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A Level H1 Physics Energy Power Quiz

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A Level H1 Physics From Real Exams Generated by Gemma 4 31B Updated 2026-06-03

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Questions

A-Level Physics H1 Quiz - Energy Power

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 55

Duration: 60 Minutes
Total Marks: 55

Instructions:

Answer all questions.
Show all necessary working for calculation questions.
Use $g = 9.81 \text{ m s}^{-2}$ where applicable.

Section A: Fundamental Concepts (Short Answer)

Define the term work done by a force. [2]

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State the relationship between power, work done, and time. [1]

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A force of $15.0 \text{ N}$ acts on a body, displacing it by $2.50 \text{ m}$ in the direction of the force. Calculate the work done. [2]

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Explain why the work done by a force is zero if the displacement is perpendicular to the force. [2]

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State the SI unit of power and define it in terms of base units. [2]

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Section B: Energy and Power Applications (Calculation & Proof)

A motor lifts a mass of $20.0 \text{ kg}$ vertically through a height of $5.00 \text{ m}$ in $10.0 \text{ s}$ . Calculate the useful power output of the motor. [3]

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The motor in Question 6 has an efficiency of $65\%$ . Calculate the total electrical power input. [3]

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A car of mass $1200 \text{ kg}$ accelerates from rest to $25.0 \text{ m s}^{-1}$ in $8.00 \text{ s}$ . Calculate the average power delivered by the engine, ignoring air resistance. [3]

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A block of mass $0.500 \text{ kg}$ slides down a rough inclined plane. It starts from rest at a height of $2.00 \text{ m}$ and reaches the bottom with a speed of $4.00 \text{ m s}^{-1}$ . Calculate the energy lost to friction. [3]

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A pump delivers water at a rate of $0.120 \text{ kg s}^{-1}$ to a tank $15.0 \text{ m}$ above the pump. Calculate the minimum power required. [3]

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A constant force of $50.0 \text{ N}$ acts at an angle of $30.0^\circ$ to the horizontal. Calculate the work done when the object is moved $4.00 \text{ m}$ horizontally. [3]

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An electric heater is rated at $2.50 \text{ kW}$ . Calculate the energy it transfers to the surroundings in $15.0$ minutes. [2]

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A ball of mass $0.200 \text{ kg}$ is dropped from a height of $3.00 \text{ m}$ . Calculate its speed just before it hits the ground, assuming no air resistance. [3]

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A spring with force constant $k = 500 \text{ N m}^{-1}$ is compressed by $0.0400 \text{ m}$ . Calculate the elastic potential energy stored. [2]

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A crane lifts a $500 \text{ kg}$ crate at a constant speed of $0.200 \text{ m s}^{-1}$ . Calculate the power output of the crane. [3]

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Section C: Structured Analysis (Data & Diagram Interpretation)

A car of mass $1500 \text{ kg}$ travels at a constant speed of $20.0 \text{ m s}^{-1}$ . The total resistive force (air resistance and friction) is $600 \text{ N}$ . (a) Calculate the power required to maintain this constant speed. [2]

(b) If the car accelerates to $30.0 \text{ m s}^{-1}$ , explain qualitatively how the power required changes, assuming the resistive force increases with speed. [3]
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A pendulum bob of mass $0.100 \text{ kg}$ is released from a point where it is $0.150 \text{ m}$ above its equilibrium position. (a) Calculate the maximum kinetic energy of the bob. [2]

(b) Determine the maximum speed of the bob. [3]
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A battery with EMF $\varepsilon$ and internal resistance $r$ is connected to a load resistor $R$ . (a) State the expression for the power dissipated in the load resistor $R$ . [2]

(b) Explain why the power delivered to the load is lower when the internal resistance $r$ is increased. [3]
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A $1.0 \text{ kg}$ object is pushed up a slope of angle $20^\circ$ at a constant speed of $0.5 \text{ m s}^{-1}$ . The coefficient of friction is $0.1$ . (a) Calculate the work done against gravity per second. [3]

(b) Calculate the total power required to move the object. [4]
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A graph of Force $F$ vs Displacement $s$ is provided for a spring (linear through origin). (a) Explain how the area under an $F-s$ graph relates to energy. [2]

(b) If the gradient of the graph is $200 \text{ N m}^{-1}$ , calculate the work done to stretch the spring from $0.1 \text{ m}$ to $0.2 \text{ m}$ . [3]
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Answers

A-Level Physics H1 Quiz - Energy Power (Answer Key)

Definition: The product of the force applied to an object and the displacement of the object in the direction of the force. [2]
Relationship: $\text{Power} = \frac{\text{Work Done}}{\text{Time}}$ or $P = \frac{W}{t}$ . [1]
$W = Fs = 15.0 \times 2.50 = 37.5 \text{ J}$ . [2]
Explanation: Work done is $W = Fs \cos \theta$ . If $\theta = 90^\circ$ , $\cos 90^\circ = 0$ , therefore $W = 0$ . No component of the force acts in the direction of displacement. [2]
SI Unit: Watt (W). Base units: $\text{kg m}^2 \text{ s}^{-3}$ (from $\text{J/s} = (\text{kg m}^2 \text{ s}^{-2}) / \text{s}$ ). [2]
$P = \frac{mgh}{t} = \frac{20.0 \times 9.81 \times 5.00}{10.0} = 98.1 \text{ W}$ . [3]
$\text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}} = \frac{98.1}{0.65} = 150.9 \text{ W}$ (or $151 \text{ W}$ ). [3]
$\Delta KE = \frac{1}{2}mv^2 = 0.5 \times 1200 \times 25^2 = 375,000 \text{ J}$ . $P = \frac{W}{t} = \frac{375,000}{8.00} = 46,875 \text{ W}$ or $46.9 \text{ kW}$ . [3]
$PE_{\text{initial}} = mgh = 0.500 \times 9.81 \times 2.00 = 9.81 \text{ J}$ . $KE_{\text{final}} = \frac{1}{2}mv^2 = 0.5 \times 0.500 \times 4^2 = 4.00 \text{ J}$ . $\text{Energy lost} = 9.81 - 4.00 = 5.81 \text{ J}$ . [3]
$P = \frac{mgh}{t} = \frac{dm}{dt}gh = 0.120 \times 9.81 \times 15.0 = 17.66 \text{ W}$ . [3]
$W = Fs \cos \theta = 50.0 \times 4.00 \times \cos(30^\circ) = 173.2 \text{ J}$ . [3]
$E = P \times t = 2500 \times (15 \times 60) = 2,250,000 \text{ J}$ or $2.25 \text{ MJ}$ . [2]
$mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 3.00} = 7.67 \text{ m s}^{-1}$ . [3]
$E = \frac{1}{2}kx^2 = 0.5 \times 500 \times (0.04)^2 = 0.400 \text{ J}$ . [2]
$P = Fv = (mg)v = (500 \times 9.81) \times 0.200 = 981 \text{ W}$ . [3]
(a) $P = Fv = 600 \times 20.0 = 12,000 \text{ W}$ or $12 \text{ kW}$ . [2] (b) Power increases. Both $F$ (resistive force) and $v$ (velocity) increase. Since $P = Fv$ , the product increases significantly. [3]
(a) $KE_{\max} = PE_{\max} = mgh = 0.100 \times 9.81 \times 0.150 = 0.147 \text{ J}$ . [2] (b) $0.147 = 0.5 \times 0.100 \times v^2 \Rightarrow v = \sqrt{2.94} = 1.71 \text{ m s}^{-1}$ . [3]
(a) $P = I^2R$ or $P = \frac{V^2}{R}$ (where $V$ is terminal voltage). [2] (b) Internal resistance $r$ acts as a potential divider. Increasing $r$ increases the voltage drop across the internal part of the battery, reducing the terminal voltage $V$ available to the load $R$ . Since $P = \frac{V^2}{R}$ , power decreases. [3]
(a) $P_{\text{grav}} = mgh/t = mg \sin\theta \times v = 1.0 \times 9.81 \times \sin(20^\circ) \times 0.5 = 1.68 \text{ W}$ . [3] (b) $F_{\text{friction}} = \mu mg \cos\theta = 0.1 \times 1.0 \times 9.81 \times \cos(20^\circ) = 0.922 \text{ N}$ . $P_{\text{friction}} = F_{\text{friction}} \times v = 0.922 \times 0.5 = 0.461 \text{ W}$ . $P_{\text{total}} = 1.68 + 0.461 = 2.14 \text{ W}$ . [4]
(a) The area under a Force-Displacement graph represents the work done by the force. [2] (b) $W = \int F ds = \frac{1}{2}k(x_2^2 - x_1^2) = 0.5 \times 200 \times (0.2^2 - 0.1^2) = 100 \times (0.04 - 0.01) = 3.0 \text{ J}$ . [3]