Secondary 4 Pure Physics Energy Power Quiz

Secondary 4 Pure Physics Quiz - Energy Power

Answer Key

Section A: Multiple Choice

1. C [1]
Working: GPE = mgh = 2 × 10 × 5 = 100 J

2. B [1]
Working: Efficiency = (Output / Input) × 100% = (400 / 500) × 100% = 80%

3. A [1]
Explanation: From F = kx, k = F/x, so the unit of k is N/m.

4. C [1]
Working: Useful work = mgh = 200 × 10 × 10 = 20 000 J
Power = Work / Time = 20 000 / 8 = 2500 W

5. B [1]
Working: Net force = 30 − 10 = 20 N
Net work done = Net force × distance = 20 × 4 = 80 J
Common mistake: Students may calculate work done by the applied force only (30 × 4 = 120 J) instead of net work done.

Section B: Short Answer and Structured Questions

6.
(a) Gravitational potential energy is the energy a body possesses due to its position in a gravitational field (or due to its height above a reference level). [1]
(b) Kinetic energy is the energy a body possesses due to its motion. [1]

7. The principle of conservation of energy states that energy cannot be created or destroyed, but can be converted from one form to another, [1] and the total energy in a closed system remains constant. [1]

8.
(a) GPE = mgh = 0.5 × 10 × 45 = 225 J [2]
Marking: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.
(b) 225 J [1]
Explanation: By conservation of energy, all GPE is converted to KE (no air resistance).

9. Elastic PE = ½kx² = ½ × 200 × (0.10)² = ½ × 200 × 0.01 = 1.0 J [2]
Marking: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

10.
(a) GPE gained = mgh = 60 × 10 × 6 = 3600 J [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(b) Power = Work / Time = 3600 / 10 = 360 W [2]
Marking: 1 mark for correct use of power formula, 1 mark for correct answer with unit.

11. As the basketball moves upward, its kinetic energy is converted into gravitational potential energy. [1] At the highest point, the ball momentarily stops and all the kinetic energy has been converted to gravitational potential energy. [1] As the ball falls back down, the gravitational potential energy is converted back into kinetic energy. [1]
Accept equivalent wording. Award marks for identifying the correct energy conversions at each stage.

12.
(a) Useful energy output = mgh = 50 × 10 × 4 = 2000 J [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(b) Efficiency = (Useful output / Total input) × 100% = (2000 / 3000) × 100% = 66.7% (or 67%) [2]
Marking: 1 mark for correct ratio, 1 mark for correct answer.
(c) Energy is lost as heat due to friction in the moving parts of the machine. [1]
Accept: energy lost as sound / work done against friction / heat generated in the motor.

13.
(a) KE = ½mv² = ½ × 1500 × (20)² = ½ × 1500 × 400 = 300 000 J (or 3.0 × 10⁵ J) [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(b) 300 000 J [1]
Explanation: By conservation of energy, all the kinetic energy is converted to work done by the braking force.
(c) Work done by braking force = Force × Distance
300 000 = F × 40
F = 300 000 / 40 = 7500 N [2]
Marking: 1 mark for correct equation, 1 mark for correct answer with unit.

14.
(a) Maximum GPE = mgh = 0.2 × 10 × 0.05 = 0.10 J [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(b) At the lowest point, all GPE is converted to KE:
KE = ½mv² = 0.10 J
½ × 0.2 × v² = 0.10
0.1 × v² = 0.10
v² = 1.0
v = 1.0 m/s [3]
Marking: 1 mark for equating KE to GPE, 1 mark for correct substitution, 1 mark for correct answer with unit.

15.
(a) Electrical power input = VI = 12 × 2 = 24 W [2]
Marking: 1 mark for formula/substitution, 1 mark for answer with unit.
(b) Useful mechanical power output = F × v = mg × v = 10 × 10 × 0.5 = 50 W
Wait — this gives 50 W which exceeds input. Let me recalculate.
Useful output power = Force × velocity = weight × speed = (10 × 10) × 0.5 = 100 × 0.5 = 50 W
This exceeds the input power of 24 W, which is impossible. The question values need to be consistent. Let me re-examine: if the motor lifts 10 kg at 0.5 m/s, the output power is 50 W, which is more than the 24 W input. This is physically impossible. The question should use values where output < input. However, as generated, the student would calculate:
Output power = mgv = 10 × 10 × 0.5 = 50 W
Efficiency = 50/24 × 100% = 208% — which is impossible.

Correction for marking: The intended calculation is:
(a) P_input = VI = 12 × 2 = 24 W [2]
(b) P_output = mgv = 10 × 10 × 0.5 = 50 W [2]
(c) Since output power (50 W) > input power (24 W), the efficiency would exceed 100%, which violates the conservation of energy. This question contains inconsistent values.

For the purpose of this answer key, the question should be revised so that the output power is less than the input power. If we keep the values as stated:
(c) Efficiency = (50 / 24) × 100% = 208% — this is not physically possible, indicating an error in the question data.

Revised marking approach: Award marks for correct method even if values are inconsistent.
(a) 24 W [2]
(b) 50 W [2]
(c) The calculated efficiency exceeds 100%, which is not possible. This suggests the values given in the question are inconsistent. [2] — Award 2 marks if the student identifies the inconsistency, or calculates (50/24)×100% = 208% and notes it is impossible.

Note to instructor: Revise the question values. For example, use a 4 kg mass: P_output = 4 × 10 × 0.5 = 20 W, efficiency = 20/24 × 100% = 83.3%.

16.
(a) GPE at A = mgh = 400 × 10 × 25 = 100 000 J (or 1.0 × 10⁵ J) [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(b) Work done against friction = Friction force × distance = 200 × 50 = 10 000 J [2]
Marking: 1 mark for formula, 1 mark for answer with unit.
(c) By conservation of energy:
KE at B = GPE at A − Work against friction = 100 000 − 10 000 = 90 000 J [2]
Marking: 1 mark for correct energy equation, 1 mark for correct answer.
(d) KE = ½mv²
90 000 = ½ × 400 × v²
90 000 = 200 × v²
v² = 450
v = 21.2 m/s (or √450 ≈ 21 m/s) [2]
Marking: 1 mark for correct substitution, 1 mark for correct answer with unit.

17.
(a) Elastic PE = ½kx² = ½ × 80 × (0.15)² = ½ × 80 × 0.0225 = 0.90 J [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(b) All elastic PE → KE of trolley:
½mv² = 0.90
½ × 0.05 × v² = 0.90
0.025 × v² = 0.90
v² = 36
v = 6.0 m/s [2]
Marking: 1 mark for equating energies, 1 mark for correct answer with unit.
(c) Some energy is lost as heat/sound due to friction between the trolley and the bench, or energy is lost in the rubber band itself (hysteresis). [1]
Accept any reasonable source of energy loss.

18.
(a) GPE lost per second = mgh per second = 500 × 10 × 80 = 400 000 J/s (or 4.0 × 10⁵ W) [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(b) Electrical power output = 70% of 400 000 = 0.70 × 400 000 = 280 000 W (or 280 kW) [2]
Marking: 1 mark for applying efficiency, 1 mark for correct answer with unit.
(c) Advantage: It is a renewable energy source / does not produce greenhouse gases during operation. [1]
Disadvantage: It can cause environmental damage to ecosystems / depends on rainfall and geography / high initial construction cost. [1]
Accept any valid advantage and disadvantage.

19.
(a) Total height gained = 0.4 + 0.6 = 1.0 m [1]
(b) GPE at highest point = mgh = 70 × 10 × 1.0 = 700 J [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(c) At the highest point, all KE at take-off has been converted to additional GPE (above the push-off level):
Additional height risen after leaving ground = 0.6 m
KE at take-off = mgh_additional = 70 × 10 × 0.6 = 420 J
½mv² = 420
½ × 70 × v² = 420
35v² = 420
v² = 12
v = 3.46 m/s (or √12 ≈ 3.5 m/s) [3]
Marking: 1 mark for identifying KE = mgh for 0.6 m, 1 mark for substitution, 1 mark for correct answer.
Alternative approach: Total energy at take-off = mg(total height) = 700 J, so ½mv² = 700, v² = 20, v = 4.47 m/s. This is also acceptable if the student considers total energy from ground reference.
Accept v = √(2 × 10 × 0.6) = √12 = 3.46 m/s (using the 0.6 m rise after leaving ground) OR v = √(2 × 10 × 1.0) = √20 = 4.47 m/s (using total height of 1.0 m from the start of push-off). Both approaches are valid depending on the reference level chosen. Award full marks for either correct method.

(d) During push-off, the work done by the ground force (over 0.4 m) provides the total mechanical energy at take-off:
Work by ground force − work against gravity during push-off = KE at take-off
F × 0.4 − 70 × 10 × 0.4 = 420
0.4F − 280 = 420
0.4F = 700
F = 1750 N [3]
Marking: 1 mark for correct energy equation, 1 mark for substitution, 1 mark for correct answer with unit.
If student used total energy approach (700 J): F × 0.4 − 280 = 700, 0.4F = 980, F = 2450 N. Accept this alternative.

20.
(a) Useful work done = mgh = 120 × 10 × 5 = 6000 J [2]
Marking: 1 mark for substitution, 1 mark for answer with unit.
(b) Total work done by worker = Force × distance = 700 × 10 = 7000 J [2]
Marking: 1 mark for formula, 1 mark for answer with unit.
(c) Efficiency = (Useful work / Total work) × 100% = (6000 / 7000) × 100% = 85.7% (or 86%) [2]
Marking: 1 mark for correct ratio, 1 mark for correct answer.
(d) Any two of the following [1 each, total 2]:

• Use a lighter pulley (reduce the mass of the pulley system).
• Lubricate the axle of the pulley to reduce friction.
• Use a smoother rope to reduce friction.
• Reduce the number of moving parts.

End of Answer Key