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Secondary 4 Pure Physics Energy Power Quiz
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Questions
Secondary 4 Pure Physics Quiz - Energy Power
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 40
Duration: 45 minutes Total Marks: 40
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working for calculation questions.
- Use g = 10 m/s² unless otherwise stated.
- Specific heat capacity of water = 4200 J/(kg·K).
- Give final answers to 2 or 3 significant figures where appropriate.
Section A: Structured Questions (10 marks)
Answer all questions in this section.
1. State the Principle of Conservation of Energy.
[2 marks]
2. A student lifts a 5.0 kg box vertically through a height of 2.0 m at constant speed.
(a) Calculate the work done by the student in lifting the box.
[2 marks]
(b) The student takes 4.0 s to lift the box. Calculate the power developed.
[1 mark]
3. A car of mass 1200 kg accelerates from rest to 20 m/s on a horizontal road.
(a) Calculate the kinetic energy of the car at 20 m/s.
[2 marks]
(b) The car's engine provides a useful power output of 30 kW. Calculate the minimum time needed to reach this speed, assuming no energy losses.
[2 marks]
4. State what is meant by the term efficiency in the context of energy transfers.
[1 mark]
5. A hot cup of tea of mass 0.30 kg cools from 85°C to 25°C. The specific heat capacity of the tea is the same as water.
Calculate the thermal energy lost by the tea.
[2 marks]
Section B: Calculation and Application (18 marks)
Answer all questions in this section.
6. A hydroelectric power station has water falling from a height of 80 m. The mass of water flowing per second is 2500 kg.
(a) Calculate the gravitational potential energy lost by the water each second.
[2 marks]
(b) The power station has an efficiency of 65%. Calculate the useful electrical power output.
[2 marks]
7. An electric motor is used to raise a 15 kg load at a constant speed of 0.50 m/s.
(a) Calculate the gain in gravitational potential energy of the load each second.
[2 marks]
(b) The motor has an efficiency of 80%. Calculate the input power to the motor.
[2 marks]
8. A bullet of mass 12 g is fired from a rifle at a speed of 400 m/s.
(a) Calculate the kinetic energy of the bullet as it leaves the rifle.
[2 marks]
(b) The bullet embeds itself in a wooden target and comes to rest. State the main energy conversion that occurs as the bullet stops.
[1 mark]
9. A student of mass 50 kg runs up a flight of stairs of vertical height 12 m in 8.0 s.
(a) Calculate the work done by the student against gravity.
[2 marks]
(b) Calculate the average power developed by the student.
[1 mark]
10. An electric heater of power 2000 W is used to heat 2.5 kg of water. The water temperature rises from 20°C to 60°C.
(a) Calculate the thermal energy gained by the water.
[2 marks]
(b) The heater is switched on for 250 s. Calculate the total electrical energy supplied.
[1 mark]
(c) Hence, calculate the efficiency of the heating process.
[1 mark]
Section C: Data Interpretation and Explanation (12 marks)
Answer all questions in this section.
11. Figure 11.1 shows a simplified energy transfer diagram for a coal-fired power station.
Chemical energy → [Boiler] → Thermal energy → [Turbine] → Kinetic energy → [Generator] → Electrical energy
(coal) |
Thermal energy
(waste heat)
(a) State the main energy store of the coal before it is burned.
[1 mark]
(b) Explain why the overall efficiency of the power station is less than 100%.
[2 marks]
12. A ball of mass 0.50 kg is dropped from a height of 8.0 m above the ground. Air resistance is negligible.
(a) Calculate the gravitational potential energy of the ball before it is dropped.
[2 marks]
(b) Using the Principle of Conservation of Energy, determine the speed of the ball just before it hits the ground.
[3 marks]
13. A solar panel receives 500 J of light energy from the Sun. It converts 75 J into electrical energy.
(a) Calculate the efficiency of the solar panel.
[2 marks]
(b) State what happens to the energy that is not converted into electrical energy.
[1 mark]
14. A pendulum bob of mass 0.20 kg is released from rest at point A, which is 0.15 m above the lowest point B. Assume no energy losses.
(a) Calculate the gravitational potential energy of the bob at point A relative to point B.
[2 marks]
(b) Determine the maximum speed of the bob as it passes through point B.
[2 marks]
15. A wind turbine has blades that capture kinetic energy from the wind. The wind provides 8000 J of kinetic energy per second to the turbine. The turbine generates 2400 W of electrical power.
(a) Calculate the efficiency of the wind turbine.
[2 marks]
(b) Suggest one reason why the efficiency is less than 100%.
[1 mark]
Section D: Challenging Problems (10 marks)
Answer all questions in this section.
16. A roller coaster car of mass 500 kg is pulled to the top of a hill 30 m high. It then descends and goes over a second hill 10 m high. Assume no energy losses due to friction or air resistance.
(a) Calculate the work done to pull the car to the top of the first hill.
[2 marks]
(b) Determine the speed of the car at the top of the second hill.
[3 marks]
17. An electric kettle has a power rating of 2200 W. It is used to heat 1.5 kg of water from 25°C to 100°C. The kettle has an efficiency of 85%.
(a) Calculate the thermal energy required to heat the water.
[2 marks]
(b) Calculate the time taken to heat the water.
[3 marks]
18. A crane lifts a 200 kg load vertically upwards at a constant speed of 0.80 m/s for 15 s.
(a) Calculate the height the load is raised.
[1 mark]
(b) Calculate the work done by the crane on the load.
[2 marks]
(c) The crane's motor has an input power of 2500 W. Calculate the efficiency of the crane during this lift.
[2 marks]
19. A stone of mass 0.40 kg is thrown vertically upwards with an initial speed of 15 m/s. Air resistance is negligible.
(a) Calculate the initial kinetic energy of the stone.
[2 marks]
(b) Determine the maximum height reached by the stone.
[3 marks]
20. A solar water heater uses energy from the Sun to heat water. The solar panel receives 2.5 × 10⁷ J of solar energy over a period of 5.0 hours. The water absorbs 1.5 × 10⁷ J of thermal energy.
(a) Calculate the efficiency of the solar water heater.
[2 marks]
(b) The heated water (mass 100 kg) experiences a temperature rise. Calculate this temperature rise. (Specific heat capacity of water = 4200 J/(kg·K))
[3 marks]
Answers
Secondary 4 Pure Physics Quiz - Energy Power
ANSWER KEY AND MARKING SCHEME
Total Marks: 40
Section A: Structured Questions (10 marks)
1. State the Principle of Conservation of Energy.
[2 marks]
Answer: Energy cannot be created or destroyed. [1] Energy can only be converted/transferred from one form to another. [1] OR: The total energy in an isolated system remains constant. [1] Energy can be transferred between stores but the total amount is unchanged. [1]
Marking notes: Award 1 mark for "cannot be created or destroyed" and 1 mark for "converted/transferred" or "total energy constant in isolated system".
2. (a) Calculate the work done by the student in lifting the box.
[2 marks]
Answer: Work done = Force × distance [1] Force = weight = mg = 5.0 × 10 = 50 N Work done = 50 × 2.0 = 100 J [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 100 J.
(b) Calculate the power developed.
[1 mark]
Answer: Power = Work done / time = 100 / 4.0 = 25 W [1]
Marking notes: Award 1 mark for correct answer with units. Accept 25 W.
3. (a) Calculate the kinetic energy of the car at 20 m/s.
[2 marks]
Answer: KE = ½mv² [1] KE = ½ × 1200 × (20)² = ½ × 1200 × 400 = 240,000 J = 240 kJ [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 240,000 J or 240 kJ.
(b) Calculate the minimum time needed to reach this speed.
[2 marks]
Answer: Power = Energy / time, so time = Energy / Power [1] time = 240,000 / 30,000 = 8.0 s [1]
Marking notes: Award 1 mark for correct rearrangement, 1 mark for correct answer with units. Accept 8.0 s.
4. State what is meant by the term efficiency in the context of energy transfers.
[1 mark]
Answer: Efficiency = (useful energy output / total energy input) × 100% [1] OR: Efficiency is the ratio/percentage of useful energy output to total energy input. [1]
Marking notes: Award 1 mark for correct definition. Must mention useful output and total input.
5. Calculate the thermal energy lost by the tea.
[2 marks]
Answer: Q = mcΔθ [1] Q = 0.30 × 4200 × (85 - 25) = 0.30 × 4200 × 60 = 75,600 J = 75.6 kJ [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 75,600 J or 75.6 kJ.
Section B: Calculation and Application (18 marks)
6. (a) Calculate the gravitational potential energy lost by the water each second.
[2 marks]
Answer: GPE = mgh [1] GPE = 2500 × 10 × 80 = 2,000,000 J = 2.0 × 10⁶ J = 2.0 MJ [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 2.0 MJ or 2,000,000 J.
(b) Calculate the useful electrical power output.
[2 marks]
Answer: Power input = Energy per second = 2.0 × 10⁶ W [1] Useful power output = Efficiency × Power input = 0.65 × 2.0 × 10⁶ = 1.3 × 10⁶ W = 1.3 MW [1]
Marking notes: Award 1 mark for identifying power input as energy per second, 1 mark for correct calculation with units. Accept 1.3 MW or 1,300,000 W.
7. (a) Calculate the gain in gravitational potential energy of the load each second.
[2 marks]
Answer: Height gained per second = speed = 0.50 m [1] GPE gained per second = mgh = 15 × 10 × 0.50 = 75 J [1]
Marking notes: Award 1 mark for identifying height gained per second, 1 mark for correct answer with units. Accept 75 J.
(b) Calculate the input power to the motor.
[2 marks]
Answer: Useful power output = 75 W (from part a) [1] Input power = Useful power output / Efficiency = 75 / 0.80 = 93.75 W ≈ 94 W [1]
Marking notes: Award 1 mark for identifying useful power output, 1 mark for correct calculation with units. Accept 93.75 W or 94 W.
8. (a) Calculate the kinetic energy of the bullet as it leaves the rifle.
[2 marks]
Answer: Mass = 12 g = 0.012 kg [1] KE = ½mv² = ½ × 0.012 × (400)² = ½ × 0.012 × 160,000 = 960 J [1]
Marking notes: Award 1 mark for correct conversion of mass to kg and substitution, 1 mark for correct answer with units. Accept 960 J.
(b) State the main energy conversion that occurs as the bullet stops.
[1 mark]
Answer: Kinetic energy is converted to thermal energy (heat) / internal energy of the bullet and target. [1]
Marking notes: Award 1 mark for identifying kinetic to thermal/internal energy conversion. Accept "kinetic energy to heat energy".
9. (a) Calculate the work done by the student against gravity.
[2 marks]
Answer: Work done = Force × distance = Weight × height = mgh [1] Work done = 50 × 10 × 12 = 6000 J = 6.0 kJ [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 6000 J or 6.0 kJ.
(b) Calculate the average power developed by the student.
[1 mark]
Answer: Power = Work done / time = 6000 / 8.0 = 750 W [1]
Marking notes: Award 1 mark for correct answer with units. Accept 750 W.
10. (a) Calculate the thermal energy gained by the water.
[2 marks]
Answer: Q = mcΔθ [1] Q = 2.5 × 4200 × (60 - 20) = 2.5 × 4200 × 40 = 420,000 J = 420 kJ [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 420,000 J or 420 kJ.
(b) Calculate the total electrical energy supplied.
[1 mark]
Answer: E = Pt = 2000 × 250 = 500,000 J = 500 kJ [1]
Marking notes: Award 1 mark for correct answer with units. Accept 500,000 J or 500 kJ.
(c) Hence, calculate the efficiency of the heating process.
[1 mark]
Answer: Efficiency = (Useful energy output / Total energy input) × 100% = (420,000 / 500,000) × 100% = 84% [1]
Marking notes: Award 1 mark for correct answer. Accept 84% or 0.84.
Section C: Data Interpretation and Explanation (12 marks)
11. (a) State the main energy store of the coal before it is burned.
[1 mark]
Answer: Chemical energy (store). [1]
Marking notes: Award 1 mark for "chemical energy".
(b) Explain why the overall efficiency of the power station is less than 100%.
[2 marks]
Answer: Energy is dissipated/wasted as thermal energy (heat) to the surroundings at various stages. [1] This wasted energy is not converted into useful electrical energy, so the useful output is always less than the total input. [1]
Marking notes: Award 1 mark for identifying thermal energy loss/dissipation, 1 mark for linking this to efficiency being less than 100%. Accept references to friction, sound, or heat losses in boiler/turbine/generator.
12. (a) Calculate the gravitational potential energy of the ball before it is dropped.
[2 marks]
Answer: GPE = mgh [1] GPE = 0.50 × 10 × 8.0 = 40 J [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 40 J.
(b) Using the Principle of Conservation of Energy, determine the speed of the ball just before it hits the ground.
[3 marks]
Answer: By conservation of energy: GPE at top = KE at bottom [1] 40 = ½ × 0.50 × v² [1] v² = 40 / (0.25) = 160 v = √160 = 12.65 m/s ≈ 12.6 m/s or 13 m/s [1]
Marking notes: Award 1 mark for stating energy conservation (GPE = KE), 1 mark for correct substitution, 1 mark for correct answer with units. Accept 12.6 m/s or 13 m/s.
13. (a) Calculate the efficiency of the solar panel.
[2 marks]
Answer: Efficiency = (Useful energy output / Total energy input) × 100% [1] Efficiency = (75 / 500) × 100% = 15% [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer. Accept 15% or 0.15.
(b) State what happens to the energy that is not converted into electrical energy.
[1 mark]
Answer: It is dissipated/transferred as thermal energy (heat) to the surroundings. [1]
Marking notes: Award 1 mark for identifying thermal energy dissipation/heat transfer to surroundings.
14. (a) Calculate the gravitational potential energy of the bob at point A relative to point B.
[2 marks]
Answer: GPE = mgh [1] GPE = 0.20 × 10 × 0.15 = 0.30 J [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 0.30 J.
(b) Determine the maximum speed of the bob as it passes through point B.
[2 marks]
Answer: By conservation of energy: GPE at A = KE at B [1] 0.30 = ½ × 0.20 × v² v² = 0.30 / 0.10 = 3.0 v = √3.0 = 1.73 m/s ≈ 1.7 m/s [1]
Marking notes: Award 1 mark for equating GPE and KE, 1 mark for correct answer with units. Accept 1.7 m/s or 1.73 m/s.
15. (a) Calculate the efficiency of the wind turbine.
[2 marks]
Answer: Power input = 8000 J/s = 8000 W [1] Efficiency = (Useful power output / Power input) × 100% = (2400 / 8000) × 100% = 30% [1]
Marking notes: Award 1 mark for identifying power input, 1 mark for correct answer. Accept 30% or 0.30.
(b) Suggest one reason why the efficiency is less than 100%.
[1 mark]
Answer: Energy is lost as thermal energy (heat) due to friction in the moving parts. [1] OR: Some kinetic energy of the wind is not captured by the blades. [1] OR: Energy is lost as sound energy. [1]
Marking notes: Award 1 mark for any valid reason related to energy dissipation/losses.
Section D: Challenging Problems (10 marks)
16. (a) Calculate the work done to pull the car to the top of the first hill.
[2 marks]
Answer: Work done = Gain in GPE = mgh [1] Work done = 500 × 10 × 30 = 150,000 J = 150 kJ [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 150,000 J or 150 kJ.
(b) Determine the speed of the car at the top of the second hill.
[3 marks]
Answer: By conservation of energy: GPE lost = KE gained + GPE at second hill [1] GPE at top of first hill = 500 × 10 × 30 = 150,000 J GPE at top of second hill = 500 × 10 × 10 = 50,000 J KE at second hill = 150,000 - 50,000 = 100,000 J [1] ½ × 500 × v² = 100,000 v² = 100,000 / 250 = 400 v = 20 m/s [1]
Marking notes: Award 1 mark for applying conservation of energy, 1 mark for correct energy difference, 1 mark for correct answer with units. Accept 20 m/s.
17. (a) Calculate the thermal energy required to heat the water.
[2 marks]
Answer: Q = mcΔθ [1] Q = 1.5 × 4200 × (100 - 25) = 1.5 × 4200 × 75 = 472,500 J = 472.5 kJ [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 472,500 J or 472.5 kJ.
(b) Calculate the time taken to heat the water.
[3 marks]
Answer: Useful power output of kettle = Efficiency × Power input = 0.85 × 2200 = 1870 W [1] Energy required = 472,500 J [1] Time = Energy / Power = 472,500 / 1870 = 252.7 s ≈ 250 s (or 253 s) [1]
Marking notes: Award 1 mark for calculating useful power, 1 mark for using correct energy, 1 mark for correct answer with units. Accept 250 s to 253 s.
18. (a) Calculate the height the load is raised.
[1 mark]
Answer: Height = speed × time = 0.80 × 15 = 12 m [1]
Marking notes: Award 1 mark for correct answer with units. Accept 12 m.
(b) Calculate the work done by the crane on the load.
[2 marks]
Answer: Work done = Force × distance = Weight × height = mgh [1] Work done = 200 × 10 × 12 = 24,000 J = 24 kJ [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 24,000 J or 24 kJ.
(c) Calculate the efficiency of the crane during this lift.
[2 marks]
Answer: Useful power output = Work done / time = 24,000 / 15 = 1600 W [1] Efficiency = (Useful power output / Input power) × 100% = (1600 / 2500) × 100% = 64% [1]
Marking notes: Award 1 mark for calculating useful power output, 1 mark for correct efficiency. Accept 64% or 0.64.
19. (a) Calculate the initial kinetic energy of the stone.
[2 marks]
Answer: KE = ½mv² [1] KE = ½ × 0.40 × (15)² = ½ × 0.40 × 225 = 45 J [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 45 J.
(b) Determine the maximum height reached by the stone.
[3 marks]
Answer: By conservation of energy: Initial KE = GPE at maximum height [1] 45 = mgh = 0.40 × 10 × h [1] h = 45 / (0.40 × 10) = 45 / 4.0 = 11.25 m ≈ 11.3 m [1]
Marking notes: Award 1 mark for equating KE and GPE, 1 mark for correct substitution, 1 mark for correct answer with units. Accept 11.3 m or 11.25 m.
20. (a) Calculate the efficiency of the solar water heater.
[2 marks]
Answer: Efficiency = (Useful energy output / Total energy input) × 100% [1] Efficiency = (1.5 × 10⁷ / 2.5 × 10⁷) × 100% = 60% [1]
Marking notes: Award 1 mark for correct formula/substitution, 1 mark for correct answer. Accept 60% or 0.60.
(b) Calculate this temperature rise.
[3 marks]
Answer: Q = mcΔθ [1] 1.5 × 10⁷ = 100 × 4200 × Δθ [1] Δθ = 1.5 × 10⁷ / (100 × 4200) = 1.5 × 10⁷ / 420,000 = 35.7°C ≈ 36°C [1]
Marking notes: Award 1 mark for correct formula, 1 mark for correct substitution, 1 mark for correct answer with units. Accept 35.7°C or 36°C.