/h runs onto a 30° sand ramp; sand exerts 50000 N. Find the ramp length to stop it (g = 10, rises 1 m per 2 m along ramp).

v = 72 km/h = 20 m/s. Initial KE = ½ × 10000 × 20² = 2000000 J.
Along distance d: height = d/2, so final PE = mg(d/2) = 10000 × 10 × d/2 = 50000d.
Work by sand = −50000d = final energy − initial energy = 50000d − 2000000.
2000000 = 100000d → d = 20 m.
📝 Check Your Concepts
QDefine potential energy and give the formula for gravitational PE.
Energy stored due to deformation or relative position. Gravitational PE = mgh.
QState the law of conservation of mechanical energy.
When only gravity acts (no friction), the total mechanical energy (KE + PE) remains constant.
QWhy does a real pendulum eventually stop?
Energy is gradually lost to friction at the support and air resistance.
🧮 Numerical Practice — try, then tap to check
NFind the PE of a 5 kg object raised to 4 m (g = 10 m/s²).
U = mgh = 5 × 10 × 4 = 200 J.
NA body falls from height 20 m. Find its speed just before hitting the ground (g = 10 m/s²).
v = √(2gh) = √(2 × 10 × 20) = √400 = 20 m/s.
PART 4
Power & Simple Machines
🧠 Section at a Glance Power & Machines
Power
  • Rate of doing work
  • P = W/t
  • Unit: watt (W)
Simple Machines
  • Make work easier
  • MA = load/effort
  • Don’t reduce total work
Pulley & Incline
  • Fixed pulley: MA = 1
  • Incline: MA = L/h
Lever
  • Fulcrum, load, effort
  • MA = effort arm/load arm
  • 3 classes
7.5 Power
  • Power = the rate at which work is done.
  • Running upstairs in 1 minute vs walking in 5 minutes — same work, but different power.
Power EXAM · 1 mark
The rate of doing work: P = W / t. SI unit = watt (W); 1 W = 1 J s⁻¹.
Power P = W / t 1 watt = 1 joule per second · 1 horsepower (hp) = 746 W
✍️ Example 7.10 (method — study this)

A weightlifter lifts 75 kg by 2 m in 5 s. Find the power (g = 10).

Work = mgh = 75 × 10 × 2 = 1500 J. Power = W/t = 1500 ÷ 5 = 300 W.
✍️ Example 7.11 (method — study this)

A 1000 kg car reaches 72 km/h from rest in 10 s. Find the engine power.

v = 20 m/s. Work = ΔKE = ½ × 1000 × 20² − 0 = 200000 J.
Power = W/t = 200000 ÷ 10 = 20000 W.
7.6 Simple Machines
  • Total work can’t be reduced, but a simple machine makes a task easier by changing the magnitude or direction of the force.
  • The force we apply = effort; the force to overcome = load.
Mechanical Advantage EXAM · formula
The ratio of load to effort: MA = load / effort. It shows how much a machine multiplies the applied force.
7.6.1 Pulley
  • A pulley is a grooved wheel that guides a rope.
  • A fixed pulley only changes the direction of the force (pull down to lift up) — easier, but MA = 1 (load = effort).
  • Movable/compound pulleys can give MA > 1 (used in cranes, elevators).
Pulley - direct lift vs using a pulley
👀 LOOK HERE: A fixed pulley lets you pull down to lift a load up — same force, easier direction.
7.6.2 Inclined Plane
  • An inclined plane (ramp) helps move a heavy load to a higher level with less force.
  • Longer/shallower ramp → smaller effort, but you push over a larger distance.
Inclined plane - three cases of force vs distance
👀 LOOK HERE: Longer ramp (larger L) → smaller force needed. Work stays the same.
Inclined Plane — Mechanical Advantage MA = L / h L = length of the ramp · h = height. Since L > h, MA > 1
✍️ Example 7.12 (method — study this)

A ramp raises an object over a 30 cm step; ramp width 40 cm, length 50 cm. Find the MA.

MA = L/h = 50/30 = 1.67.
7.6.3 Lever
  • A lever is a rigid bar that rotates about a fixed point (fulcrum).
  • Three parts: fulcrum (pivot), load (force to overcome), effort (force applied).
  • Distance of load from fulcrum = load arm; distance of effort from fulcrum = effort arm.
  • Balance condition: effort × effort arm = load × load arm.
Lever showing fulcrum, load, effort, load arm, effort arm
👀 LOOK HERE: A longer effort arm lets a small effort lift a big load. Fulcrum is the pivot.
Lever — Mechanical Advantage MA = effort arm / load arm A longer effort arm → larger MA → smaller effort needed
✍️ Example 7.13 (method — study this)

A seesaw has seats A, B, D, E; fulcrum C; AC = EC = 2 m, BC = DC = 1 m. Where should 15 kg and 30 kg children sit to balance it?

Put 15 kg at A (2 m from C). Balance: 15 × 2 = 30 × L → L = 1 m.
So the 30 kg child sits at seat D (1 m from the fulcrum).
Three Classes of Levers
Three classes of levers - Class I, II, III
👀 LOOK HERE: Class I — fulcrum in middle (scissors). Class II — load in middle (wheelbarrow). Class III — effort in middle (tongs).
📋 At a Glance — Summary
  • Work is done when a force displaces an object in the direction of the force.
  • An object able to do work possesses energy.
  • Work-energy theorem: work done = change in energy.
  • Kinetic energy = ½mv² (motion); potential energy = mgh (position).
  • Power = rate of doing work = W/t.
  • Simple machines make work easier by changing the size or direction of force — they don’t reduce total work.
🌙 Night-Before Quick Recap
  • Work: W = F × s; unit joule (J); zero if F=0, s=0, or F⊥s.
  • Positive work: force & displacement same way; negative: opposite.
  • Work-energy theorem: work = change in energy.
  • Kinetic energy: K = ½mv²; double v → 4× KE.
  • Potential energy: U = mgh.
  • Conservation: KE + PE = constant (no friction); falling body ME = mgh.
  • Power: P = W/t; unit watt (W); 1 hp = 746 W.
  • MA: load/effort. Fixed pulley MA = 1; incline MA = L/h; lever MA = effort arm/load arm.
  • Lever balance: effort × effort arm = load × load arm.
  • 3 lever classes: I fulcrum-middle, II load-middle, III effort-middle.
⚠️ Common Mistakes
  • Work ≠ effort/tiredness: pushing a wall = zero work (no displacement).
  • KE ∝ v² not v: double the speed → four times the KE.
  • joule vs watt: joule = energy/work; watt = power (J/s).
  • Simple machines don’t reduce work — they only change the force’s size/direction.
  • Negative work sign: remember the minus when force opposes displacement.
  • MA of a fixed pulley = 1 (it only changes direction, not force).
📝 Check Your Concepts
QDefine power and give its SI unit.
Power is the rate of doing work, P = W/t. SI unit = watt (W).
QWhat is mechanical advantage? Give the formula for a lever.
MA = load/effort. For a lever, MA = effort arm / load arm.
QWhy is the mechanical advantage of a fixed pulley 1?
Because the effort equals the load — it only changes the direction of the force, not its size.
🧮 Numerical Practice — try, then tap to check
NA machine does 600 J of work in 4 s. Find its power.
P = W/t = 600 ÷ 4 = 150 W.
NA ramp of length 3 m raises a load to a height of 1 m. Find its mechanical advantage.
MA = L/h = 3/1 = 3.
PART 5
Revise, Reflect, Refine — All Questions Solved

Every NCERT end-of-chapter question. Try first, then tap to reveal the answer.

1State True or False.
(i) Work done when a force is applied even if the object doesn’t move — False (no displacement, no work).
(ii) Lifting a bucket vertically upward = positive work — True.
(iii) SI unit for both work and energy is joule (J) — True.
(iv) A motionless stretched rubber band has kinetic energy — False (it has potential energy).
(v) Energy can change from one form to another — True.
2Fill in the blanks.
(i) Work done = force × displacement (in the direction of force).
(ii) 1 joule when a force of 1 newton displaces an object by 1 metre.
(iii) Kinetic energy of a body = ½mv².
(iv) Potential energy at height h = mgh.
(v) Power is the rate at which work is done.
3When a ball thrown upward reaches its highest point, which statements are correct?
At the highest point the ball is momentarily at rest. (iii) Its kinetic energy is zero and (iv) its potential energy is maximum are correct. (Force of gravity still acts, and acceleration = g, so (i) and (ii) are false.)
4Identify the energy transformation in each situation.
(i) Truck moving uphill: chemical → kinetic + potential.
(ii) Unwinding watch spring: potential (elastic) → kinetic.
(iii) Photosynthesis: light → chemical.
(iv) Water flowing from a dam: potential → kinetic.
(v) Burning matchstick: chemical → heat + light.
(vi) Firecracker explosion: chemical → heat + light + sound.
(vii) Speaking into a microphone: sound → electrical.
(viii) Glowing bulb: electrical → light + heat.
(ix) Solar panel: light → electrical.
5A 50 kg student is lifted (elevator) and later climbs stairs to height h = 72.5 m (g = 10). Find PE gains and the conclusion.
(i) Lifted straight up: PE = mgh = 50 × 10 × 72.5 = 36250 J.
(ii) Climbing stairs to the same top: PE = mgh = 36250 J (same).
(iii) Conclusion: gravitational PE depends only on the height, not on the path taken.
6A crane lifts mass m to the 10th floor in time t, then to the 20th floor in double the time. How much more energy and power?
20th floor = double the height → double the work/energy. Energy: twice as much.
Power = work/time. Double work in double time → power = same. Power: unchanged.
7Which factors decide the energy to raise a flag by pulley? Does raising it slowly/quickly change the work? If speed doubles, how does power change?
Work depends on the flag’s weight (mg) and the height — not on speed. So raising slowly or quickly does the same work.
Power = work/time. If speed doubles, time halves → power doubles.
8A 60 kg man rides a 100 kg scooter to speed v. Next day his 40 kg son joins. Same speed, same time. Ratio of fuel used on the two days?
Fuel ∝ kinetic energy = ½mv² (same v).
Day 1 mass = 60 + 100 = 160 kg. Day 2 mass = 160 + 40 = 200 kg.
Ratio = 160 : 200 = 4 : 5.
9On a seesaw, an adult weighs twice the child, yet it balances. Draw the distances from the fulcrum.
Balance: load × load arm = effort × effort arm. Since the adult’s weight is double, the adult must sit at half the distance from the fulcrum compared to the child. (e.g. child at 2 m, adult at 1 m.)
10A 2 kg ball is thrown up at 20 m/s.
(i) Going up: gravity opposes motion → negative work. Coming down: gravity aids motion → positive work.
(ii) Ideal height (no air) = v²/2g = 20²/(2×10) = 20 m. Actual = 19.4 m, so 0.6 m less.
Work by air resistance = −mg × (height lost as KE) ≈ energy difference = mg × 20 − mg × 19.4 = 2 × 10 × 0.6 = −12 J (approx, energy lost to air).
11A 10 kg block, KE = 180 J at 0 m; force applied per Fig 7.37 from 0 to 4 m. Find speed at 0 m and at 4 m; any negative acceleration?
Q11 force-displacement graph
👀 Work = area under the force–displacement graph.
(i) At 0 m: ½ × 10 × v² = 180 → v² = 36 → v = 6 m/s.
Work = area of trapezoid = ½ × (sum of parallel sides) × height. Here area ≈ ½ × (4 + 2) × 50 = 150 J (using the graph’s shape).
(ii) KE at 4 m = 180 + work done = 180 + 150 = 330 J → v = √(2×330/10) = √66 ≈ 8.1 m/s.
The force is always in the direction of motion (positive), so there is no negative acceleration in this region.
12Moon’s gravity ≈ 1/6 of Earth’s. A ball thrown up reaches 8 m on Earth. How high on the Moon (same speed)?
Height ∝ 1/g (since h = v²/2g). Moon’s g is 1/6 → height is 6 times.
Height on Moon = 8 × 6 = 48 m.
13A 1000 kg car: speed-time graph (Fig 7.38). (i) motion A→B, (ii) KE at A, (iii) work by brakes B→C, (iv) what KE transforms into.
Q13 speed-time graph
👀 A→B flat (constant speed); B→C slopes down (braking).
(i) A→B: the car moves at constant speed (35 m/s).
(ii) KE at A = ½ × 1000 × 35² = 612500 J.
(iii) Work by brakes = change in KE = 0 − 612500 = −612500 J.
(iv) The car’s KE transforms mainly into heat (in the brakes) and some sound.
14PE-displacement graph of a 0.5 kg ball (Fig 7.39). At O, v = 0 and PE = 30 J. Find v at P, Q, R.
Q14 PE-displacement graph
👀 Total energy = 30 J everywhere (frictionless). KE = 30 − PE at each point.
Total mechanical energy = 30 J (since at O, KE = 0 and PE = 30).
At P (PE = 20 J): KE = 10 J → v = √(2×10/0.5) = √40 ≈ 6.3 m/s.
At Q (PE = 30 J): KE = 0 → v = 0 m/s.
At R (PE = 40 J): KE would be −10 J — impossible, so the ball cannot reach R.
15A 1.5 kg coconut falls from a 10 m tree onto sand (g = 10).
(i) v = √(2gh) = √(2 × 10 × 10) = √200 ≈ 14.1 m/s.
(ii) All energy = mgh = 1.5 × 10 × 10 = 150 J goes into the depression. Sand force 3000 N × depth d = 150 → d = 150/3000 = 0.05 m (5 cm).
🧪 Test Yourself — 25-Question Quiz