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A Level H1 Physics Energy Power Quiz

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Questions

A-Level Physics H1 Quiz - Energy Power

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 45 minutes Total Marks: 50

Instructions:

Answer ALL questions in the spaces provided.
Show all working for calculation questions.
Use appropriate units in all final answers.
Take g = 9.81 m s⁻² unless otherwise stated.

Section A: Short Answer (10 marks)

Answer all questions in this section.

1. State the principle of conservation of energy.

[2 marks]

2. Define power in terms of work done and time.

[1 mark]

3. A machine does 2400 J of useful work in 30 seconds. Calculate its useful power output.

[2 marks]

4. State the SI unit of power and express it in base SI units.

[2 marks]

5. Explain what is meant by the efficiency of a device.

[1 mark]

Section B: Structured Questions (10 marks)

Answer all questions in this section.

6. A student lifts a 5.0 kg mass vertically through a height of 2.0 m at constant speed. Calculate the work done by the student.

[2 marks]

7. A car of mass 1200 kg accelerates uniformly from rest to 25 m s⁻¹ in 8.0 s along a horizontal road.

(a) Calculate the kinetic energy gained by the car.

[2 marks]

(b) Calculate the average power developed by the car's engine during this acceleration, assuming no energy losses.

[2 marks]

(c) In practice, the engine must supply more power than the value calculated in (b). Suggest one reason for this.

[1 mark]

8. A pump is used to raise water from a well. In one minute, the pump raises 300 kg of water through a vertical height of 12 m.

(a) Calculate the work done by the pump in one minute.

[2 marks]

(b) Calculate the minimum power rating of the pump.

[1 mark]

Section C: Structured Questions (10 marks)

Answer all questions in this section.

9. The pump in question 8 has an efficiency of 65%. Calculate the input power required.

[2 marks]

10. A cyclist travels at a constant speed of 8.0 m s⁻¹ along a horizontal road. The total resistive force acting on the cyclist and bicycle is 45 N.

(a) State the magnitude of the forward force exerted by the cyclist.

[1 mark]

(b) Calculate the power output of the cyclist.

[2 marks]

11. The cyclist now travels up a slope inclined at 5.0° to the horizontal at the same speed. The total mass of the cyclist and bicycle is 85 kg. The resistive force remains 45 N.

(a) Calculate the additional force required to overcome the component of weight acting down the slope.

[2 marks]

(b) Determine the new power output required from the cyclist.

[2 marks]

12. A wind turbine has blades of length 22 m. The wind speed is 12 m s⁻¹ and the density of air is 1.2 kg m⁻³.

(a) Show that the mass of air passing through the area swept by the blades per second is approximately 2.2 × 10⁴ kg.

[1 mark]

Section D: Data-Based and Application Questions (20 marks)

Answer all questions in this section.

13. Refer to the wind turbine in question 12.

(a) Calculate the kinetic energy per second of the air passing through the turbine.

[2 marks]

(b) The turbine converts 40% of this kinetic energy into electrical power. Calculate the electrical power output.

[2 marks]

14. Figure 14.1 shows a simplified energy flow diagram for a coal-fired power station.

[Chemical energy in coal: 1000 MJ]
              ↓
        [Boiler: 85% efficient]
              ↓
    [Turbine: 40% efficient]
              ↓
      [Generator: 95% efficient]
              ↓
   [Electrical energy output]

(a) Calculate the energy output from the boiler.

[1 mark]

(b) Calculate the energy output from the turbine.

[1 mark]

(c) Calculate the electrical energy output from the generator.

[1 mark]

(d) Determine the overall efficiency of the power station.

[2 marks]

(e) State where most of the energy is lost in this process and suggest what form this lost energy takes.

[2 marks]

15. A hydroelectric power station uses water falling from a reservoir at a height of 180 m. The flow rate of water is 450 m³ s⁻¹. The density of water is 1000 kg m⁻³.

(a) Calculate the mass of water falling per second.

[1 mark]

(b) Calculate the gravitational potential energy lost by the water per second.

[2 marks]

(c) The power station has an efficiency of 85%. Calculate the electrical power output.

[2 marks]

(d) During a drought, the flow rate decreases to 150 m³ s⁻¹. Calculate the new electrical power output, assuming the same efficiency.

[1 mark]

16. A household electric kettle has a power rating of 2200 W. It takes 2.5 minutes to heat 1.0 kg of water from 25°C to 100°C. The specific heat capacity of water is 4200 J kg⁻¹ °C⁻¹.

(a) Calculate the useful energy required to heat the water.

[2 marks]

(b) Calculate the electrical energy supplied to the kettle during this time.

[2 marks]

(c) Determine the efficiency of the kettle.

[2 marks]

17. A solar panel with an area of 2.0 m² receives solar radiation at an intensity of 800 W m⁻². The panel has an efficiency of 18%.

(a) Calculate the total solar power incident on the panel.

[1 mark]

(b) Calculate the electrical power output of the panel.

[1 mark]

18. An electric motor lifts a 50 kg load vertically at a constant speed of 0.40 m s⁻¹. The motor operates at an efficiency of 75%.

(a) Calculate the useful power output of the motor.

[2 marks]

(b) Calculate the electrical power input to the motor.

[2 marks]

19. A ball of mass 0.50 kg is dropped from rest at a height of 10 m above the ground. Air resistance is negligible.

(a) Calculate the gravitational potential energy of the ball before it is dropped.

[1 mark]

(b) Using energy considerations, calculate the speed of the ball just before it hits the ground.

[2 marks]

20. A light bulb converts 60 J of electrical energy into 54 J of thermal energy and 6 J of light energy every second.

(a) State the useful power output of the light bulb.

[1 mark]

(b) Calculate the efficiency of the light bulb.

[1 mark]

END OF QUIZ

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Answers

A-Level Physics H1 Quiz - Energy Power: ANSWER KEY

Total Marks: 50

Section A: Short Answer (10 marks)

1. State the principle of conservation of energy. [2 marks]

Answer: Energy cannot be created or destroyed [B1]. It can only be transferred/converted from one form to another [B1]. The total energy of an isolated/closed system remains constant [B1 - accept as alternative phrasing].

2. Define power in terms of work done and time. [1 mark]

Answer: Power is the rate of doing work / Power = work done ÷ time taken [B1].

3. A machine does 2400 J of useful work in 30 seconds. Calculate its useful power output. [2 marks]

Answer: P = W / t [M1] P = 2400 / 30 = 80 W [A1]

4. State the SI unit of power and express it in base SI units. [2 marks]

Answer: SI unit: watt (W) [B1] In base units: kg m² s⁻³ [B1] (Accept J s⁻¹ for 1 mark if base units not given)

5. Explain what is meant by the efficiency of a device. [1 mark]

Answer: Efficiency = (useful energy output / total energy input) × 100% [B1] OR Efficiency = (useful power output / total power input) × 100% [B1]

Section B: Structured Questions (10 marks)

6. A student lifts a 5.0 kg mass vertically through a height of 2.0 m at constant speed. Calculate the work done by the student. [2 marks]

Answer: Work done = mgh [M1] W = 5.0 × 9.81 × 2.0 = 98.1 J [A1] (Accept 98 J or 100 J if g = 10 m s⁻² used)

7. Car acceleration problem.

(a) Calculate the kinetic energy gained by the car. [2 marks]

Answer: KE = ½mv² [M1] KE = ½ × 1200 × (25)² = 375,000 J = 3.75 × 10⁵ J [A1]

(b) Calculate the average power developed by the car's engine during this acceleration, assuming no energy losses. [2 marks]

Answer: P = work done / time = KE gained / time [M1] P = 375,000 / 8.0 = 46,875 W ≈ 4.69 × 10⁴ W [A1]

(c) In practice, the engine must supply more power than the value calculated in (b). Suggest one reason for this. [1 mark]

Answer: Any one of:

Work done against air resistance / drag [B1]
Work done against friction in the engine/transmission [B1]
Energy lost as heat/sound [B1]
Not all chemical energy in fuel converted to kinetic energy [B1]

8. Water pump problem.

(a) Calculate the work done by the pump in one minute. [2 marks]

Answer: W = mgh [M1] W = 300 × 9.81 × 12 = 35,316 J ≈ 3.53 × 10⁴ J [A1]

(b) Calculate the minimum power rating of the pump. [1 mark]

Answer: P = W / t [M1] P = 35,316 / 60 = 588.6 W ≈ 589 W [A1]

Section C: Structured Questions (10 marks)

9. The pump in question 8 has an efficiency of 65%. Calculate the input power required. [2 marks]

Answer: Efficiency = useful power output / input power [M1] 0.65 = 589 / P_input P_input = 589 / 0.65 = 906 W ≈ 910 W [A1]

10. Cyclist problem.

(a) State the magnitude of the forward force exerted by the cyclist. [1 mark]

Answer: 45 N [B1] (At constant speed, forward force = resistive force)

(b) Calculate the power output of the cyclist. [2 marks]

Answer: P = Fv [M1] P = 45 × 8.0 = 360 W [A1]

11. Cyclist on a slope.

(a) Calculate the additional force required to overcome the component of weight acting down the slope. [2 marks]

Answer: Component of weight down slope = mg sin θ [M1] = 85 × 9.81 × sin 5.0° = 85 × 9.81 × 0.0872 = 72.7 N ≈ 73 N [A1]

(b) Determine the new power output required from the cyclist. [2 marks]

Answer: Total force required = resistive force + component of weight [M1] = 45 + 72.7 = 117.7 N P = Fv = 117.7 × 8.0 = 942 W ≈ 940 W [A1]

12. Wind turbine problem.

(a) Show that the mass of air passing through the area swept by the blades per second is approximately 2.2 × 10⁴ kg. [1 mark]

Answer: Area swept = πr² = π × (22)² = 1520.5 m² [M1] Volume per second = area × speed = 1520.5 × 12 = 18,246 m³ s⁻¹ [M1] Mass per second = density × volume per second = 1.2 × 18,246 = 21,895 kg s⁻¹ ≈ 2.2 × 10⁴ kg s⁻¹ [A1]

Section D: Data-Based and Application Questions (20 marks)

13. Wind turbine (continued).

(a) Calculate the kinetic energy per second of the air passing through the turbine. [2 marks]

Answer: KE per second = ½ × (mass per second) × v² [M1] = ½ × 21,895 × (12)² = ½ × 21,895 × 144 = 1,576,440 J s⁻¹ ≈ 1.58 × 10⁶ W [A1]

(b) The turbine converts 40% of this kinetic energy into electrical power. Calculate the electrical power output. [2 marks]

Answer: Electrical power = 0.40 × 1.58 × 10⁶ [M1] = 6.32 × 10⁵ W = 632 kW [A1]

14. Coal-fired power station problem.

(a) Calculate the energy output from the boiler. [1 mark]

Answer: Energy output = 0.85 × 1000 = 850 MJ [A1]

(b) Calculate the energy output from the turbine. [1 mark]

Answer: Energy output = 0.40 × 850 = 340 MJ [A1]

(c) Calculate the electrical energy output from the generator. [1 mark]

Answer: Energy output = 0.95 × 340 = 323 MJ [A1]

(d) Determine the overall efficiency of the power station. [2 marks]

Answer: Overall efficiency = (useful output / total input) × 100% [M1] = (323 / 1000) × 100% = 32.3% [A1] (Accept 32% or calculation via product of efficiencies: 0.85 × 0.40 × 0.95 = 0.323 = 32.3%)

(e) State where most of the energy is lost in this process and suggest what form this lost energy takes. [2 marks]

Answer: Most energy is lost at the turbine stage (60% loss) [B1]. The lost energy is primarily in the form of thermal energy / heat [B1]. (Accept: lost as heat to the environment / cooling water / exhaust gases)

15. Hydroelectric power station problem.

(a) Calculate the mass of water falling per second. [1 mark]

Answer: Mass per second = density × volume flow rate [M1 implied] = 1000 × 450 = 450,000 kg s⁻¹ = 4.5 × 10⁵ kg s⁻¹ [A1]

(b) Calculate the gravitational potential energy lost by the water per second. [2 marks]

Answer: GPE lost per second = (mass per second) × g × h [M1] = 450,000 × 9.81 × 180 = 7.95 × 10⁸ J s⁻¹ = 795 MW [A1]

(c) The power station has an efficiency of 85%. Calculate the electrical power output. [2 marks]

Answer: Electrical power = efficiency × input power [M1] = 0.85 × 7.95 × 10⁸ = 6.76 × 10⁸ W = 676 MW [A1]

(d) During a drought, the flow rate decreases to 150 m³ s⁻¹. Calculate the new electrical power output, assuming the same efficiency. [1 mark]

Answer: New mass flow rate = 1000 × 150 = 150,000 kg s⁻¹ New input power = 150,000 × 9.81 × 180 = 2.65 × 10⁸ W New electrical output = 0.85 × 2.65 × 10⁸ = 2.25 × 10⁸ W = 225 MW [A1] (Accept proportional reasoning: output = 676 × (150/450) = 225 MW)

16. Electric kettle problem.

(a) Calculate the useful energy required to heat the water. [2 marks]

Answer: Q = mcΔθ [M1] Q = 1.0 × 4200 × (100 - 25) = 315,000 J = 3.15 × 10⁵ J [A1]

(b) Calculate the electrical energy supplied to the kettle during this time. [2 marks]

Answer: t = 2.5 × 60 = 150 s [M1] E = Pt = 2200 × 150 = 330,000 J = 3.30 × 10⁵ J [A1]

(c) Determine the efficiency of the kettle. [2 marks]

Answer: Efficiency = (useful energy output / total energy input) × 100% [M1] = (315,000 / 330,000) × 100% = 95.5% ≈ 95% [A1]

17. Solar panel problem.

(a) Calculate the total solar power incident on the panel. [1 mark]

Answer: P_incident = intensity × area = 800 × 2.0 = 1600 W [A1]

(b) Calculate the electrical power output of the panel. [1 mark]

Answer: P_output = efficiency × P_incident = 0.18 × 1600 = 288 W [A1]

18. Electric motor problem.

(a) Calculate the useful power output of the motor. [2 marks]

Answer: Force to lift load = mg = 50 × 9.81 = 490.5 N [M1] Useful power = Fv = 490.5 × 0.40 = 196.2 W ≈ 196 W [A1]

(b) Calculate the electrical power input to the motor. [2 marks]

Answer: Efficiency = useful power output / input power [M1] 0.75 = 196.2 / P_input P_input = 196.2 / 0.75 = 261.6 W ≈ 262 W [A1]

19. Falling ball problem.

(a) Calculate the gravitational potential energy of the ball before it is dropped. [1 mark]

Answer: GPE = mgh = 0.50 × 9.81 × 10 = 49.05 J ≈ 49 J [A1]

(b) Using energy considerations, calculate the speed of the ball just before it hits the ground. [2 marks]

Answer: Loss in GPE = gain in KE [M1] mgh = ½mv² v = √(2gh) = √(2 × 9.81 × 10) = √196.2 = 14.0 m s⁻¹ [A1]

20. Light bulb problem.

(a) State the useful power output of the light bulb. [1 mark]

Answer: Useful power output = 6 W [B1]

(b) Calculate the efficiency of the light bulb. [1 mark]

Answer: Efficiency = (useful power output / total power input) × 100% [M1] = (6 / 60) × 100% = 10% [A1]

END OF ANSWER KEY