Answers · Energy, work and power

Work and power — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. EQ12 · Practice

Model answer. The force is horizontal and the sledge moves 3.0 m horizontally, so W = F d = 40 N × 3.0 m = 120 J. The child does 120 J of work and transfers 120 J of energy.

Checked automatically

Work done: 120 J (exact value)
- Common wrong answer 13.3 : Multiply the force by the distance, do not divide.
- Common wrong answer 43 : Multiply, do not add.

A full answer includes

Method: I multiplied the force, 40 N, by the distance, 3.0 m.
Answer: I wrote 120 J.
I did not divide the force by the distance.

Also accept

120 J

Watch for

“Work done = force ÷ distance.” — Work done is force multiplied by distance: a bigger force or a longer distance means more work.

2. EQ13 · Practice

Model answer. A force does work on the wheelbarrow only when the wheelbarrow moves in the direction of the force. Pushed 15 m along the path, it moves in the direction of her push, so her push does work on it and transfers energy. Held still, stuck in the mud, or resting on the grass, the wheelbarrow does not move, so the distance is zero and these forces do no work on it, even though holding and pushing are tiring.

A full answer includes

I chose: she pushes it 15 m along a level garden path.
I did not choose holding it still: it does not move, so no work is done on it.

Also accept

she pushes it 15 m along a level garden path

Watch for

“She holds its handles up while she stands still.” — The wheelbarrow does not move, so the distance is zero and her force does no work on it. Her arms tire because energy is transferred inside her muscles.
“She pushes hard, but its wheel is stuck in the mud.” — Her push does not move the wheelbarrow, so the distance moved is zero and her push does no work on it.
“It rests on the grass, which pushes up on it.” — The grass pushes up, but the wheelbarrow does not move, so this force does no work on it.

3. EQ14 · Practice

Model answer. The push acts along the ramp, and the trolley moves 6.0 m along the ramp, so that is the distance moved in the direction of the force. W = F d = 120 N × 6.0 m = 720 J. The 1.5 m is the height gained, which is not in the direction of the push.

Checked automatically

Work done: 720 J (exact value)
- Common wrong answer 180 : That uses the height. The push is along the ramp, so use 6.0 m.
- Common wrong answer 20 : Multiply the force by the distance.

A full answer includes

Method: I used the distance along the ramp, 6.0 m, because the push is along the ramp.
Method: I multiplied 120 N by 6.0 m.
Answer: I wrote 720 J.
I did not use the height, 1.5 m.

Also accept

720 J

Watch for

“Use the height, 1.5 m, because the trolley is being raised.” — The force here is the push along the ramp, so use the distance along the ramp. The height goes with the weight, not with the push.

4. EQ15 · Practice

Model answer. (a) Electrical work done is equal to the energy transferred, so the battery transfers 450 J. At most, all of it could do work on the weight. The motor pulls up with a force equal to the weight, 15 N, so W = F d gives d = W ÷ F = 450 J ÷ 15 N = 30 m. (b) A real motor gets warm and hums: some of the 450 J heats the motor and the wires, and some leaves as sound. Less than 450 J of work is done on the weight, so it is lifted less than 30 m.

Checked automatically

(a) Greatest height: 30 m (within ±0.5 m)
- Common wrong answer 6750 : Divide the work by the force, do not multiply: d = W ÷ F.
- Common wrong answer 0.033 : Divide the work by the force, not the force by the work.

A full answer includes

(a) Method: I used 450 J as the most work the motor could do on the weight.
(a) Method: I divided 450 J by the 15 N weight.
(a) Answer: I wrote 30 m.
(a) I did not multiply 450 J by 15 N.
(b) I wrote that some of the 450 J heats the motor and the wires.
(b) I wrote that less than 450 J of work is done on the weight, so it rises less than 30 m.

Also accept

(a) 30 m (b) some of the energy heats the motor and the wires, so less work is done on the weight

Watch for

“Electrical work is not the same as energy transferred.” — Mechanical or electrical, work done is always equal to the energy transferred. The battery did 450 J of work, so it transferred 450 J.
“All the battery’s energy goes into lifting the weight.” — A motor gets warm and hums. Some of the energy heats the motor and the wires, or leaves as sound, so less goes to the weight.

5. EQ25 · Practice

Model answer. To stop her, the braking force must transfer all 1800 J out of the kinetic store, so it does 1800 J of work, transferring the energy to the internal store of the brakes, the wheels and the air. W = F d, so d = W ÷ F = 1800 J ÷ 240 N = 7.5 m.

Checked automatically

Braking distance: 7.5 m (within ±0.05 m)
- Common wrong answer 432000 : Divide the work by the force, do not multiply: d = W ÷ F.
- Common wrong answer 0.13 : Divide the work by the force, not the force by the work.
- Common wrong answer 15 : There is no 2 here. W = F d, so divide the 1800 J by the 240 N once.
- Common wrong answer 3.75 : There is no ½ here. All 1800 J is transferred by the braking force: d = W ÷ F.

A full answer includes

Method: I wrote that the brakes must do 1800 J of work to stop her.
Method: I divided 1800 J by 240 N.
Answer: I wrote 7.5 m.
I did not multiply 1800 J by 240 N.

Also accept

7.5 m

Watch for

“A bigger braking force means a longer braking distance.” — The work needed to stop her is fixed, 1800 J. With d = W ÷ F, a bigger force means a shorter distance.
“The energy is destroyed when she brakes.” — The friction in the brakes transfers it to the internal store of the brakes, the wheels and the air.

6. EQ16 · Practice

Model answer. Power is the work done per unit time: P = W / t = 6000 J ÷ 25 s = 240 W. The motor does 240 J of work every second.

Checked automatically

Power: 240 W (exact value)
- Common wrong answer 150000 : Divide the work by the time, do not multiply.
- Common wrong answer 0.0042 : Divide the work by the time, not the time by the work.

A full answer includes

Method: I divided the work, 6000 J, by the time, 25 s.
Answer: I wrote 240 W.
I did not multiply the work by the time.

Also accept

240 W

Watch for

“Power = work done × time.” — Power is how much work is done each second, so divide the work by the time.

7. EQ17 · Practice

Model answer. Both hoists lift the same weight through the same height, so they do the same work: W = F d. Power is the work done per unit time, so the time is t = W ÷ P. With the same W and three times the P, hoist B takes a third of the time.

A full answer includes

I chose: both do the same work, and hoist B takes a third of the time.
I did not choose an option in which hoist B does more work.

Also accept

both do the same work, and hoist B takes a third of the time

Watch for

“Hoist B does three times the work, and takes the same time.” — The work depends only on the weight and the height, which are the same for both. A greater power means the same work in less time.
“Hoist B does three times the work, in a third of the time.” — The work is the same for both, because the weight and the height are the same. Only the time changes.
“Both do the same work, and hoist B takes three times as long.” — A greater power means the work is done faster, not slower: t = W ÷ P, so three times the power means a third of the time.

8. EQ18 · Practice

Model answer. First change the units: 1.5 kW = 1500 W, so the hairdryer transfers 1500 J every second, and 4.0 minutes = 4.0 × 60 s = 240 s. Rearranging P = ΔE / t gives ΔE = P × t = 1500 W × 240 s = 360 000 J (360 kJ).

Checked automatically

Energy transferred: 360000 J (within ±1%)
- Common wrong answer 6 : Change 1.5 kW to 1500 W and 4.0 minutes to 240 s before you multiply.
- Common wrong answer 6000 : Change 4.0 minutes to seconds as well: 4.0 × 60 s = 240 s.
- Common wrong answer 360 J: That is the energy in kilojoules. Change 1.5 kW to 1500 W first, and the answer comes out in joules.
- Common wrong answer 6.25 : Multiply the power by the time, do not divide.

A full answer includes

Method: I changed 1.5 kW to 1500 W.
Method: I changed 4.0 minutes to 240 s.
Method: I multiplied 1500 W by 240 s.
Answer: I wrote 360 000 J.
I did not leave the time in minutes.

Also accept

360 000 J (360 kJ)

Watch for

“1.5 kW × 4.0 min = 6.0 J.” — Change the power to watts and the time to seconds first: 1500 W × 240 s = 360 000 J.

9. EQ19 · Practice

Model answer. First change the work to joules: 36 kJ = 36 000 J. P = W / t, so t = W ÷ P = 36 000 J ÷ 1200 W = 30 s.

Checked automatically

Time: 30 s (within ±0.5 s)
- Common wrong answer 0.03 : Change 36 kJ to 36 000 J first.
- Common wrong answer 43200000 : Divide the work by the power, do not multiply.
- Common wrong answer 0.033 : Divide the work by the power, not the power by the work.

A full answer includes

Method: I changed 36 kJ to 36 000 J.
Method: I divided 36 000 J by 1200 W.
Answer: I wrote 30 s.
I did not divide 36 by 1200.

Also accept

30 s

Watch for

“Time = power ÷ work.” — P = W / t rearranges to t = W ÷ P: the work divided by the power.

10. EQ20 · Practice

Model answer. Ek = ½ m v² = ½ × 0.50 kg × (4.0 m/s)² = ½ × 0.50 × 16 = 4.0 J.

Checked automatically

Kinetic energy: 4.0 J (within ±0.05 J)
- Common wrong answer 1.0 : Square the speed: 4.0² = 16.
- Common wrong answer 8.0 : Remember the ½.
- Common wrong answer 2.0 : Square the whole speed, then halve: ½ × 0.50 × 16.

A full answer includes

Method: I squared the speed: (4.0 m/s)² = 16 m²/s².
Method: I calculated ½ × 0.50 kg × 16 m²/s².
Answer: I wrote 4.0 J.
I did not use 4.0 m/s without squaring it.

Also accept

4.0 J

Watch for

“Ek = ½ × m × v, without squaring.” — The speed is squared: (4.0 m/s)² = 16 m²/s².

11. EQ21 · Practice

Model answer. ΔEp = m g Δh = 2.0 kg × 9.8 N/kg × 1.5 m = 29.4 J (about 29 J).

Checked automatically

Change in gravitational potential energy: 29.4 J (within ±0.5 J)
- Common wrong answer 3.0 : Multiply by g = 9.8 N/kg as well.
- Common wrong answer 13.3 : Multiply all three: m × g × Δh.

A full answer includes

Method: I multiplied 2.0 kg × 9.8 N/kg × 1.5 m.
Answer: I wrote 29.4 J (or 29 J).
I did not leave out g.

Also accept

29.4 J (29 J)

Watch for

“ΔEp = m × Δh.” — Multiply by g as well: m g is the weight in newtons, 19.6 N here.

12. EQ22 · Practice

Model answer. With no air resistance, the kinetic energy gained equals the gravitational potential energy lost: ½ m v² = m g Δh. The mass cancels, so v² = 2 g Δh = 2 × 9.8 N/kg × 5.0 m = 98 m²/s², and v = √98 = 9.9 m/s.

Checked automatically

Speed: 9.9 m/s (within ±0.1 m/s)
- Common wrong answer 98 : That is v². Take the square root.
- Common wrong answer 7.0 : v² = 2 g Δh: remember the 2.
- Common wrong answer 49 : v² = 2 × 9.8 × 5.0 = 98, then take the square root.

A full answer includes

Method: I set ½ m v² equal to m g Δh, and the mass cancelled.
Method: I calculated v² = 2 × 9.8 × 5.0 = 98 m²/s².
Answer: I wrote 9.9 m/s.
I did not stop at v² = 98.

Also accept

9.9 m/s

Watch for

“You need the mass of the ball to find its speed.” — The mass is on both sides of ½ m v² = m g Δh, so it cancels: any ball dropped from 5.0 m reaches the same speed.

13. EQ23 · Practice

Model answer. Call car Q’s mass m and car P’s speed v. Car P: Ek = ½ × 2m × v² = m v². Car Q: Ek = ½ × m × (2v)² = ½ × m × 4v² = 2 m v². So car Q has twice the kinetic energy of car P: doubling the speed multiplies Ek by 4, but having half the mass only halves it.

A full answer includes

I chose: car Q has twice the kinetic energy of car P.
Method: I found that car P has ½ × 2m × v² = m v².
Method: I found that car Q has ½ × m × (2v)² = 2 m v².
I did not choose “the same”: the speed is squared, the mass is not.

Also accept

car Q has twice the kinetic energy of car P

Watch for

“Car P has twice the kinetic energy of car Q.” — That would be true if the mass counted more than the speed. The speed is squared, so car Q’s doubled speed counts four times, more than car P’s doubled mass.
“Car Q has four times the kinetic energy of car P.” — Doubling the speed gives four times as much, but car Q also has half the mass of car P, which halves it: 4 ÷ 2 = 2.
“The two cars have the same kinetic energy.” — The two doublings do not cancel, because the speed is squared and the mass is not.

14. EQ24 · Practice

Model answer. (a) Ek = ½ m v², so m = 2 Ek ÷ v² = 2 × 12 000 J ÷ (20 m/s)² = 24 000 ÷ 400 = 60 kg. (b) ΔEp = m g Δh = 60 kg × 9.8 N/kg × 100 m = 58 800 J. (c) She lost 58 800 J of gravitational potential energy but has only 12 000 J of kinetic energy. The rest, 58 800 J − 12 000 J = 46 800 J, was transferred by friction and air resistance to the internal store of the snow, her skis and the air.

Checked automatically

(a) Mass of the skier: 60 kg (within ±0.5 kg)
- Common wrong answer 30 : Ek = ½ m v², so m = 2 Ek ÷ v²: remember the 2.
- Common wrong answer 1200 : Square the speed: v² = 400 m²/s².
(b) Gravitational potential energy lost: 58800 J (within ±1%)
- Common wrong answer 6000 : Multiply by g = 9.8 N/kg as well.
(c) Energy to the internal store: 46800 J (within ±1%)
- Common wrong answer 70800 : Subtract the kinetic energy from the potential energy lost, do not add.

A full answer includes

(a) Method: I rearranged Ek = ½ m v² to m = 2 Ek ÷ v².
(a) Answer: I wrote 60 kg.
(a) I did not leave out the 2.
(b) Method: I multiplied 60 kg × 9.8 N/kg × 100 m.
(b) Answer: I wrote 58 800 J.
(c) Method: I took the kinetic energy, 12 000 J, from the 58 800 J she lost.
(c) Answer: I wrote 46 800 J.

Also accept

(a) 60 kg (b) 58 800 J (c) 46 800 J

Watch for

“The difference has been destroyed by friction.” — Friction transfers it to the internal store of the snow, the skis and the air. Energy is never destroyed.

15. ET3 · Apply

Model answer. (a) To raise the car steadily, the motor pulls up with a force equal to the weight, 8000 N, through 15 m: W = F d = 8000 N × 15 m = 120 000 J. (b) P = W / t = 120 000 J ÷ 12 s = 10 000 W (10 kW). (c) The work is the same, 120 000 J, and 15 kW = 15 000 W, so t = W ÷ P = 120 000 J ÷ 15 000 W = 8.0 s. (d) The energy transferred equals the work done, which depends only on the force and the distance moved in its direction. Both are the same, so both motors transfer 120 000 J to the car’s gravitational potential store; the newer one just transfers it in less time, because its power is greater.

Checked automatically

(a) Work done: 120000 J (within ±1%)
- Common wrong answer 533 : Multiply the weight by the height, do not divide.
(b) Power of the first motor: 10000 W (within ±1%)
- Common wrong answer 1440000 : Divide the work by the time, do not multiply.
(c) Time for the newer motor: 8.0 s (within ±0.05 s)
- Common wrong answer 8000 : Change 15 kW to 15 000 W first.
- Common wrong answer 0.125 : Divide the work by the power, not the power by the work.

A full answer includes

(a) Method: I multiplied the weight, 8000 N, by the height, 15 m.
(a) Answer: I wrote 120 000 J.
(b) Method: I divided 120 000 J by 12 s.
(b) Answer: I wrote 10 000 W.
(c) Method: I changed 15 kW to 15 000 W.
(c) Method: I divided the same work, 120 000 J, by 15 000 W.
(c) Answer: I wrote 8.0 s.
(d) I wrote that the work depends only on the force and the distance, which are the same.
(d) I wrote that the newer motor transfers the same energy in less time.
(d) I did not write that a more powerful motor does more work.

Also accept

(a) 120 000 J (b) 10 000 W (c) 8.0 s (d) the same force and distance, so the same work; it is only done faster

Watch for

“A more powerful motor gives the car more energy.” — A more powerful motor transfers the same energy faster. The energy depends on the weight and the height only.

16. ET4 · Apply

Model answer. (a) ΔEp = m g Δh = 50 kg × 9.8 N/kg × 10 m = 4900 J. (b) With no air resistance, all of it goes to her kinetic store: ½ m v² = 4900 J, so v² = 2 × 4900 J ÷ 50 kg = 196 m²/s², and v = √196 = 14 m/s.

Checked automatically

(a) Gravitational potential energy lost: 4900 J (within ±1%)
- Common wrong answer 500 : Multiply by g = 9.8 N/kg as well.
(b) Speed at the water: 14 m/s (within ±0.1 m/s)
- Common wrong answer 196 : That is v². Take the square root.
- Common wrong answer 9.9 : v² = 2 × 4900 ÷ 50 = 196: remember the 2.

A full answer includes

(a) Method: I multiplied 50 kg × 9.8 N/kg × 10 m.
(a) Answer: I wrote 4900 J.
(b) Method: I set ½ × 50 kg × v² equal to 4900 J.
(b) Method: I calculated v² = 2 × 4900 ÷ 50 = 196 m²/s².
(b) Answer: I wrote 14 m/s.
(b) I did not leave out the ½ in ½ m v².

Also accept

(a) 4900 J (b) 14 m/s

Watch for

“A heavier diver would hit the water faster.” — The mass cancels in ½ m v² = m g Δh, so every diver falling 10 m reaches the same speed (with no air resistance).