Class 9 Science · Chapter 7

7. Work, Energy, and Simple Machines

Complete solutions to Revise, Reflect, Refine. CBSE Class 9 Science (NCERT Exploration).

Revise, Reflect, Refine

1State whether True or False.

(i) Work is said to be done when a force is applied, even if the object does not move. — False

Work is done only when the object is displaced.

(ii) Lifting a bucket vertically upward results in positive work done on the bucket. — True

(iii) The SI unit for both work and energy is joule (J). — True

(iv) A motionless stretched rubber band has kinetic energy. — False

It is not moving. It has elastic 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 of work is done when a force of 1 newton displaces an object by 1 metre in the direction of the force.

(iii) The expression for kinetic energy of a body of mass m and velocity v is ½mv².

(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is mgh.

(v) Power is defined as the rate at which work is done.

3When a ball thrown upwards reaches its highest point, which of the following statement(s) are correct?

Answer: (iii) Its kinetic energy is zero. and (iv) Its potential energy is maximum.

At the highest point the ball stops for a moment, so its kinetic energy is zero. Gravity still acts on it, so the force and acceleration (g) are not zero.

4Identify the energy transformation that takes place in each situation.

Situation	Energy transformation
(i) A truck moving uphill	Chemical (fuel) → kinetic + potential energy (and some heat)
(ii) Unwinding of a watch spring	Elastic potential → kinetic energy
(iii) Photosynthesis in green leaves	Light (solar) → chemical energy
(iv) Water flowing from a dam	Potential → kinetic energy
(v) Burning of a matchstick	Chemical → heat + light energy
(vi) Explosion of a fire cracker	Chemical → heat, light, sound and kinetic energy
(vii) Speaking into a microphone	Sound → electrical energy
(viii) A glowing electric bulb	Electrical → light + heat energy
(ix) A solar panel	Light (solar) → electrical energy

5A student (m = 50 kg) is lifted in an elevator to the top of a building (h = 72.5 m). Later, she climbs the stairs to the top. Take g = 10 m s⁻².

(i) Gain in PE (elevator) = mgh = 50 × 10 × 72.5 = 36,250 J
(ii) Gain in PE (stairs) = mgh = 50 × 10 × 72.5 = 36,250 J

(iii) The gain in potential energy is the same in both cases. So potential energy does not depend on the path taken. It depends only on the vertical height.

6A crane lifts a mass m to the 10th floor in a certain time. It then raises the same mass to the 20th floor in double the time. How much more energy and power are required?

Height doubles ⇒ energy = mg(2h) = 2 × initial energy
Power = energy/time = 2E/2t = E/t = same as initial power

So twice the energy is needed, but the power remains the same.

7Which factors determine the energy required to raise a flag to the top of a flagpole using a pulley? Does raising the flag slowly or quickly change the work done? If the speed is doubled, how does the power requirement change?

Factors: the mass (weight) of the flag, the height of the pole, and the friction in the pulley and rope.

Speed and work: no change. Work = mgh, which depends on weight and height, not on speed.

Power: if the speed is doubled, the time becomes half. Power = work/time, so the power doubles.

8A man (60 kg) rides a scooter (100 kg) and accelerates to velocity v. Next day, his son (40 kg) joins him. If the scooter reaches the same speed in the same time on both days, what is the ratio of fuel used on the two days?

Day 1 mass = 60 + 100 = 160 kg; Day 2 mass = 160 + 40 = 200 kg
Fuel used ∝ kinetic energy = ½mv²
Ratio = ½ × 160 × v² : ½ × 200 × v² = 160 : 200 = 4 : 5

9On a seesaw with sliding seats, a child sits on one side and an adult on the other. The adult weighs twice the child. The seesaw is balanced. Draw a figure showing the distances from the fulcrum.

For balance, weight × distance must be equal on both sides.

W × (2d) = 2W × d

So the child must sit at twice the distance from the fulcrum compared to the adult.

The adult (2W) sits at distance d and the child (W) sits at distance 2d from the fulcrum.

10A ball of mass 2 kg is thrown up with a velocity of 20 m s⁻¹.

(i) Identify the sign of the work done by gravity during its upward and downward motion.

Upward motion: negative (gravity acts downward, opposite to displacement).
Downward motion: positive (gravity and displacement are in the same direction).

(ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance? (g = 10 m s⁻²)

Initial KE = ½ × 2 × 20² = 400 J
PE at top = mgh = 2 × 10 × 19.4 = 388 J
Work done by air resistance = 388 − 400 = −12 J

11A 10.0 kg block moves on a frictionless floor. A variable force acts in its direction of motion from 0 m to 4 m (Fig. 7.37). Its KE at 0 m was 180 J. Find its speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion?

(i) ½ × 10 × v² = 180 ⇒ v² = 36 ⇒ v = 6 m s⁻¹
Work done = area under F-x graph
= ½ × 1 × 50 + 2 × 50 + ½ × 1 × 50 = 25 + 100 + 25 = 150 J
KE at 4 m = 180 + 150 = 330 J
(ii) ½ × 10 × v² = 330 ⇒ v² = 66 ⇒ v = √66 m s⁻¹ ≈ 8.1 m s⁻¹

No. The force always acts in the direction of motion (it is never negative). So the block speeds up all the way.

12Gravity on the Moon is about 1/6th of that on the Earth. An astronaut can throw a ball up to 8 m on the Earth. How high will the ball go on the Moon with the same upward velocity?

½mv² = mgh ⇒ h = v²/2g, so h ∝ 1/g
h_Moon = 8 × 6 = 48 m

13A 1000 kg car is moving at constant speed. The driver applies the brakes and stops. Its motion is shown in Fig. 7.38.

(i) Describe how the car moves between A and B.

It moves with a constant speed of 35 m s⁻¹ for 1 s. This is the driver’s reaction time.

(ii) Calculate the kinetic energy of the car at A.

KE = ½ × 1000 × 35² = 500 × 1225 = 6,12,500 J

(iii) State the work done by the brakes in bringing the car to a halt between B and C.

Work done = change in KE = 0 − 612500 = −6,12,500 J

(iv) What does the kinetic energy of the car transform into?

Mostly into heat (thermal) energy in the brakes and tyres due to friction, and a little into sound.

14The potential energy-displacement graph of a 0.5 kg ball on a frictionless track is shown in Fig. 7.39. At O, the velocity is 0 and PE is 30 J. Calculate the velocity at P, Q and R.

Total mechanical energy = 30 J (KE at O is zero). It stays constant on a frictionless track.

At P (PE = 20 J): KE = 10 J ⇒ ½ × 0.5 × v² = 10 ⇒ v² = 40 ⇒ v = 2√10 m s⁻¹
At Q (PE = 30 J): KE = 0 ⇒ v = 0 m s⁻¹
At R (PE = 40 J): needs more than the total energy (30 J)

So the ball cannot reach R. It stops at Q and rolls back.

15A coconut of mass 1.5 kg falls from a 10 m high tree onto wet sand and comes to rest, making a depression. (g = 10 m s⁻²)

(i) Calculate the velocity of the coconut just before it hits the sand.

½mv² = mgh ⇒ v = √(2gh) = √(2 × 10 × 10) = √200 = 10√2 m s⁻¹ ≈ 14.1 m s⁻¹

(ii) If the average resistive force of sand is 3000 N, calculate the depth of the depression.

Energy of coconut = mgh = 1.5 × 10 × 10 = 150 J
Force × depth = 150 ⇒ depth = 150/3000 = 0.05 m (5 cm)

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