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O Level Physics Energy Power Quiz

Free O Level Physics Energy Power quiz, HY3 Exam version, with questions, answers, and O Level-style practice for Singapore students.

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Answers

O-Level Physics Quiz - Energy Power (Answer Key)

Total Marks: 40
Topic: Energy Power

Section A

1. [1] B. Joule
Teaching note: Energy is measured in joules (J). Watt is unit of power, Newton is force, Ampere is current.

2. [1] Ep=mghE_p = mgh
Teaching note: Gravitational potential energy = mass × gravitational field strength × height.

3. [2]
Ep=mgh=2×10×5=100 JE_p = mgh = 2 \times 10 \times 5 = 100\ \text{J}
Marks: 1 for correct substitution, 1 for answer with unit.

4. [1] Power is the rate of energy transfer (or work done per unit time): P=EtP = \frac{E}{t}.

5. [2]
P=Et=60020=30 WP = \frac{E}{t} = \frac{600}{20} = 30\ \text{W}
Marks: 1 for formula, 1 for answer.

Section B

6. [3]
(a) Ek=12mv2=12×1000×202=200 000 JE_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 20^2 = 200\,000\ \text{J} [2]
(b) Converted to heat and sound at brakes / dissipated [1]

7. [4]
(a) Work = mgh=500×10×8=40 000 Jmgh = 500 \times 10 \times 8 = 40\,000\ \text{J} [2]
(b) Power = Wt=40 00010=4000 W\frac{W}{t} = \frac{40\,000}{10} = 4000\ \text{W} [2]

8. [2]
Useful output = 0.40×2000=800 J0.40 \times 2000 = 800\ \text{J}
Marks: 1 for efficiency conversion, 1 for answer.

9. [2]
Incorrect [1]. Law of conservation of energy: energy cannot be created or destroyed, only transferred/transformed [1].

10. [4]
(a) E=VIt=12×3×(5×60)=12×3×300=10 800 JE = VIt = 12 \times 3 \times (5 \times 60) = 12 \times 3 \times 300 = 10\,800\ \text{J} [2]
(b) Useful = 0.80×10 800=8640 J0.80 \times 10\,800 = 8640\ \text{J} [2]

11. [2]
Device C has highest output (700 J of 1000 J).
Efficiency = 7001000×100%=70%\frac{700}{1000} \times 100\% = 70\% [2]

12. [3]
(a) Ep=mgh=0.5×10×4=20 JE_p = mgh = 0.5 \times 10 \times 4 = 20\ \text{J} [2]
(b) 20 J [1] (all GPE → KE)

13. [3]
Terminal velocity occurs when air resistance equals weight, resultant force zero, constant speed [1]. GPE decreases and is transferred to internal energy of air/object via friction [2].

14. [2]
Efficiency = 10005000×100%=20%\frac{1000}{5000} \times 100\% = 20\% [2]

15. [3]
Total input = 400+100=500 J400 + 100 = 500\ \text{J} [1]
Efficiency = 400500×100%=80%\frac{400}{500} \times 100\% = 80\% [2]

Section C

16. [5]
(a) Vertical work = mgh=2×10×1.5=30 Jmgh = 2 \times 10 \times 1.5 = 30\ \text{J} [2]
(b) Along ramp = F×l=15×3=45 JF \times l = 15 \times 3 = 45\ \text{J} [2]
(c) Extra work done against friction [1]

17. [4]
Renewable: lower environmental impact (e.g. solar/wind no emissions) [1], less reliable (weather-dependent) [1].
Non-renewable: higher impact (CO₂, pollution) [1], more reliable (constant supply) [1].

18. [4]
Chain: GPE of water → KE of falling water → KE of turbine → electrical energy [2].
Advantage: clean renewable [1]; Disadvantage: habitat disruption / high cost [1].

19. [3]
Fan: 0.060×5=0.30 kWh0.060 \times 5 = 0.30\ \text{kWh}
Fridge: 0.150×24=3.60 kWh0.150 \times 24 = 3.60\ \text{kWh}
Lamp: 0.020×4=0.08 kWh0.020 \times 4 = 0.08\ \text{kWh}
Total = 3.98 kWh3.98\ \text{kWh} [3]

20. [5]
(a) Ep1=300×10×20=60 000 JE_{p1} = 300 \times 10 \times 20 = 60\,000\ \text{J} [2]
(b) Ep2=300×10×5=15 000 JE_{p2} = 300 \times 10 \times 5 = 15\,000\ \text{J} [2]
(c) KE = 60 000−15 000=45 000 J60\,000 - 15\,000 = 45\,000\ \text{J} [1]