A Level H1 Physics Energy Power Quiz

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A-Level Physics H1 Quiz - Energy Power

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ___________ / 40

Duration: 60 minutes
Total Marks: 40

Instructions:

• Answer all 20 questions.
• Show your working clearly for calculation questions.
• Use the provided spaces for your answers.
• Section A: Short Answer (Q1–5), Section B: Calculation (Q6–13), Section C: Structured & Data-Based (Q14–20).

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Section A: Short Answer (1–5)

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

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2. Define power in terms of energy transfer. [2]

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3. Write down the expression for kinetic energy in terms of mass $m$ and speed $v$. [1]

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4. A device has an efficiency of 0.40. State what this means in terms of energy input and useful output. [2]

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5. State the work done by a constant force $F$ acting through a displacement $s$ in the direction of the force. [1]

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Section B: Calculation (6–13)

6. A block of mass $4.0\ \text{kg}$ slides down a frictionless slope from rest and reaches a speed of $6.0\ \text{m s}^{-1}$ at the bottom. Calculate its kinetic energy at the bottom. [2]

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7. A crane lifts a load of weight $800\ \text{N}$ vertically through $12\ \text{m}$ in $20\ \text{s}$. Calculate the useful power output of the crane. [3]

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8. A electric heater uses $2.0\ \text{kJ}$ of electrical energy and delivers $1.5\ \text{kJ}$ as useful heat. Calculate its efficiency as a percentage. [2]

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9. A car of mass $1000\ \text{kg}$ accelerates from $10\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$. Calculate the increase in its kinetic energy. [3]

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10. A pump transfers $500\ \text{kg}$ of water to a height of $8.0\ \text{m}$ in $10\ \text{s}$. Given $g = 9.8\ \text{N kg}^{-1}$, calculate the minimum power required. [3]

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11. A spring of spring constant $k = 200\ \text{N m}^{-1}$ is stretched by $0.10\ \text{m}$. Calculate the elastic potential energy stored. [2]

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12. A machine has an input power of $1.2\ \text{kW}$ and an efficiency of $25\%$. Calculate the useful output power in watts. [2]

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13. A ball of mass $0.50\ \text{kg}$ is dropped from a height of $20\ \text{m}$. Using $g = 9.8\ \text{N kg}^{-1}$, calculate the kinetic energy just before impact (assuming no air resistance). [3]

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Section C: Structured & Data-Based (14–20)

14. (a) Define gravitational potential energy. [2]
(b) A $5.0\ \text{kg}$ mass is raised by $4.0\ \text{m}$. Calculate the increase in gravitational potential energy ($g = 9.8\ \text{N kg}^{-1}$). [2]

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15. A student claims: "In a closed system, kinetic energy is always conserved." Explain why this statement is not correct, using the concept of energy conservation. [3]

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16. The table below shows energy input and useful output for three devices.

Device	Input (J)	Useful Output (J)
X	100	40
Y	200	90
Z	150	75

Calculate the efficiency of each device and state which is most efficient. [4]

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17. A wind turbine generates $3.0\ \text{kW}$ of electrical power from a wind energy input of $15\ \text{kW}$.
(a) Calculate the efficiency. [2]
(b) Suggest one reason why the efficiency is less than 100%. [1]

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18.

Image pending generation: graph for Q18.

The graph shows the force-extension relationship for a spring. Determine the elastic potential energy stored when the extension is $0.20\ \text{m}$ using the graph. [3]

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19. A hydroelectric plant uses water falling through $50\ \text{m}$. In one second, $2.0 \times 10^4\ \text{kg}$ of water passes through.
(a) Calculate the gravitational potential energy lost per second ($g = 9.8\ \text{N kg}^{-1}$). [2]
(b) If the plant converts 80% of this to electrical power, calculate the output power. [2]

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20. (a) State the work-energy theorem. [2]
(b) A constant force of $50\ \text{N}$ pushes a box $4.0\ \text{m}$ in the direction of the force. Calculate the work done. [1]
(c) If the box started from rest and has mass $10\ \text{kg}$, find its final speed. [3]

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A-Level Physics H1 Quiz - Energy Power (Answers)

Total Marks: 40
Topic: Energy & Power

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Section A: Short Answer

1. [2 marks]
Principle of conservation of energy: Energy cannot be created or destroyed; it can only be transferred from one store to another or transformed from one form to another. Total energy of an isolated system is constant.
Marking: [B1] total energy constant / not created destroyed; [B1] transferred/transformed between forms.

2. [2 marks]
Power is the rate of energy transfer (or rate of doing work). $P = \frac{E}{t}$ where $E$ is energy transferred and $t$ is time.
Marking: [B1] rate of energy transfer; [B1] $P = E/t$ or equivalent.

3. [1 mark]
$E_k = \frac{1}{2}mv^2$

4. [2 marks]
Efficiency 0.40 means that 40% of the input energy is converted to useful output energy; 60% is wasted (e.g. as heat).
Marking: [B1] 40% input becomes useful; [B1] rest wasted.

5. [1 mark]
$W = Fs$ (or $W = F \times s$)

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Section B: Calculation

6. [2 marks]
$E_k = \frac{1}{2}mv^2 = \frac{1}{2}(4.0)(6.0)^2 = \frac{1}{2} \times 4.0 \times 36 = 72\ \text{J}$
M1 substitution, A1 answer 72 J.

7. [3 marks]
Work done = $Fs = 800 \times 12 = 9600\ \text{J}$
Power = $\frac{W}{t} = \frac{9600}{20} = 480\ \text{W}$
M1 work, M1 power formula, A1 480 W.

8. [2 marks]
Efficiency = $\frac{\text{useful}}{\text{input}} \times 100\% = \frac{1.5}{2.0} \times 100\% = 75\%$
M1 ratio, A1 75%.

9. [3 marks]
$\Delta E_k = \frac{1}{2}m(v_f^2 - v_i^2) = \frac{1}{2}(1000)(20^2 - 10^2) = 500(400 - 100) = 500 \times 300 = 1.5 \times 10^5\ \text{J}$
M1 formula, M1 substitution, A1 $1.5 \times 10^5$ J.

10. [3 marks]
Weight = $mg = 500 \times 9.8 = 4900\ \text{N}$
Work = $mgh = 4900 \times 8.0 = 39200\ \text{J}$
Power = $\frac{39200}{10} = 3920\ \text{W}$
M1 weight, M1 work, A1 3920 W (or 3.92 kW).

11. [2 marks]
$E = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.10)^2 = 100 \times 0.01 = 1.0\ \text{J}$
M1 sub, A1 1.0 J.

12. [2 marks]
Output = $0.25 \times 1200\ \text{W} = 300\ \text{W}$
M1 conversion 1.2 kW = 1200 W; A1 300 W.

13. [3 marks]
Loss in GPE = gain in KE: $mgh = 0.50 \times 9.8 \times 20 = 98\ \text{J}$
M1 formula, M1 sub, A1 98 J.

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Section C: Structured & Data-Based

14. [4 marks]
(a) [2] Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. $E_p = mgh$.
(b) [2] $\Delta E_p = mgh = 5.0 \times 9.8 \times 4.0 = 196\ \text{J}$.
Marking: (a) [B1] definition, [B1] formula; (b) [M1] sub, [A1] 196 J.

15. [3 marks]
Conservation of energy states total energy is conserved, not specifically kinetic. In inelastic collisions, kinetic energy converts to thermal/sound. A closed system can have KE not conserved while total energy is.
Marking: [B1] total not KE conserved; [B1] example transformation; [B1] closed system total constant.

16. [4 marks]
X: $40/100 = 0.40$ (40%)
Y: $90/200 = 0.45$ (45%)
Z: $75/150 = 0.50$ (50%)
Most efficient: Z.
Marking: [M1] each ratio (max 3), [A1] Z identified.

17. [3 marks]
(a) [2] Eff = $3.0/15 \times 100 = 20\%$.
(b) [1] Some wind energy not captured / friction / turbulence.
Marking: (a) [M1] calc, [A1] 20%; (b) [B1] valid reason.

18. [3 marks]
Area under graph = $\frac{1}{2} \times 0.20 \times 40 = 4.0\ \text{J}$.
Marking: [M1] triangle area, [A1] 4.0 J, [B1] from graph shaded.

19. [4 marks]
(a) [2] GPE lost/s = $mgh = (2.0\times10^4)(9.8)(50) = 9.8\times10^6\ \text{J s}^{-1} = 9.8\ \text{MW}$.
(b) [2] Output = $0.80 \times 9.8\times10^6 = 7.84\times10^6\ \text{W}$.
Marking: (a) [M1] calc, [A1]; (b) [M1] 80%, [A1].

20. [6 marks]
(a) [2] Work-energy theorem: net work done on object equals change in its kinetic energy.
(b) [1] $W = 50 \times 4.0 = 200\ \text{J}$.
(c) [3] $\frac{1}{2}mv^2 = 200 \Rightarrow v^2 = \frac{400}{10} = 40 \Rightarrow v = 6.32\ \text{m s}^{-1}$.
Marking: (a) [B1]×2; (b) [A1]; (c) [M1] formula, [M1] solve, [A1] 6.32.