Class 9 Science

Chapter 7

Work, Energy and Simple Machines

Detailed solutions to all worked examples, conceptual questions and back exercises

Formula bank used below
Work: W = F × s | Kinetic energy: KE = ½mv² Potential energy: PE = mgh | Power: P = W/t Mechanical advantage: MA = Load/Effort = Effort arm/Load arm Unless stated otherwise, take g = 10 m s⁻². Always convert km h⁻¹ into m s⁻¹ before using KE.

Think It Over – Slide

What is the speed of a child at the bottom of a smooth slide of height h? Do mass and shape of the slide matter?

Solution

At the top, the child has gravitational potential energy mgh. At the bottom, this becomes kinetic energy, if friction is neglected.

mgh = ½mv²v² = 2ghv = √(2gh)

The mass m cancels. Therefore, children of different masses reach with the same speed, and slides with the same vertical height give the same speed. A larger height gives a larger speed.

Worked Examples – Fully Explained

Example 7.1 – Lifting and lowering a dumbbell

When does the girl do positive work, and when does she do negative work?

While lifting, her force and the dumbbell’s displacement are both upward.

Force and displacement in same direction ⟹ work done is positive.

While lowering slowly, she still applies an upward force to control the dumbbell, but its displacement is downward.

Force and displacement in opposite directions ⟹ work done is negative.

Example 7.2 – Goalkeeper stopping a ball

Full question: While saving a goal, a goalkeeper’s hand moved back by 15 cm as she stopped a ball while applying a force of 200 N. How much work did the goalkeeper do on the ball in stopping it?

Given: F = 200 NThe hand and ball move opposite to the goalkeeper’s applied force.Therefore, s = −15 cm = −0.15 mW = F × sW = 200 N × (−0.15 m)W = −30 J

The negative sign shows that the goalkeeper removes kinetic energy from the ball.

Example 7.3 – Carrom collision

Identify work and energy changes when a striker hits a white coin which hits a black coin.

The moving striker exerts force in the direction in which the white coin moves. It does positive work on the white coin; the white coin gains kinetic energy.
By Newton’s third law, the white coin exerts an opposite force on the striker. It does negative work on the striker; the striker loses kinetic energy.
The white coin then does positive work on the black coin; the black coin gains kinetic energy and moves towards the pocket.
The black coin does negative work on the white coin; the white coin loses some kinetic energy.

Example 7.4 – Effect of doubling speed

Initial KE = ½mv²New speed = 2vNew KE = ½m(2v)² = ½m × 4v² = 4(½mv²)The kinetic energy becomes 4 times.

Example 7.5 – Kinetic energy of a cricket ball

Full question: In one of their fastest deliveries, an Indian cricketer bowled a cricket ball of approximate mass 0.2 kg at a velocity of about 154.8 km h⁻¹. Calculate the kinetic energy of the ball at the time of delivery.

Given: m = 0.2 kgv = 154.8 km h⁻¹v = 154.8 × 5/18 = 43 m s⁻¹KE = ½mv²KE = ½ × 0.2 kg × (43 m s⁻¹)²KE = 184.9 J

Example 7.6 – Aircraft stopped by a wire

Full question: A jet aircraft of mass 15,000 kg lands on the deck of an aircraft carrier. A hook on its tail catches a wire stretched across the deck. The wire exerts an approximately constant backward force of 367,500 N and stops the jet within 100 m. What was the velocity of the aircraft just before the wire caught the hook?

Given: m = 15,000 kg, F = 367,500 N, s = −100 mWork done by wire = F × sW = 367,500 × (−100) = −36,750,000 JΔKE = KE_final − KE_initial−36,750,000 = 0 − ½ × 15,000 × v²36,750,000 = 7,500v²v² = 4,900 m² s⁻²v = √4,900v = 70 m s⁻¹ = 252 km h⁻¹

Example 7.7 – Potential energy of a cricket ball

Full question: After taking a catch, a fielder throws a cricket ball of mass 200 g high up in the air, about 10 m above the ground, in celebration. How much potential energy does the ball have when it reaches its maximum height? Assume g = 10 m s⁻².

Given: m = 200 g = 0.2 kgh = 10 mg = 10 m s⁻²PE = mghPE = 0.2 × 10 × 10PE = 20 J

Example 7.8 – Child reaching the bottom of a slide

PE at top = mghKE at bottom = ½mv²mgh = ½mv²v = √(2gh)

The answer does not contain mass. For smooth slides of the same height, the final speed is independent of mass and slide shape.

Example 7.9 – Escape ramp

Full question: Escape ramps are inclined planes filled with sand or gravel that help stop trucks when brakes fail. A truck of mass 10,000 kg is moving at 72 km h⁻¹ when its brakes fail. The driver steers it onto an escape ramp inclined at 30°, where it comes to rest. If sand exerts a force of 50,000 N opposite to the truck’s motion, what is the minimum length of the ramp needed to stop the truck? Take g = 10 m s⁻². (For a 30° incline, the truck rises 1 m vertically for every 2 m travelled along the ramp.)

Given: m = 10,000 kg; u = 72 km h⁻¹; F_sand = 50,000 Nu = 72 × 5/18 = 20 m s⁻¹Initial KE = ½mu²Initial KE = ½ × 10,000 × (20)² = 2,000,000 JLet ramp length = d mHeight gained = d/2 mGain in PE = mg(d/2)= 10,000 × 10 × d/2 = 50,000d JWork done by sand = −50,000d JWork by sand = change in mechanical energy−50,000d = (50,000d) − 2,000,000100,000d = 2,000,000d = 20 m

Example 7.10 – Power of a weightlifter

Full question: A weightlifter lifts a 75 kg mass by 2 m in 5 seconds. How much power does she require for this task?

Given: m = 75 kg; h = 2 m; t = 5 sW = mghW = 75 × 10 × 2 = 1,500 JP = W/tP = 1,500/5P = 300 W

Example 7.11 – Power of a car engine

Full question: A car of mass 1,000 kg starts from rest and reaches a speed of 72 km h⁻¹ in 10 seconds. Calculate the power of the engine required to achieve this start.

Given: m = 1,000 kg; u = 0; v = 72 km h⁻¹; t = 10 sv = 72 × 5/18 = 20 m s⁻¹W = change in KEW = ½mv² − ½mu²W = ½ × 1,000 × (20)² − 0W = 200,000 JP = W/tP = 200,000/10P = 20,000 W = 20 kW

Example 7.12 – Inclined ramp

Full question: A person uses an inclined ramp to raise an object over a step 30 cm high. The ramp has a width of 40 cm. What is the mechanical advantage of the ramp that helps the person achieve the task?

Given: height, h = 30 cm; width = 40 cmRamp length, L = √(30² + 40²)L = √(900 + 1,600) = √2,500 = 50 cmMA = length of inclined plane/vertical heightMA = L/h = 50/30MA = 1.67 (approximately)

Example 7.13 – Seesaw balance

Full question: For a seesaw having four seats A, B, D and E with fulcrum at C, AC = EC = 2 m and BC = DC = 1 m. On which seats should children of masses 15 kg and 30 kg sit to make the seesaw balanced?

Let the 15 kg child sit at A, 2 m from C.Let the 30 kg child sit at distance L from C on the other side.For balance: 15 × 2 = 30 × L30 = 30LL = 1 mThe 15 kg child sits at A and the 30 kg child sits at D.

Pause and Ponder – Answer Key

1. Is a weightlifter doing work on a barbell held steady?

W = F × s; s = 0

No work is done on the barbell because it has no displacement, though the lifter uses energy internally and feels tired.

2. What is the work done by friction on coins moving over a rough surface?

Friction acts opposite to displacement.

Work done by friction is negative.

3. What forms does muscular energy take while cycling on a flat road?

It appears mainly as kinetic energy of the bicycle and rider. Some becomes thermal energy due to friction in tyres, chain and air resistance, and a small amount becomes sound energy.

4. Masses m and 4m have equal KE. Find velocity ratio.

½m v_A² = ½(4m)v_B²v_A² = 4v_B²v_A : v_B = 2 : 1

5. Does KE change with position for constant velocity?

No. Since KE = ½mv², it depends on mass and speed. With constant mass and velocity, KE remains constant at every position.

6. Does potential energy change during horizontal and vertical motion near Earth?

For horizontal motion at constant height, h is unchanged; therefore PE = mgh is unchanged. During gradual upward vertical motion, h increases; hence PE increases.

7. Show the mechanical energy just before a falling ball hits the ground.

At the ground, h = 0. Therefore:

PE = mgh = 0From v² = 2gh, KE = ½mv² = ½m(2gh) = mghMechanical energy = KE + PE = mgh + 0 = mgh

8. Explain energy changes in a roller coaster at A, B and C.

At A (highest point), PE is maximum and KE is minimum. At B (lowest point), PE is minimum and KE is maximum. At C, the ball has gained height again, so PE increases and KE decreases compared with B. Later peaks D and E are lower because friction and air resistance change some mechanical energy into heat and sound.

9. Why are hill roads built with gentle bends?

A winding road gives a larger distance L for the same height h.

MA = L/h

A larger L gives greater mechanical advantage, so less force is needed to go uphill.

10. Why is an inclined ladder easier to climb than a vertical one?

An inclined ladder increases the distance over which the same height is gained. The same work mgh is spread over a longer path, so a smaller force is required at each step.

11. Why does a spoon make opening a can lid easier?

The spoon acts as a lever. The handle provides a long effort arm; a small effort produces a larger turning effect about the fulcrum.

12. Why cut a hard object near the scissors fulcrum?

Moving the object closer makes the load arm smaller.

MA = effort arm/load arm

Smaller load arm ⟹ larger MA ⟹ less effort needed.

13. Why do real perpetual-motion machines stop?

Friction, air resistance and deformation do negative work. Useful mechanical energy continually changes into heat and sound. As no machine creates energy, its available mechanical energy decreases until it stops.

Back Exercise – Detailed Solutions

1. True or False

Statement	Answer	Reason
(i) Force applied but object does not move.	False	Displacement is zero, so work is zero.
(ii) Lifting a bucket vertically upward.	True	Force and displacement are upward; work is positive.
(iii) SI unit of both work and energy is joule.	True	Both are measured in joule (J).
(iv) A motionless stretched rubber band has kinetic energy.	False	It has elastic potential energy, not kinetic energy.
(v) Energy changes form.	True	For example, electrical energy changes into light and heat in a bulb.

2. Fill in the blanks

Work done = force × displacement (in direction of force).
1 newton.
½mv².
mgh.
rate.

3. Ball at its highest point

At the highest point its instantaneous velocity is zero; hence its kinetic energy is zero. Gravity still acts downward, so force and acceleration are not zero. Its height is greatest, so PE is maximum.

Correct statements: (iii) and (iv).

4. Energy transformations

Situation	Transformation
(i) Truck moving uphill	Chemical energy of fuel ⟶ kinetic energy and gravitational potential energy (with heat/sound losses).
(ii) Unwinding watch spring	Elastic potential energy ⟶ kinetic energy of gears.
(iii) Photosynthesis	Light energy ⟶ chemical energy in food.
(iv) Water from a dam	Gravitational potential energy ⟶ kinetic energy.
(v) Burning matchstick	Chemical energy ⟶ heat and light energy.
(vi) Firecracker explosion	Chemical energy ⟶ heat, light, sound and kinetic energy.
(vii) Speaking into microphone	Sound energy ⟶ electrical energy.
(viii) Glowing electric bulb	Electrical energy ⟶ light and heat energy.
(ix) Solar panel	Light energy ⟶ electrical energy.

5. Student reaching the top floor

Full question: A student is slowly lifted straight up in an elevator from ground level to the top floor of a building. Later, the same student climbs the staircase all the way to the top. The height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m s⁻², and the student’s mass is m = 50 kg. (i) Find the gain in potential energy when the student is lifted straight up. (ii) Find the gain in potential energy when the student climbs the stairs to the same top. (iii) What do you conclude about dependence of potential energy on the path taken?

Given: m = 50 kg; g = 10 m s⁻²; h = 72.5 mGain in PE = mgh= 50 × 10 × 72.5= 36,250 J

(i) Elevator: gain in PE = 36,250 J.

(ii) Stairs: same mass reaches the same height, so gain in PE = 36,250 J.

(iii) Gravitational potential energy depends only on vertical height change, not on the path taken.

6. Crane at 10th and 20th floors

Full question: A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that all floors have equal height.

Let the height of one floor be H. The 10th floor corresponds to height 10H and the 20th to 20H.

Energy for 10th floor = mg(10H)Energy for 20th floor = mg(20H) = 2 × mg(10H)

So, the 20th-floor lift needs twice the energy, i.e. one extra amount equal to the first energy.

P₁₀ = mg(10H)/tP₂₀ = mg(20H)/(2t) = mg(10H)/tPower required is the same.

7. Raising a flag with a pulley

Full question: Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the work done? If its speed is doubled, how does the power requirement change? Explain.

Energy required depends on the flag’s mass m, vertical height h and gravitational acceleration g.

Work done = mgh

Raising the flag slowly or quickly through the same height does not change the work done. A fixed pulley changes the direction of the effort, not the ideal amount of work.

P = W/t

If the raising speed is doubled, the time becomes half while work stays same.

Power requirement doubles.

8. Fuel used by a scooter

Full question: A man of mass 60 kg rides a scooter of mass 100 kg and accelerates it to velocity v. The next day, his son of mass 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of fuel used on the two days? Assume energy transfer to the scooter is entirely due to fuel, with no losses due to air resistance or friction.

Day 1 total mass:

m₁ = 60 + 100 = 160 kg

Day 2 total mass:

m₂ = 60 + 40 + 100 = 200 kg

Both reach the same speed, so fuel used is proportional to the kinetic energy required.

Fuel₁ : Fuel₂ = KE₁ : KE₂= ½m₁v² : ½m₂v²= 160 : 200Fuel used on the two days = 4 : 5.

9. Balanced seesaw: adult weighs twice the child

Full question: On a seesaw with sliding seats, a child is sitting on one side and an adult on the other. The adult weighs twice as much as the child. The seesaw is balanced. Draw a figure showing the distances from the fulcrum where the child and adult are seated.

For balance, clockwise turning effect = anticlockwise turning effect.

Child’s weight × child’s distance = adult’s weight × adult’s distanceW × d_child = 2W × d_adultd_child = 2d_adult

The child must sit twice as far from the fulcrum as the adult. One possible labelled diagram is below.

Child
weight = W

Adult
weight = 2W

2d from fulcrum

d from fulcrum

Balanced because W × 2d = 2W × d

10. Ball thrown upward

Full question: A ball of mass 2 kg is thrown upward with a velocity of 20 m s⁻¹. (i) Identify the sign of the work done by gravity during upward motion and downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance? Assume g = 10 m s⁻².

(i) During upward motion, gravity is downward but displacement is upward. Work by gravity is negative. During downward motion, gravity and displacement are both downward. Work by gravity is positive.

(ii) Given:

m = 2 kg; u = 20 m s⁻¹; h = 19.4 m; g = 10 m s⁻²KE_initial = ½mu²= ½ × 2 × 20² = 400 JPE_initial = 0 J

At maximum height, v = 0, so KE_final = 0.

PE_final = mgh= 2 × 10 × 19.4 = 388 JWork by air resistance = final mechanical energy − initial mechanical energy= 388 − 400= −12 J

Air resistance removes 12 J of mechanical energy.

11. Block acted on by a variable force

m = 10 kg; initial KE = 180 J. Force rises linearly from 0 to 50 N over 0-1 m, stays at 50 N from 1-3 m, then falls to 0 from 3-4 m.

The work is the area under the force-displacement graph.

Force (N)

Displacement (m)

Area = 2 triangles + 1 rectangle

Work from 0 to 1 m = ½ × 1 × 50 = 25 JWork from 1 to 3 m = 2 × 50 = 100 JWork from 3 to 4 m = ½ × 1 × 50 = 25 JTotal work = 25 + 100 + 25 = 150 J

(i) Speed at 0 m

KE = ½mv²180 = ½ × 10 × v²v = 6 m s⁻¹

(ii) Speed at 4 m

KE_final = 180 + 150 = 330 J330 = ½ × 10 × v²v² = 66v = √66 ≈ 8.12 m s⁻¹

There is no negative acceleration. The force always acts in the direction of motion; it becomes smaller after 3 m but never reverses direction.

12. Ball thrown on the Moon

Full question: Gravitational attraction on the Moon’s surface is about 1/6 of that on Earth. An astronaut can throw a ball up to a height of 8 m from Earth’s surface. How far up will a ball thrown with the same upward velocity travel from the Moon’s surface?

For a given initial speed, maximum height is inversely proportional to g.

h = u²/(2g)g_Moon = g_Earth/6h_Moon = 6 × h_Earth = 6 × 8h_Moon = 48 m

13. Braking car

m = 1,000 kg. From the speed-time graph, speed at A and B is 35 m s⁻¹; from B to C it falls uniformly to zero.

Speed (m s⁻¹)

Time (s)

A to B: constant speed; B to C: uniform deceleration

(i) From A to B, the car moves with constant speed 35 m s⁻¹. Its acceleration is zero.

(ii)

KE at A = ½mv² = ½ × 1,000 × 35²KE at A = 612,500 J

(iii) At C the car is at rest, so final KE = 0.

Work by brakes = ΔKE = 0 − 612,500Work done by brakes = −612,500 J

(iv) The car’s kinetic energy transforms mainly into heat energy in the brake pads, tyres and road, with a small amount of sound energy.

14. Potential energy-displacement graph

m = 0.5 kg. At O, v = 0 and PE = 30 J. The track is frictionless.

Total mechanical energy at O:

E = PE + KE = 30 + 0 = 30 J

From the graph: PE at P = 20 J, at Q = 30 J and at R = 40 J.

At P

KE = 30 − 20 = 10 J10 = ½ × 0.5 × v²v² = 40v_P = √40 ≈ 6.32 m s⁻¹

At Q

KE = 30 − 30 = 0v_Q = 0 m s⁻¹

At R

Required PE at R = 40 J, but total energy available = 30 JThe ball cannot reach R; it turns back at Q.

15. Coconut falling into wet sand

m = 1.5 kg; h = 10 m; resistive force = 3,000 N; g = 10 m s⁻².

(i) Speed just before impact

mgh = ½mv²v = √(2gh) = √(2 × 10 × 10)v = √200 ≈ 14.14 m s⁻¹

(ii) Depth of depression

Energy just before impact = mgh = 1.5 × 10 × 10 = 150 JWork done against sand = F × d3,000 × d = 150d = 150/3,000 = 0.05 mDepth of depression = 0.05 m = 5 cm

Full question wording for the graph-based numericals

Q11: A 10.0 kg block is moving on a horizontal floor with negligible friction. A variable force is applied in the direction of motion from 0 m to 4 m. The force increases linearly from 0 N to 50 N from 0 m to 1 m, stays 50 N from 1 m to 3 m, and decreases linearly to 0 N from 3 m to 4 m. The block has kinetic energy 180 J at 0 m. Find its speed at 0 m and 4 m. Does it have negative acceleration at any stage?

Q13: A 1000 kg car moves at constant speed. The driver applies brakes after noticing an obstruction. Its speed-time graph shows 35 m s⁻¹ from A to B and a uniform decrease to zero from B to C. Describe its motion from A to B, calculate KE at A, state work done by the brakes from B to C, and state the final form of the kinetic energy.

Q14: A 0.5 kg ball moves on a frictionless track. At O, its velocity is 0 m s⁻¹ and potential energy is 30 J. Using the potential-energy graph, calculate the velocities at P, Q and R.

Q15: A 1.5 kg coconut falls from a height of 10 m onto wet beach sand and comes to rest after making a depression. Calculate its speed just before impact. If the average sand force is 3000 N and all energy makes the depression, calculate the depth. Take g = 10 m s⁻².

Exam reminder: Write the formula first, substitute SI units on the next line, show each calculation, and box the final answer with its unit. In work questions, state the direction to justify the sign.