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A Level H2 Physics Energy Power Quiz
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Questions
A-Level Physics H2 Quiz - Energy Power
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 60
Duration: 1 hour 15 minutes Total Marks: 60
Instructions:
- This quiz contains 20 questions on the topic of Energy & Power.
- Answer ALL questions in the spaces provided.
- Show all working clearly for calculation questions.
- Use g = 9.81 m s⁻² unless otherwise stated.
- The number of marks is given in brackets [ ] at the end of each question or part question.
Section A: Structured Questions (15 marks)
Answer all questions in this section.
1. State the principle of conservation of energy. [2 marks]
2. Define power and give its SI unit. [2 marks]
3. A force of 50 N pushes a box through a distance of 8.0 m along a horizontal surface. The force acts at an angle of 30° to the horizontal. Calculate the work done by the force. [2 marks]
4. Explain why the work done by a centripetal force on an object moving in uniform circular motion is zero. [2 marks]
5. A machine lifts a mass of 200 kg vertically upwards at a constant speed of 0.50 m s⁻¹. Calculate the power output of the machine. [2 marks]
Section B: Structured Questions (15 marks)
Answer all questions in this section.
6. Distinguish between elastic potential energy and gravitational potential energy. [2 marks]
7. A spring of spring constant k = 500 N m⁻¹ is compressed by 0.040 m. Calculate the elastic potential energy stored in the spring. [2 marks]
8. State one situation in which the work done by a force is negative, and explain your answer. [1 mark]
9. A car of mass 1200 kg accelerates uniformly from rest to 25 m s⁻¹ in 8.0 s along a straight, level road. The total resistive force acting on the car is 600 N.
(a) Calculate the acceleration of the car. [2 marks]
(b) Calculate the net force required to produce this acceleration. [2 marks]
(c) Calculate the force exerted by the engine. [2 marks]
10. A pump raises water from a well 30 m deep at a rate of 15 kg s⁻¹. The water emerges from the pipe at a speed of 4.0 m s⁻¹.
(a) Calculate the gain in gravitational potential energy per second of the water raised. [2 marks]
(b) Calculate the gain in kinetic energy per second of the water. [2 marks]
Section C: Calculations and Applications (15 marks)
Answer all questions in this section.
11. A block of mass 5.0 kg slides down a frictionless incline of height 3.0 m. At the bottom, it continues along a rough horizontal surface with coefficient of kinetic friction 0.25.
(a) Calculate the speed of the block at the bottom of the incline. [2 marks]
(b) Calculate the kinetic energy of the block at the bottom of the incline. [1 mark]
(c) Calculate the distance the block travels on the rough surface before coming to rest. [3 marks]
12. Refer to the car in question 9.
(d) Calculate the work done by the engine during the acceleration. [3 marks]
(e) Calculate the average power output of the engine during this time. [2 marks]
13. Refer to the pump in question 10.
(c) Hence, calculate the minimum power required by the pump. [2 marks]
(d) The pump has an efficiency of 65%. Calculate the electrical power input required. [2 marks]
Section D: Data Analysis and Extended Response (15 marks)
Answer all questions in this section.
14. A student investigates the relationship between the power P dissipated in a resistor and the current I flowing through it. The following data are obtained:
| Current I / A | 0.50 | 1.00 | 1.50 | 2.00 | 2.50 | 3.00 |
|---|---|---|---|---|---|---|
| Power P / W | 0.55 | 2.20 | 4.95 | 8.80 | 13.8 | 19.8 |
It is believed that P and I are related by a power law of the form P = kIⁿ, where k and n are constants.
(a) Explain how the data can be used to verify this relationship and determine the values of k and n. [3 marks]
(b) Use the data for I = 1.00 A and I = 3.00 A to estimate the value of n. [3 marks]
(c) Hence, determine the value of k, stating its units. [2 marks]
15. A hydroelectric power station uses water stored in a reservoir at a height of 120 m above the turbines. Water flows through the turbines at a rate of 500 kg s⁻¹. The efficiency of the turbine-generator system is 80%.
(a) Calculate the gravitational potential energy lost per second by the water. [2 marks]
(b) Calculate the theoretical maximum power that could be generated. [1 mark]
(c) Calculate the actual electrical power output of the station. [2 marks]
16. Refer to the hydroelectric power station in question 15.
(d) The electrical power is transmitted at a potential difference of 25 kV. Calculate the current in the transmission cables. [2 marks]
17. Refer to the hydroelectric power station in question 15.
(e) The transmission cables have a total resistance of 4.0 Ω. Calculate the power loss in the cables. [2 marks]
18. Refer to the hydroelectric power station in question 15.
(f) Suggest and explain one method of reducing the power loss in the transmission cables. [2 marks]
19. A pendulum bob of mass 0.50 kg is released from rest at a point where the string makes an angle of 60° with the vertical. The length of the string is 1.2 m. Calculate the speed of the bob at the lowest point of its swing. [3 marks]
20. A force F = (3.0i + 4.0j) N acts on a particle as it moves from position r₁ = (2.0i + 1.0j) m to r₂ = (5.0i + 5.0j) m. Calculate the work done by the force. [3 marks]
END OF QUIZ
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Answers
A-Level Physics H2 Quiz - Energy Power: Answer Key
Total Marks: 60
Section A: Structured Questions (15 marks)
1. State the principle of conservation of energy. [2 marks]
Answer: Energy cannot be created or destroyed; it can only be transferred or transformed from one form to another. [1 mark] The total energy of an isolated system remains constant. [1 mark]
Marking notes: Award 1 mark for "cannot be created or destroyed" and 1 mark for "total energy constant" or "transferred/transformed". Accept equivalent phrasing.
2. Define power and give its SI unit. [2 marks]
Answer: Power is the rate of doing work, or the rate at which energy is transferred or converted. [1 mark] SI unit: watt (W), equivalent to J s⁻¹. [1 mark]
Marking notes: Award 1 mark for correct definition (rate of work done or energy transfer) and 1 mark for watt (W).
3. A force of 50 N pushes a box through a distance of 8.0 m along a horizontal surface. The force acts at an angle of 30° to the horizontal. Calculate the work done by the force. [2 marks]
Answer: W = F s cos θ [1 mark] W = 50 × 8.0 × cos 30° W = 50 × 8.0 × 0.866 W = 346.4 J ≈ 350 J [1 mark]
Marking notes: Award 1 mark for correct formula with cos θ, 1 mark for correct answer with unit. Accept 346–350 J.
4. Explain why the work done by a centripetal force on an object moving in uniform circular motion is zero. [2 marks]
Answer: The centripetal force is always directed towards the centre of the circle, perpendicular to the direction of motion (velocity) of the object. [1 mark] Since work done W = F s cos θ, and θ = 90°, cos 90° = 0, so W = 0. No work is done. [1 mark]
Marking notes: Award 1 mark for stating force is perpendicular to displacement/velocity, 1 mark for linking to W = Fs cos 90° = 0 or explaining that there is no displacement in the direction of the force.
5. A machine lifts a mass of 200 kg vertically upwards at a constant speed of 0.50 m s⁻¹. Calculate the power output of the machine. [2 marks]
Answer: Force required = weight = mg = 200 × 9.81 = 1962 N [1 mark] Power = F v = 1962 × 0.50 = 981 W ≈ 980 W [1 mark]
Marking notes: Award 1 mark for correct force (weight), 1 mark for correct power with unit. Accept 980–981 W. Alternative method using P = mgh/t also acceptable.
Section B: Structured Questions (15 marks)
6. Distinguish 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. [1 mark] Gravitational potential energy is the energy possessed by an object due to its position in a gravitational field, typically relative to a reference level. [1 mark]
Marking notes: Award 1 mark for each correct description. Accept: elastic PE depends on deformation/extension; gravitational PE depends on height/mass/field strength.
7. A spring of spring constant k = 500 N m⁻¹ is compressed by 0.040 m. Calculate the elastic potential energy stored in the spring. [2 marks]
Answer: E = ½ k x² [1 mark] E = ½ × 500 × (0.040)² E = 250 × 0.0016 E = 0.40 J [1 mark]
Marking notes: Award 1 mark for correct formula, 1 mark for correct answer with unit.
8. State one situation in which the work done by a force is negative, and explain your answer. [1 mark]
Answer: When friction acts on a moving object, the friction force opposes the direction of motion. The work done by friction is negative because the force and displacement are in opposite directions (θ = 180°, cos 180° = -1). [1 mark]
Marking notes: Award 1 mark for any valid example with correct explanation. Accept: lowering an object (gravity does positive work, applied force does negative work); deceleration where net force opposes motion.
9. A car of mass 1200 kg accelerates uniformly from rest to 25 m s⁻¹ in 8.0 s along a straight, level road. The total resistive force acting on the car is 600 N.
(a) Calculate the acceleration of the car. [2 marks]
Answer: a = (v - u) / t [1 mark] a = (25 - 0) / 8.0 = 3.125 m s⁻² ≈ 3.1 m s⁻² [1 mark]
(b) Calculate the net force required to produce this acceleration. [2 marks]
Answer: F_net = m a [1 mark] F_net = 1200 × 3.125 = 3750 N ≈ 3800 N [1 mark]
(c) Calculate the force exerted by the engine. [2 marks]
Answer: F_engine - F_resistive = F_net [1 mark] F_engine = F_net + F_resistive = 3750 + 600 = 4350 N ≈ 4400 N [1 mark]
10. A pump raises water from a well 30 m deep at a rate of 15 kg s⁻¹. The water emerges from the pipe at a speed of 4.0 m s⁻¹.
(a) Calculate the gain in gravitational potential energy per second of the water raised. [2 marks]
Answer: GPE gained per kg = mgh = 1 × 9.81 × 30 = 294.3 J kg⁻¹ [1 mark] GPE gained per second = 15 × 294.3 = 4414.5 W ≈ 4.4 × 10³ W [1 mark]
(b) Calculate the gain in kinetic energy per second of the water. [2 marks]
Answer: KE gained per kg = ½ m v² = ½ × 1 × (4.0)² = 8.0 J kg⁻¹ [1 mark] KE gained per second = 15 × 8.0 = 120 W [1 mark]
Section C: Calculations and Applications (15 marks)
11. A block of mass 5.0 kg slides down a frictionless incline of height 3.0 m. At the bottom, it continues along a rough horizontal surface with coefficient of kinetic friction 0.25.
(a) Calculate the speed of the block at the bottom of the incline. [2 marks]
Answer: By conservation of energy: mgh = ½ m v² [1 mark] v = √(2gh) = √(2 × 9.81 × 3.0) = √58.86 = 7.67 m s⁻¹ ≈ 7.7 m s⁻¹ [1 mark]
(b) Calculate the kinetic energy of the block at the bottom of the incline. [1 mark]
Answer: KE = ½ m v² = ½ × 5.0 × (7.67)² = 147 J [1 mark] Or: KE = mgh = 5.0 × 9.81 × 3.0 = 147 J
(c) Calculate the distance the block travels on the rough surface before coming to rest. [3 marks]
Answer: Work done by friction = loss in KE [1 mark] Frictional force f = μ N = μ m g = 0.25 × 5.0 × 9.81 = 12.26 N [1 mark] f × d = KE → d = KE / f = 147 / 12.26 = 12.0 m [1 mark]
Alternative: μ m g d = ½ m v² → d = v²/(2μg) = 7.67²/(2 × 0.25 × 9.81) = 12.0 m.
12. Refer to the car in question 9.
(d) Calculate the work done by the engine during the acceleration. [3 marks]
Answer: Distance travelled: s = ½ (u + v) t = ½ × (0 + 25) × 8.0 = 100 m [1 mark] Work done by engine = F_engine × s [1 mark] W = 4350 × 100 = 435 000 J = 4.35 × 10⁵ J [1 mark]
Alternative: W = gain in KE + work against resistance = ½ × 1200 × 25² + 600 × 100 = 375 000 + 60 000 = 435 000 J.
(e) Calculate the average power output of the engine during this time. [2 marks]
Answer: P = W / t [1 mark] P = 435 000 / 8.0 = 54 375 W ≈ 5.4 × 10⁴ W (or 54 kW) [1 mark]
Alternative: P = F_engine × v_average = 4350 × 12.5 = 54 375 W.
13. Refer to the pump in question 10.
(c) Hence, calculate the minimum power required by the pump. [2 marks]
Answer: Minimum power = GPE gain per second + KE gain per second [1 mark] P_min = 4414.5 + 120 = 4534.5 W ≈ 4.5 × 10³ W [1 mark]
(d) The pump has an efficiency of 65%. Calculate the electrical power input required. [2 marks]
Answer: Efficiency = P_output / P_input [1 mark] P_input = P_output / η = 4534.5 / 0.65 = 6976 W ≈ 7.0 × 10³ W (or 7.0 kW) [1 mark]
Section D: Data Analysis and Extended Response (15 marks)
14. Power law relationship P = kIⁿ
(a) Explain how the data can be used to verify this relationship and determine the values of k and n. [3 marks]
Answer: Take logarithms of both sides: log P = log k + n log I. [1 mark] Plot a graph of log P (y-axis) against log I (x-axis). [1 mark] If the relationship holds, the graph will be a straight line. The gradient gives n, and the y-intercept gives log k, from which k can be found. [1 mark]
(b) Use the data for I = 1.00 A and I = 3.00 A to estimate the value of n. [3 marks]
Answer: P₁ = 2.20 W at I₁ = 1.00 A; P₂ = 19.8 W at I₂ = 3.00 A [1 mark] P₂/P₁ = (I₂/I₁)ⁿ [1 mark] 19.8/2.20 = (3.00/1.00)ⁿ → 9.0 = 3ⁿ n = log 9.0 / log 3.0 = 2.0 [1 mark]
(c) Hence, determine the value of k, stating its units. [2 marks]
Answer: Using P = k I² with I = 1.00 A, P = 2.20 W: [1 mark] k = P / I² = 2.20 / (1.00)² = 2.20 W A⁻² [1 mark]
15. A hydroelectric power station uses water stored in a reservoir at a height of 120 m above the turbines. Water flows through the turbines at a rate of 500 kg s⁻¹. The efficiency of the turbine-generator system is 80%.
(a) Calculate the gravitational potential energy lost per second by the water. [2 marks]
Answer: GPE lost per kg = mgh = 1 × 9.81 × 120 = 1177.2 J kg⁻¹ [1 mark] GPE lost per second = 500 × 1177.2 = 588 600 W = 5.89 × 10⁵ W [1 mark]
(b) Calculate the theoretical maximum power that could be generated. [1 mark]
Answer: Theoretical maximum power = GPE lost per second = 5.89 × 10⁵ W [1 mark]
(c) Calculate the actual electrical power output of the station. [2 marks]
Answer: Efficiency = P_output / P_input [1 mark] P_output = η × P_input = 0.80 × 588 600 = 470 880 W ≈ 4.7 × 10⁵ W [1 mark]
16. Refer to the hydroelectric power station in question 15.
(d) The electrical power is transmitted at a potential difference of 25 kV. Calculate the current in the transmission cables. [2 marks]
Answer: P = I V [1 mark] I = P / V = 470 880 / 25 000 = 18.84 A ≈ 19 A [1 mark]
17. Refer to the hydroelectric power station in question 15.
(e) The transmission cables have a total resistance of 4.0 Ω. Calculate the power loss in the cables. [2 marks]
Answer: P_loss = I² R [1 mark] P_loss = (18.84)² × 4.0 = 1419 W ≈ 1.4 × 10³ W [1 mark]
18. Refer to the hydroelectric power station in question 15.
(f) Suggest and explain one method of reducing the power loss in the transmission cables. [2 marks]
Answer: Increase the transmission voltage. [1 mark] Since P = I V, for the same power transmitted, a higher voltage results in a lower current. Power loss in cables is given by P_loss = I² R, so a lower current significantly reduces the power loss. [1 mark]
Accept: Use thicker cables to reduce resistance R, since P_loss = I² R. However, this is less practical for long-distance transmission.
19. A pendulum bob of mass 0.50 kg is released from rest at a point where the string makes an angle of 60° with the vertical. The length of the string is 1.2 m. Calculate the speed of the bob at the lowest point of its swing. [3 marks]
Answer: Vertical height difference: h = L - L cos 60° = 1.2 - 1.2 × 0.5 = 0.6 m [1 mark] By conservation of energy: mgh = ½ m v² [1 mark] v = √(2gh) = √(2 × 9.81 × 0.6) = √11.772 = 3.43 m s⁻¹ ≈ 3.4 m s⁻¹ [1 mark]
20. A force F = (3.0i + 4.0j) N acts on a particle as it moves from position r₁ = (2.0i + 1.0j) m to r₂ = (5.0i + 5.0j) m. Calculate the work done by the force. [3 marks]
Answer: Displacement: Δr = r₂ - r₁ = (5.0 - 2.0)i + (5.0 - 1.0)j = (3.0i + 4.0j) m [1 mark] Work done W = F · Δr [1 mark] W = (3.0)(3.0) + (4.0)(4.0) = 9.0 + 16.0 = 25 J [1 mark]
END OF ANSWER KEY