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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 g = 9.81 m s⁻² unless otherwise stated.
Section A: Short Answer (10 marks)
Answer all questions. Each question carries 2 marks.
1. State the principle of conservation of energy.
2. Define power and state its SI unit.
3. A force of 50 N pushes a box 3.0 m along a horizontal floor. Calculate the work done by the force if it acts parallel to the displacement.
4. Explain the difference between elastic potential energy and gravitational potential energy.
5. State two factors that affect the efficiency of an electrical appliance.
Section B: Structured Questions (10 marks)
Answer all questions. Marks are indicated in brackets.
6. A student of mass 60 kg runs up a flight of stairs of vertical height 5.0 m in 4.0 s.
(a) Calculate the work done by the student against gravity. [2 marks]
(b) Calculate the average power developed by the student. [2 marks]
(c) Explain why the actual power output of the student's muscles is greater than the value calculated in (b). [1 mark]
7. A crane lifts a 500 kg concrete block vertically upward at a constant speed of 0.80 m s⁻¹.
(a) Calculate the tension in the cable. [2 marks]
(b) Calculate the power output of the crane's motor. [2 marks]
(c) The crane motor has an efficiency of 75%. Calculate the electrical power input to the motor. [2 marks]
8. A spring of spring constant k = 200 N m⁻¹ is compressed by 0.15 m from its natural length.
(a) Calculate the elastic potential energy stored in the spring. [2 marks]
(b) The spring is released and propels a 0.050 kg ball horizontally. Assuming all the stored energy is transferred to the ball, calculate the speed of the ball as it leaves the spring. [3 marks]
(c) In practice, the ball's speed is less than the calculated value. Suggest two reasons for this discrepancy. [2 marks]
9. A car of mass 1200 kg accelerates from rest to 25 m s⁻¹ in 8.0 s along a level road. The total resistive force acting on the car is 600 N.
(a) Calculate the kinetic energy gained by the car. [2 marks]
(b) Calculate the average useful power delivered to the car. [2 marks]
(c) Determine the average driving force produced by the engine. [3 marks]
(d) Calculate the average power output of the engine. [2 marks]
10. A pump lifts 200 kg of water per minute through a vertical height of 15 m.
(a) Calculate the work done by the pump per minute. [2 marks]
(b) Calculate the minimum power rating of the pump. [2 marks]
(c) If the pump motor has an efficiency of 80%, calculate the electrical power input required. [1 mark]
Section C: Data Analysis and Application (15 marks)
Answer all questions. Marks are indicated in brackets.
11. A wind turbine has blades of length 25 m. The wind speed is 12 m s⁻¹ and the density of air is 1.2 kg m⁻³. The theoretical maximum power available from the wind passing through the area swept by the blades is given by:
[ P = \frac{1}{2} \rho A v^3 ]
where A is the swept area.
(a) Calculate the swept area of the turbine blades. [2 marks]
(b) Calculate the theoretical maximum power available from the wind. [3 marks]
(c) The actual electrical power output of the turbine is 450 kW. Calculate the efficiency of the turbine. [2 marks]
(d) Suggest two reasons why the actual power output is less than the theoretical maximum. [2 marks]
12. A hydroelectric power station uses water from a reservoir at a height of 120 m above the turbine. Water flows through the turbine at a rate of 15 m³ s⁻¹. The density of water is 1000 kg m⁻³.
(a) Calculate the mass of water flowing through the turbine per second. [2 marks]
(b) Calculate the gravitational potential energy lost by the water per second. [2 marks]
(c) The turbine and generator have a combined efficiency of 85%. Calculate the electrical power output of the station. [2 marks]
(d) During a drought, the water flow rate drops to 5.0 m³ s⁻¹. Calculate the new electrical power output, assuming the same efficiency. [2 marks]
13. An electric kettle is rated at 2200 W, 240 V. It is used to heat 1.5 kg of water from 25°C to 100°C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
(a) Calculate the energy required to heat the water. [2 marks]
(b) Calculate the minimum time needed to heat the water if the kettle is 100% efficient. [2 marks]
(c) In practice, the kettle takes 4 minutes 30 seconds to heat the water. Calculate the actual efficiency of the kettle. [3 marks]
(d) Explain where the wasted energy goes and suggest one design improvement to increase efficiency. [2 marks]
Section D: Conceptual and Extended Problems (15 marks)
Answer all questions. Marks are indicated in brackets.
14. A roller coaster car of mass 500 kg is pulled to the top of a 40 m high hill.
(a) Calculate the gravitational potential energy gained by the car. [2 marks]
(b) Assuming no energy losses, calculate the speed of the car at the bottom of the hill. [2 marks]
(c) In reality, the speed at the bottom is 22 m s⁻¹. Calculate the work done against friction. [2 marks]
15. A 75 W filament light bulb converts only 8% of the electrical energy into light.
(a) Calculate the power wasted as heat. [1 mark]
(b) Calculate the energy wasted as heat when the bulb is on for 3 hours. Give your answer in joules. [2 marks]
(c) An LED bulb produces the same light output with a power rating of 10 W. Calculate the energy saved over 3 hours compared to the filament bulb. [2 marks]
16. A bungee jumper of mass 70 kg jumps from a bridge. The bungee cord has an unstretched length of 20 m and a spring constant of 150 N m⁻¹. The jumper falls a total distance of 45 m before coming to rest momentarily.
(a) Calculate the gravitational potential energy lost by the jumper during the fall. [2 marks]
(b) Calculate the elastic potential energy stored in the cord at the lowest point. [2 marks]
(c) Explain why the elastic potential energy is less than the gravitational potential energy lost. [1 mark]
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]
(c) The panel charges a 12 V battery. Calculate the current delivered to the battery, assuming no further losses. [2 marks]
(d) Suggest one reason why the actual charging current is lower. [1 mark]
18. A cyclist of total mass 85 kg (including bicycle) climbs a hill of gradient 1 in 10 (for every 10 m along the slope, the height increases by 1 m) at a constant speed of 4.0 m s⁻¹. The total resistive force is 50 N.
(a) Calculate the component of weight acting down the slope. [2 marks]
(b) Calculate the total force the cyclist must exert parallel to the slope. [2 marks]
(c) Calculate the power output of the cyclist. [2 marks]
19. A compressed air energy storage system uses excess electricity to compress air into an underground cavern at a pressure of 7.0 × 10⁶ Pa. The cavern has a volume of 3000 m³. When energy is needed, the air is released through a turbine. The energy stored per unit volume of compressed air is given by P × ln(P/P₀), where P₀ = 1.0 × 10⁵ Pa.
(a) Calculate the energy stored per unit volume of compressed air. [2 marks]
(b) Calculate the total energy stored in the cavern. [1 mark]
(c) If the turbine has an efficiency of 60%, calculate the electrical energy that can be recovered. [2 marks]
20. A student investigates the efficiency of an electric motor by using it to lift a 0.50 kg mass through a height of 2.0 m. The motor is connected to a 6.0 V power supply and draws a current of 0.80 A. It takes 3.5 s to lift the mass.
(a) Calculate the useful work done by the motor. [1 mark]
(b) Calculate the electrical energy supplied to the motor. [2 marks]
(c) Calculate the efficiency of the motor. [1 mark]
(d) The student repeats the experiment with a 1.0 kg mass and finds the efficiency is higher. Suggest why. [1 mark]
END OF QUIZ
Check your work carefully. Ensure all answers are in the correct units.
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; it can only be transferred from one form to another or transformed from one type to another. The total energy of an isolated system remains constant.
Marking:
- [B1] Energy cannot be created or destroyed
- [B1] Total energy of an isolated/closed system is constant OR energy is transferred/transformed
2. Define power and state its SI unit. [2 marks]
Answer: Power is the rate of doing work or the rate of transfer of energy. The SI unit of power is the watt (W), which is equivalent to one joule per second (J s⁻¹).
Marking:
- [B1] Rate of doing work / rate of energy transfer
- [B1] Watt (W) or J s⁻¹
3. A force of 50 N pushes a box 3.0 m along a horizontal floor. Calculate the work done by the force if it acts parallel to the displacement. [2 marks]
Answer: W = F × d = 50 × 3.0 = 150 J
Marking:
- [M1] Correct formula W = Fd or substitution
- [A1] 150 J (correct unit required)
4. Explain the difference between elastic potential energy and gravitational potential energy. [2 marks]
Answer: Elastic potential energy is the energy stored in an object when it is deformed (stretched or compressed) and can return to its original shape. Gravitational potential energy is the energy an object possesses due to its position in a gravitational field, typically relative to a reference level.
Marking:
- [B1] Elastic PE: energy stored due to deformation (stretching/compressing)
- [B1] Gravitational PE: energy due to position/height in a gravitational field
5. State two factors that affect the efficiency of an electrical appliance. [2 marks]
Answer: Any two from:
- Friction between moving parts
- Electrical resistance in wires/components (producing heat)
- Air resistance
- Sound production
- Heat loss to surroundings
Marking:
- [B1] One valid factor with brief explanation
- [B1] Second valid factor with brief explanation
Section B: Structured Questions (10 marks)
6. A student of mass 60 kg runs up a flight of stairs of vertical height 5.0 m in 4.0 s.
(a) Calculate the work done by the student against gravity. [2 marks]
Answer: W = mgh = 60 × 9.81 × 5.0 = 2943 J ≈ 2940 J (or 2.94 kJ)
Marking:
- [M1] W = mgh
- [A1] 2940 J / 2.94 kJ
(b) Calculate the average power developed by the student. [2 marks]
Answer: P = W/t = 2943 / 4.0 = 736 W ≈ 740 W
Marking:
- [M1] P = W/t or P = mgh/t
- [A1] 736 W / 740 W (ecf from part a)
(c) Explain why the actual power output of the student's muscles is greater than the value calculated in (b). [1 mark]
Answer: The student's muscles also do work against internal friction in joints, to move limbs horizontally, and generate heat. Not all energy from muscles is converted to gravitational potential energy. / The human body is not 100% efficient; some energy is lost as heat.
Marking:
- [B1] Valid explanation referencing inefficiency / energy losses / additional work done
7. A crane lifts a 500 kg concrete block vertically upward at a constant speed of 0.80 m s⁻¹.
(a) Calculate the tension in the cable. [2 marks]
Answer: Since speed is constant, net force = 0. T = mg = 500 × 9.81 = 4905 N ≈ 4910 N
Marking:
- [M1] Recognition that T = mg (equilibrium) or F_net = 0
- [A1] 4910 N / 4905 N
(b) Calculate the power output of the crane's motor. [2 marks]
Answer: P = Fv = T × v = 4905 × 0.80 = 3924 W ≈ 3.9 kW
Marking:
- [M1] P = Fv
- [A1] 3920 W / 3.92 kW (ecf from part a)
(c) The crane motor has an efficiency of 75%. Calculate the electrical power input to the motor. [2 marks]
Answer: Efficiency = P_out / P_in × 100% → P_in = P_out / 0.75 = 3924 / 0.75 = 5232 W ≈ 5.2 kW
Marking:
- [M1] P_in = P_out / efficiency
- [A1] 5230 W / 5.23 kW (ecf from part b)
8. A spring of spring constant k = 200 N m⁻¹ is compressed by 0.15 m from its natural length.
(a) Calculate the elastic potential energy stored in the spring. [2 marks]
Answer: EPE = ½kx² = ½ × 200 × (0.15)² = 100 × 0.0225 = 2.25 J
Marking:
- [M1] EPE = ½kx²
- [A1] 2.25 J
(b) The spring is released and propels a 0.050 kg ball horizontally. Assuming all the stored energy is transferred to the ball, calculate the speed of the ball as it leaves the spring. [3 marks]
Answer: EPE → KE → ½kx² = ½mv² → 2.25 = ½ × 0.050 × v² → v² = 2.25 / 0.025 = 90 → v = √90 = 9.49 m s⁻¹ ≈ 9.5 m s⁻¹
Marking:
- [M1] Energy conservation: ½kx² = ½mv²
- [M1] Correct substitution and rearrangement
- [A1] 9.5 m s⁻¹ (ecf from part a)
(c) In practice, the ball's speed is less than the calculated value. Suggest two reasons for this discrepancy. [2 marks]
Answer: Any two from:
- Some energy is used to overcome friction between the ball and the spring/barrel
- The spring itself has mass and retains some kinetic energy
- Energy is lost as heat in the spring (internal friction/hysteresis)
- Air resistance acts on the ball during launch
- Sound energy is produced
Marking:
- [B1] One valid reason
- [B1] Second valid reason
9. A car of mass 1200 kg accelerates from rest to 25 m s⁻¹ in 8.0 s along a level road. The total resistive force acting on the car is 600 N.
(a) Calculate the kinetic energy gained by the car. [2 marks]
Answer: KE = ½mv² = ½ × 1200 × (25)² = 600 × 625 = 375,000 J = 375 kJ
Marking:
- [M1] KE = ½mv²
- [A1] 375 kJ / 3.75 × 10⁵ J
(b) Calculate the average useful power delivered to the car. [2 marks]
Answer: P_useful = KE gained / time = 375,000 / 8.0 = 46,875 W ≈ 46.9 kW
Marking:
- [M1] P = ΔE/t or P = KE/t
- [A1] 46.9 kW (ecf from part a)
(c) Determine the average driving force produced by the engine. [3 marks]
Answer: Acceleration a = (v − u)/t = (25 − 0)/8.0 = 3.125 m s⁻²
Net force F_net = ma = 1200 × 3.125 = 3750 N
Driving force F_drive = F_net + F_resistive = 3750 + 600 = 4350 N
Alternatively: Work done by engine = KE gained + work against resistance
F_drive × s = 375,000 + 600 × s, where s = ½(u+v)t = ½ × 25 × 8.0 = 100 m
F_drive × 100 = 375,000 + 60,000 → F_drive = 4350 N
Marking:
- [M1] Calculate acceleration or displacement
- [M1] Apply F_net = ma or work-energy principle
- [A1] 4350 N
(d) Calculate the average power output of the engine. [2 marks]
Answer: P_engine = F_drive × v_avg = 4350 × (25/2) = 4350 × 12.5 = 54,375 W ≈ 54.4 kW
Or: P_engine = Work done by engine / time = (F_drive × s) / t = (4350 × 100) / 8.0 = 54,375 W
Marking:
- [M1] P = Fv_avg or P = W/t
- [A1] 54.4 kW (ecf from part c)
10. A pump lifts 200 kg of water per minute through a vertical height of 15 m.
(a) Calculate the work done by the pump per minute. [2 marks]
Answer: W = mgh = 200 × 9.81 × 15 = 29,430 J ≈ 29.4 kJ
Marking:
- [M1] W = mgh
- [A1] 29.4 kJ / 29,430 J
(b) Calculate the minimum power rating of the pump. [2 marks]
Answer: P = W/t = 29,430 / 60 = 490.5 W ≈ 491 W
Marking:
- [M1] P = W/t with t = 60 s
- [A1] 491 W (ecf from part a)
(c) If the pump motor has an efficiency of 80%, calculate the electrical power input required. [1 mark]
Answer: P_in = P_out / efficiency = 490.5 / 0.80 = 613 W ≈ 610 W
Marking:
- [B1] 613 W / 610 W (ecf from part b)
Section C: Data Analysis and Application (15 marks)
11. A wind turbine has blades of length 25 m. The wind speed is 12 m s⁻¹ and the density of air is 1.2 kg m⁻³.
(a) Calculate the swept area of the turbine blades. [2 marks]
Answer: A = πr² = π × (25)² = π × 625 = 1963.5 m² ≈ 1960 m² (or 1.96 × 10³ m²)
Marking:
- [M1] A = πr²
- [A1] 1960 m² / 1.96 × 10³ m²
(b) Calculate the theoretical maximum power available from the wind. [3 marks]
Answer: P = ½ρAv³ = ½ × 1.2 × 1963.5 × (12)³ = 0.6 × 1963.5 × 1728 = 2,036,000 W ≈ 2.04 MW
Marking:
- [M1] Correct substitution into P = ½ρAv³
- [M1] Correct calculation of v³ = 1728
- [A1] 2.04 MW / 2.04 × 10⁶ W
(c) The actual electrical power output of the turbine is 450 kW. Calculate the efficiency of the turbine. [2 marks]
Answer: Efficiency = (P_actual / P_theoretical) × 100% = (450,000 / 2,036,000) × 100% = 22.1%
Marking:
- [M1] Efficiency = P_out / P_in × 100%
- [A1] 22.1% (ecf from part b)
(d) Suggest two reasons why the actual power output is less than the theoretical maximum. [2 marks]
Answer: Any two from:
- Not all kinetic energy can be extracted from the wind (Betz limit)
- Friction in the turbine bearings and gearbox
- Electrical losses in the generator
- Turbulence and non-uniform wind speed
- Blades do not capture all the wind passing through the swept area
Marking:
- [B1] One valid reason
- [B1] Second valid reason
12. A hydroelectric power station uses water from a reservoir at a height of 120 m above the turbine. Water flows through the turbine at a rate of 15 m³ s⁻¹. The density of water is 1000 kg m⁻³.
(a) Calculate the mass of water flowing through the turbine per second. [2 marks]
Answer: Mass per second = ρ × volume flow rate = 1000 × 15 = 15,000 kg s⁻¹
Marking:
- [M1] m/t = ρ × (V/t)
- [A1] 15,000 kg s⁻¹
(b) Calculate the gravitational potential energy lost by the water per second. [2 marks]
Answer: GPE lost per second = (m/t) × g × h = 15,000 × 9.81 × 120 = 17,658,000 W ≈ 17.7 MW
Marking:
- [M1] P = (m/t)gh
- [A1] 17.7 MW (ecf from part a)
(c) The turbine and generator have a combined efficiency of 85%. Calculate the electrical power output of the station. [2 marks]
Answer: P_elec = 0.85 × 17,658,000 = 15,009,300 W ≈ 15.0 MW
Marking:
- [M1] P_elec = efficiency × P_input
- [A1] 15.0 MW (ecf from part b)
(d) During a drought, the water flow rate drops to 5.0 m³ s⁻¹. Calculate the new electrical power output, assuming the same efficiency. [2 marks]
Answer: New mass flow = 1000 × 5.0 = 5000 kg s⁻¹
New GPE per second = 5000 × 9.81 × 120 = 5,886,000 W
New P_elec = 0.85 × 5,886,000 = 5,003,100 W ≈ 5.0 MW
Marking:
- [M1] Correct calculation of new mass flow and power input
- [A1] 5.0 MW
13. An electric kettle is rated at 2200 W, 240 V. It is used to heat 1.5 kg of water from 25°C to 100°C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.
(a) Calculate the energy required to heat the water. [2 marks]
Answer: E = mcΔθ = 1.5 × 4200 × (100 − 25) = 1.5 × 4200 × 75 = 472,500 J = 472.5 kJ
Marking:
- [M1] E = mcΔθ
- [A1] 472.5 kJ / 4.73 × 10⁵ J
(b) Calculate the minimum time needed to heat the water if the kettle is 100% efficient. [2 marks]
Answer: t = E / P = 472,500 / 2200 = 214.8 s ≈ 215 s (3 min 35 s)
Marking:
- [M1] t = E/P
- [A1] 215 s (ecf from part a)
(c) In practice, the kettle takes 4 minutes 30 seconds to heat the water. Calculate the actual efficiency of the kettle. [3 marks]
Answer: Actual time = 4 × 60 + 30 = 270 s
Energy supplied = P × t = 2200 × 270 = 594,000 J
Efficiency = (Useful energy / Energy supplied) × 100% = (472,500 / 594,000) × 100% = 79.5% ≈ 80%
Marking:
- [M1] Convert time to seconds (270 s)
- [M1] Calculate energy supplied (E = Pt)
- [A1] 79.5% / 80% (ecf from part a)
(d) Explain where the wasted energy goes and suggest one design improvement to increase efficiency. [2 marks]
Answer: Wasted energy is lost as heat to the surroundings (kettle body, air) and as sound.
Design improvement: Use better insulation (double-walled kettle) / Use a lid to reduce heat loss / Use a more efficient heating element with better thermal contact.
Marking:
- [B1] Valid explanation of energy loss (heat to surroundings)
- [B1] Valid design improvement
Section D: Conceptual and Extended Problems (15 marks)
14. A roller coaster car of mass 500 kg is pulled to the top of a 40 m high hill.
(a) Calculate the gravitational potential energy gained by the car. [2 marks]
Answer: GPE = mgh = 500 × 9.81 × 40 = 196,200 J ≈ 196 kJ
Marking:
- [M1] GPE = mgh
- [A1] 196 kJ
(b) Assuming no energy losses, calculate the speed of the car at the bottom of the hill. [2 marks]
Answer: GPE → KE → mgh = ½mv² → v = √(2gh) = √(2 × 9.81 × 40) = √784.8 = 28.0 m s⁻¹
Marking:
- [M1] Energy conservation: mgh = ½mv²
- [A1] 28.0 m s⁻¹
(c) In reality, the speed at the bottom is 22 m s⁻¹. Calculate the work done against friction. [2 marks]
Answer: Actual KE = ½ × 500 × (22)² = 250 × 484 = 121,000 J
Work against friction = GPE − Actual KE = 196,200 − 121,000 = 75,200 J ≈ 75.2 kJ
Marking:
- [M1] Calculate actual KE or energy difference
- [A1] 75.2 kJ (ecf from part a)
15. A 75 W filament light bulb converts only 8% of the electrical energy into light.
(a) Calculate the power wasted as heat. [1 mark]
Answer: P_wasted = 75 × (1 − 0.08) = 75 × 0.92 = 69 W
Marking:
- [B1] 69 W
(b) Calculate the energy wasted as heat when the bulb is on for 3 hours. Give your answer in joules. [2 marks]
Answer: t = 3 × 3600 = 10,800 s
E_wasted = P_wasted × t = 69 × 10,800 = 745,200 J ≈ 745 kJ
Marking:
- [M1] Convert time to seconds or use E = Pt
- [A1] 745 kJ (ecf from part a)
(c) An LED bulb produces the same light output with a power rating of 10 W. Calculate the energy saved over 3 hours compared to the filament bulb. [2 marks]
Answer: Energy saved = (75 − 10) × 3 × 3600 = 65 × 10,800 = 702,000 J = 702 kJ
Marking:
- [M1] Power difference × time
- [A1] 702 kJ
16. A bungee jumper of mass 70 kg jumps from a bridge. The bungee cord has an unstretched length of 20 m and a spring constant of 150 N m⁻¹. The jumper falls a total distance of 45 m before coming to rest momentarily.
(a) Calculate the gravitational potential energy lost by the jumper during the fall. [2 marks]
Answer: GPE lost = mgh = 70 × 9.81 × 45 = 30,901.5 J ≈ 30.9 kJ
Marking:
- [M1] GPE = mgh with h = 45 m
- [A1] 30.9 kJ
(b) Calculate the elastic potential energy stored in the cord at the lowest point. [2 marks]
Answer: Extension x = 45 − 20 = 25 m
EPE = ½kx² = ½ × 150 × (25)² = 75 × 625 = 46,875 J ≈ 46.9 kJ
Marking:
- [M1] Calculate extension (25 m) and use EPE = ½kx²
- [A1] 46.9 kJ
(c) Explain why the elastic potential energy is less than the gravitational potential energy lost. [1 mark]
Answer: Some energy is dissipated as heat due to air resistance and internal friction in the cord. / Not all GPE is converted to EPE; some is converted to other forms.
Marking:
- [B1] Valid explanation referencing energy dissipation
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]
Answer: P_incident = Intensity × Area = 800 × 2.0 = 1600 W
Marking:
- [B1] 1600 W
(b) Calculate the electrical power output of the panel. [1 mark]
Answer: P_out = 0.18 × 1600 = 288 W
Marking:
- [B1] 288 W (ecf from part a)
(c) The panel charges a 12 V battery. Calculate the current delivered to the battery, assuming no further losses. [2 marks]
Answer: I = P / V = 288 / 12 = 24 A
Marking:
- [M1] I = P/V
- [A1] 24 A (ecf from part b)
(d) Suggest one reason why the actual charging current is lower. [1 mark]
Answer: Any one from:
- Losses in wiring/connections
- Battery internal resistance
- Panel not operating at maximum power point
- Temperature effects reducing efficiency
- Dust or dirt on panel surface
Marking:
- [B1] Valid reason
18. A cyclist of total mass 85 kg (including bicycle) climbs a hill of gradient 1 in 10 (for every 10 m along the slope, the height increases by 1 m) at a constant speed of 4.0 m s⁻¹. The total resistive force is 50 N.
(a) Calculate the component of weight acting down the slope. [2 marks]
Answer: sin θ = 1/10 = 0.1
Component = mg sin θ = 85 × 9.81 × 0.1 = 83.385 N ≈ 83.4 N
Marking:
- [M1] sin θ = 0.1 or correct trigonometric approach
- [A1] 83.4 N
(b) Calculate the total force the cyclist must exert parallel to the slope. [2 marks]
Answer: Total force = Component of weight + Resistive force = 83.4 + 50 = 133.4 N ≈ 133 N
Marking:
- [M1] Add forces (equilibrium at constant speed)
- [A1] 133 N (ecf from part a)
(c) Calculate the power output of the cyclist. [2 marks]
Answer: P = Fv = 133.4 × 4.0 = 533.6 W ≈ 534 W
Marking:
- [M1] P = Fv
- [A1] 534 W (ecf from part b)
19. A compressed air energy storage system uses excess electricity to compress air into an underground cavern at a pressure of 7.0 × 10⁶ Pa. The cavern has a volume of 3000 m³. When energy is needed, the air is released through a turbine. The energy stored per unit volume of compressed air is given by P × ln(P/P₀), where P₀ = 1.0 × 10⁵ Pa.
(a) Calculate the energy stored per unit volume of compressed air. [2 marks]
Answer: P/P₀ = (7.0 × 10⁶) / (1.0 × 10⁵) = 70
ln(70) = 4.248
Energy per unit volume = P × ln(P/P₀) = 7.0 × 10⁶ × 4.248 = 2.974 × 10⁷ J m⁻³ ≈ 2.97 × 10⁷ J m⁻³
Marking:
- [M1] Correct calculation of P/P₀ and ln(P/P₀)
- [A1] 2.97 × 10⁷ J m⁻³
(b) Calculate the total energy stored in the cavern. [1 mark]
Answer: Total energy = Energy per unit volume × Volume = 2.974 × 10⁷ × 3000 = 8.922 × 10¹⁰ J ≈ 8.9 × 10¹⁰ J
Marking:
- [B1] 8.9 × 10¹⁰ J (ecf from part a)
(c) If the turbine has an efficiency of 60%, calculate the electrical energy that can be recovered. [2 marks]
Answer: E_elec = 0.60 × 8.922 × 10¹⁰ = 5.353 × 10¹⁰ J ≈ 5.4 × 10¹⁰ J
Marking:
- [M1] E_elec = efficiency × total energy
- [A1] 5.4 × 10¹⁰ J (ecf from part b)
20. A student investigates the efficiency of an electric motor by using it to lift a 0.50 kg mass through a height of 2.0 m. The motor is connected to a 6.0 V power supply and draws a current of 0.80 A. It takes 3.5 s to lift the mass.
(a) Calculate the useful work done by the motor. [1 mark]
Answer: W_useful = mgh = 0.50 × 9.81 × 2.0 = 9.81 J
Marking:
- [B1] 9.81 J
(b) Calculate the electrical energy supplied to the motor. [2 marks]
Answer: E_elec = VIt = 6.0 × 0.80 × 3.5 = 16.8 J
Marking:
- [M1] E = VIt
- [A1] 16.8 J
(c) Calculate the efficiency of the motor. [1 mark]
Answer: Efficiency = (W_useful / E_elec) × 100% = (9.81 / 16.8) × 100% = 58.4% ≈ 58%
Marking:
- [B1] 58% (ecf from parts a and b)
(d) The student repeats the experiment with a 1.0 kg mass and finds the efficiency is higher. Suggest why. [1 mark]
Answer: The motor's fixed losses (e.g., friction, internal resistance) become a smaller fraction of the total energy when lifting a heavier load, so a greater proportion of input energy goes to useful work. / The motor operates closer to its optimal load.
Marking:
- [B1] Valid explanation referencing proportion of fixed losses or optimal load
END OF ANSWER KEY