Energy, work and power — return questions key

1. ER1 · Return

Model answer. (a) Winding the spring twists it, so the energy is kept in its elastic (strain) store. (b) As the spring unwinds, it pushes the gears and turns the wheels, so a force does mechanical work and transfers energy to the car’s kinetic store. (c) Friction in the gears and with the floor, and air resistance, transfer energy to the internal (thermal) store of the car, the floor and the air, and the whirr carries a little away as sound. When the spring has no energy left to give, the car stops. No energy is destroyed: the total is the same, but it is now spread out where it is no longer useful.

A full answer includes

• (a) I wrote elastic (strain).
• (b) I wrote that the unwinding spring pushes on the gears and turns the wheels.
• (b) I wrote that this force does work, transferring energy to the car’s kinetic store.
• (c) I wrote that friction and air resistance act on the moving car.
• (c) I wrote that they transfer its energy to the internal (thermal) store of the car, the floor and the air.
• (c) I wrote that no energy has been destroyed.
• (c) I did not write that the energy is used up.

Also accept

• (a) elastic (strain) (b) a force does work: to kinetic (c) friction and air resistance: to internal store and sound; none destroyed

Watch for

• “The car runs out of energy because the energy is used up.” — The energy is transferred to the internal store of the surroundings, not used up.

2. ER2 · Return

Model answer. (a) The widths are drawn to scale: 40 mm stands for 800 J, so each millimetre stands for 800 J ÷ 40 mm = 20 J. The wasted arrow is 12 mm wide, so the pump wastes 12 mm × 20 J per millimetre = 240 J each second. (b) On the same scale, 160 J needs 160 J ÷ 20 J per millimetre = 8.0 mm. The useful arrow is then 32 mm wide, and 8.0 mm + 32 mm = 40 mm, the width of the arrow going in.

Checked automatically

• (a) Energy wasted each second: 240 J (exact value)
  • Common wrong answer 12 : That is the width in millimetres. Each millimetre stands for 20 J.
  • Common wrong answer 560 : That is the useful energy, for the 28 mm arrow. The wasted arrow is 12 mm wide.
• (b) Width of the better pump’s wasted arrow: 8.0 mm (within ±0.05 mm)
  • Common wrong answer 160 : Each millimetre stands for 20 J here, not 1 J.
  • Common wrong answer 32 : That is the useful arrow, for 640 J. The wasted arrow is for 160 J.

A full answer includes

• (a) Method: I found that 40 mm stands for 800 J, so each millimetre stands for 20 J.
• (a) Answer: I wrote 240 J.
• (b) Method: I divided 160 J by 20 J for each millimetre.
• (b) Answer: I wrote 8.0 mm.
• (b) I did not give the width of the useful arrow, 32 mm.

Also accept

• (a) 240 J (b) 8.0 mm

Watch for

• “The width in millimetres is the energy in joules.” — The widths are drawn to a scale. Here each millimetre stands for 20 J, so a 12 mm arrow stands for 240 J.

3. ER3 · Return

Model answer. (a) The winch pulls along the slipway and the boat moves 16 m along it, so 16 m is the distance moved in the direction of the force. W = F d, so F = W ÷ d = 4800 J ÷ 16 m = 300 N. (b) P = W / t = 4800 J ÷ 30 s = 160 W.

Checked automatically

• (a) Force of the winch: 300 N (within ±0.5 N)
  • Common wrong answer 76800 : Divide the work by the distance, do not multiply: F = W ÷ d.
  • Common wrong answer 160 : That divides by the time. For the force, divide the work by the distance, 16 m.
• (b) Power: 160 W (within ±0.5 W)
  • Common wrong answer 144000 : Divide the work by the time, do not multiply.
  • Common wrong answer 10 : That divides the force by the time. Power is the work, 4800 J, divided by the time.

A full answer includes

• (a) Method: I divided the work, 4800 J, by the distance along the slipway, 16 m.
• (a) Answer: I wrote 300 N.
• (a) I did not multiply 4800 J by 16 m.
• (b) Method: I divided 4800 J by 30 s.
• (b) Answer: I wrote 160 W.

Also accept

• (a) 300 N (b) 160 W

Watch for

• “Force = work × distance.” — W = F d rearranges to F = W ÷ d: divide the work by the distance.
• “Power = work × time.” — Power is the work done each second: divide the work by the time.

4. ER4 · Return

Model answer. (a) Ek = ½ m v² = ½ × 0.020 kg × (60 m/s)² = ½ × 0.020 × 3600 = 36 J. (b) To stop the arrow, the straw must do 36 J of work on it, transferring the energy to the internal store of the straw and the arrow. W = F d, so F = W ÷ d = 36 J ÷ 0.090 m = 400 N. (c) At the highest point the arrow stops for an instant, so with no air resistance all 36 J is in its gravitational potential store: m g Δh = 36 J, so Δh = 36 J ÷ (0.020 kg × 9.8 N/kg) = 36 ÷ 0.196 = 184 m (about 180 m).

Checked automatically

• (a) Kinetic energy of the arrow: 36 J (within ±0.5 J)
  • Common wrong answer 0.6 : Square the speed: 60² = 3600.
  • Common wrong answer 72 : Remember the ½.
• (b) Average force of the straw: 400 N (within ±1%)
  • Common wrong answer 3.24 : Divide the work by the distance, do not multiply: F = W ÷ d.
• (c) Height with no air resistance: 184 m (within ±2.5%)
  • Common wrong answer 1800 : Divide by m × g = 0.020 × 9.8 = 0.196 N, not by 0.020 alone.
  • Common wrong answer 3.7 : Divide by m × g = 0.196 N, not by g alone.

A full answer includes

• (a) Method: I calculated ½ × 0.020 kg × (60 m/s)².
• (a) Answer: I wrote 36 J.
• (a) I did not use 60 m/s without squaring it.
• (b) Method: I used 36 J as the work done by the straw and divided it by 0.090 m.
• (b) Answer: I wrote 400 N.
• (c) Method: I set m g Δh equal to 36 J.
• (c) Method: I divided 36 J by 0.020 kg × 9.8 N/kg = 0.196 N.
• (c) Answer: I wrote 184 m (or 180 m).

Also accept

• (a) 36 J (b) 400 N (c) 184 m (180 m)

Watch for

• “Ek = ½ × 0.020 × 60.” — The speed is squared: 60² = 3600.
• “Force = energy × distance.” — W = F d rearranges to F = W ÷ d: divide the 36 J by the 0.090 m.