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A Level H2 Physics Energy Power Quiz

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A Level H2 Physics AI Generated Generated by Qwen3.6 Plus Updated 2026-07-10

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Questions

A-Level Physics H2 Quiz - Energy Power

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

Duration: 45 minutes
Total Marks: 50

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working clearly. Marks may be awarded for correct working even if the final answer is incorrect.
Use $g = 9.81 \text{ m s}^{-2}$ where appropriate.

Section A: Multiple Choice & Short Concepts (10 Marks)

1. Which of the following correctly defines the efficiency of a machine?
[1]
A. $\frac{\text{Useful Power Output}}{\text{Total Power Input}} \times 100\%$
B. $\frac{\text{Total Power Input}}{\text{Useful Power Output}} \times 100\%$
C. $\frac{\text{Wasted Power}}{\text{Total Power Input}} \times 100\%$
D. $\frac{\text{Useful Energy Output}}{\text{Wasted Energy}} \times 100\%$

Answer: ______

2. A car of mass $1200 \text{ kg}$ travels at a constant speed of $25 \text{ m s}^{-1}$ on a horizontal road. The total resistive force acting on the car is $800 \text{ N}$ . What is the power developed by the engine?
[1]
A. $15 \text{ kW}$
B. $20 \text{ kW}$
C. $30 \text{ kW}$
D. $375 \text{ kW}$

Answer: ______

3. State the Principle of Conservation of Energy.
[2]

4. A ball is dropped from a height. As it falls, air resistance acts on it. Describe the energy transformations that occur from the moment it is released until it reaches terminal velocity.
[2]

5. Define the term power.
[1]

Section B: Structured Calculations (25 Marks)

6. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards from rest. The load accelerates uniformly at $0.50 \text{ m s}^{-2}$ for $4.0 \text{ s}$ .
(a) Calculate the velocity of the load after $4.0 \text{ s}$ .
[1]

(b) Calculate the gain in gravitational potential energy of the load after $4.0 \text{ s}$ .
[2]

(d) Determine the average power developed by the crane motor during the first $4.0 \text{ s}$ .
[3]

7. An electric motor is used to pump water from a well. The motor has an input power of $2.5 \text{ kW}$ and an efficiency of $60\%$ . The water is pumped vertically through a height of $12 \text{ m}$ .
(a) Calculate the useful output power of the motor.
[1]

(b) Calculate the mass of water pumped per minute. (Density of water = $1000 \text{ kg m}^{-3}$ )
[3]

8. A block of mass $2.0 \text{ kg}$ slides down a rough inclined plane. The plane is inclined at $30^\circ$ to the horizontal. The block starts from rest and travels a distance of $5.0 \text{ m}$ down the slope, reaching a speed of $4.0 \text{ m s}^{-1}$ .
(a) Calculate the loss in gravitational potential energy of the block.
[2]

(b) Calculate the gain in kinetic energy of the block.
[1]

9. A car of mass $1500 \text{ kg}$ is traveling on a horizontal road. The engine provides a constant driving force of $3000 \text{ N}$ . The resistive forces are proportional to the square of the speed ( $F_{res} = kv^2$ ). At a speed of $20 \text{ m s}^{-1}$ , the car is moving at constant velocity.
(a) Calculate the value of the constant $k$ .
[2]

(b) Calculate the instantaneous acceleration of the car when its speed is $10 \text{ m s}^{-1}$ .
[3]

10. A hydroelectric power station uses water falling from a height of $150 \text{ m}$ to drive turbines. The flow rate of water is $200 \text{ m}^3 \text{ s}^{-1}$ . The overall efficiency of the system (conversion of gravitational potential energy to electrical energy) is $85\%$ .
(a) Calculate the mass of water falling per second.
[1]

(b) Calculate the electrical power output of the station.
[3]

Section C: Advanced Applications & Analysis (15 Marks)

11. A cyclist travels up a hill inclined at $5.0^\circ$ to the horizontal at a constant speed of $8.0 \text{ m s}^{-1}$ . The total mass of the cyclist and bicycle is $80 \text{ kg}$ . The average resistive force (air resistance and friction) is $40 \text{ N}$ .
(a) Calculate the component of the weight acting down the slope.
[2]

(b) Calculate the power output required from the cyclist to maintain this speed.
[3]

12. In a physics experiment, a student investigates the relationship between the power $P$ dissipated in a resistor and the current $I$ flowing through it. The student plots a graph of $\log_{10} P$ against $\log_{10} I$ and obtains a straight line with a gradient of $2.0$ and a y-intercept of $0.60$ .
(a) State the mathematical relationship between $P$ and $I$ derived from this graph.
[2]

(b) Determine the resistance of the resistor.
[2]

13. A rocket of initial mass $M$ is launched vertically. As it burns fuel, its mass decreases. Explain, using the concepts of work and energy, why the acceleration of the rocket increases even if the thrust force remains constant. (Assume air resistance is negligible for this explanation).
[3]

14. A pendulum bob of mass $0.50 \text{ kg}$ is released from a height of $0.20 \text{ m}$ above its lowest point. At the lowest point, it collides with a stationary block of mass $1.5 \text{ kg}$ on a smooth horizontal surface. The bob and block stick together after the collision.
(a) Calculate the speed of the bob just before the collision.
[2]

(b) Calculate the kinetic energy of the combined mass immediately after the collision.
[3]

15. Solar panels are installed on a roof with a total area of $20 \text{ m}^2$ . The average solar irradiance (power per unit area) is $800 \text{ W m}^{-2}$ . The panels have an efficiency of $15\%$ .
(a) Calculate the total electrical power generated by the panels.
[2]

(b) If the average household consumes $10 \text{ kWh}$ of energy per day, estimate how many hours of sunlight are required to meet this daily demand using these panels.
[3]

16. A student pushes a box of mass $10 \text{ kg}$ across a rough horizontal floor with a constant horizontal force of $50 \text{ N}$ . The coefficient of kinetic friction between the box and the floor is $0.30$ .
(a) Calculate the frictional force acting on the box.
[2]

(b) Calculate the work done by the student in pushing the box a distance of $5.0 \text{ m}$ .
[1]

17. An elevator of mass $800 \text{ kg}$ carries passengers of total mass $200 \text{ kg}$ . It ascends at a constant speed of $3.0 \text{ m s}^{-1}$ . The frictional forces opposing the motion are $1500 \text{ N}$ .
(a) Calculate the tension in the cable supporting the elevator.
[2]

(b) Calculate the power output of the motor lifting the elevator.
[2]

18. A spring with a spring constant $k = 200 \text{ N m}^{-1}$ is compressed by $0.15 \text{ m}$ . A ball of mass $0.05 \text{ kg}$ is placed against the spring. When released, the spring launches the ball horizontally.
(a) Calculate the elastic potential energy stored in the spring.
[2]

(b) Assuming no energy losses, calculate the speed of the ball as it leaves the spring.
[2]

19. A pump lifts $500 \text{ kg}$ of water from a depth of $10 \text{ m}$ to a tank at ground level in $20 \text{ s}$ .
(a) Calculate the useful power output of the pump.
[2]

(b) If the pump is rated at $5.0 \text{ kW}$ , calculate its efficiency.
[2]

20. A car engine has a maximum power output of $100 \text{ kW}$ . The car has a mass of $1200 \text{ kg}$ .
(a) Calculate the maximum theoretical acceleration of the car when it is traveling at $20 \text{ m s}^{-1}$ , assuming no resistive forces.
[2]

(b) Explain why the actual acceleration would be lower than this value.
[1]

Answers

A-Level Physics H2 Quiz - Energy Power (Answer Key)

1. A
[1]

2. B
$P = Fv = 800 \times 25 = 20,000 \text{ W} = 20 \text{ kW}$ .
[1]

3. Energy cannot be created or destroyed; it can only be transformed from one form to another or transferred from one body to another. The total energy of an isolated system remains constant.
[2] (1 mark for "cannot be created/destroyed", 1 mark for "transformed/transferred" or "total constant")

4. Gravitational potential energy is converted into kinetic energy and internal energy (heat) due to work done against air resistance. As terminal velocity is reached, the rate of loss of GPE equals the rate of work done against air resistance, so KE remains constant.
[2] (1 mark for GPE to KE + Internal/Heat, 1 mark for mention of terminal velocity condition or constant KE)

5. Power is the rate of doing work or the rate of energy transfer.
[1]

6.
(a) $v = u + at = 0 + (0.50)(4.0) = 2.0 \text{ m s}^{-1}$ .
[1]

(b) Height $h = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(0.50)(4.0)^2 = 4.0 \text{ m}$ .
$\Delta E_p = mgh = 500 \times 9.81 \times 4.0 = 19,620 \text{ J} \approx 19.6 \text{ kJ}$ .
[2] (1 mark for height, 1 mark for correct energy)

(c) $T - mg = ma \Rightarrow T = m(g+a) = 500(9.81 + 0.50) = 500(10.31) = 5155 \text{ N} \approx 5.2 \text{ kN}$ .
[2] (1 mark for equation, 1 mark for answer)

(d) Work Done by Tension $W = T \times h = 5155 \times 4.0 = 20,620 \text{ J}$ .
Average Power $P = \frac{W}{t} = \frac{20,620}{4.0} = 5155 \text{ W} \approx 5.2 \text{ kW}$ .
Alternative: Average velocity $\bar{v} = \frac{0+2}{2} = 1.0 \text{ m s}^{-1}$ . $P = T \bar{v} = 5155 \times 1.0 = 5.2 \text{ kW}$ .
[3] (1 mark for Work/Energy or Force x Avg Vel, 1 mark for substitution, 1 mark for answer)

7.
(a) $P_{out} = \text{Efficiency} \times P_{in} = 0.60 \times 2500 = 1500 \text{ W}$ .
[1]

(b) Useful Power $P = \frac{mgh}{t}$ .
$1500 = \frac{m \times 9.81 \times 12}{60}$ (time = 60s for "per minute").
$m = \frac{1500 \times 60}{9.81 \times 12} = \frac{90000}{117.72} \approx 764.5 \text{ kg}$ .
Answer: $760 \text{ kg}$ (2 s.f.).
[3] (1 mark for formula rearrangement, 1 mark for substitution, 1 mark for answer)

8.
(a) Vertical height $h = 5.0 \sin(30^\circ) = 2.5 \text{ m}$ .
$\Delta E_p = mgh = 2.0 \times 9.81 \times 2.5 = 49.05 \text{ J} \approx 49 \text{ J}$ .
[2] (1 mark for height, 1 mark for energy)

(b) $\Delta E_k = \frac{1}{2}mv^2 - 0 = \frac{1}{2}(2.0)(4.0)^2 = 16 \text{ J}$ .
[1]

(c) Work done against friction $W_f = \Delta E_p - \Delta E_k = 49.05 - 16 = 33.05 \text{ J}$ .
$W_f = F_f \times d \Rightarrow 33.05 = F_f \times 5.0$ .
$F_f = \frac{33.05}{5.0} = 6.61 \text{ N} \approx 6.6 \text{ N}$ .
[3] (1 mark for energy difference, 1 mark for work-force relation, 1 mark for answer)

9.
(a) At constant velocity, Driving Force = Resistive Force.
$3000 = k(20)^2 \Rightarrow 3000 = 400k \Rightarrow k = \frac{3000}{400} = 7.5 \text{ N s}^2 \text{ m}^{-2}$ (or $\text{kg m}^{-1}$ ).
[2] (1 mark for equilibrium condition, 1 mark for k)

(b) At $v = 10 \text{ m s}^{-1}$ , $F_{res} = 7.5(10)^2 = 750 \text{ N}$ .
Net Force $F_{net} = F_{drive} - F_{res} = 3000 - 750 = 2250 \text{ N}$ .
$a = \frac{F_{net}}{m} = \frac{2250}{1500} = 1.5 \text{ m s}^{-2}$ .
[3] (1 mark for new resistive force, 1 mark for net force, 1 mark for acceleration)

10.
(a) Mass per second $\dot{m} = \text{Density} \times \text{Flow Rate} = 1000 \times 200 = 200,000 \text{ kg s}^{-1}$ .
[1]

(b) Input Power (Gravitational) $P_{in} = \dot{m}gh = 200,000 \times 9.81 \times 150 = 294,300,000 \text{ W}$ .
Output Power $P_{out} = 0.85 \times P_{in} = 0.85 \times 294,300,000 = 250,155,000 \text{ W}$ .
Answer: $250 \text{ MW}$ (or $2.5 \times 10^8 \text{ W}$ ).
[3] (1 mark for input power calc, 1 mark for efficiency application, 1 mark for final answer)

11.
(a) Component of weight $= mg \sin(\theta) = 80 \times 9.81 \times \sin(5.0^\circ)$ .
$\sin(5.0^\circ) \approx 0.08716$ .
$W_{parallel} = 80 \times 9.81 \times 0.08716 \approx 68.4 \text{ N}$ .
[2] (1 mark for formula, 1 mark for answer)

(b) Total force required $F = W_{parallel} + F_{resistive} = 68.4 + 40 = 108.4 \text{ N}$ .
Power $P = Fv = 108.4 \times 8.0 = 867.2 \text{ W} \approx 870 \text{ W}$ .
[3] (1 mark for total force, 1 mark for P=Fv, 1 mark for answer)

12.
(a) $\log P = 2 \log I + 0.60 \Rightarrow P = 10^{0.60} I^2$ .
Relationship: $P \propto I^2$ .
[2] (1 mark for identifying power law, 1 mark for stating proportionality)

(b) Comparing to $P = I^2 R$ , the constant $k = 10^{0.60} = R$ .
$R = 10^{0.60} \approx 3.98 \Omega \approx 4.0 \Omega$ .
[2] (1 mark for identifying intercept as log R, 1 mark for calculation)

13.
Thrust force $F$ is constant. Work done by thrust over distance $d$ is $W = Fd$ .
As mass $m$ decreases, for the same force $F$ , acceleration $a = F/m$ increases.
From energy perspective: The rate of change of kinetic energy ( $P = Fv$ ) leads to a faster increase in velocity $v$ as mass drops, because $KE = \frac{1}{2}mv^2$ . With lower $m$ , a smaller increase in energy yields a larger increase in $v$ , implying higher acceleration.
Acceptable Answer: $F=ma$ . Since $F$ is constant and $m$ decreases, $a$ must increase. The work done by the engine goes into increasing KE. As mass drops, the same work results in a larger change in velocity.
[3] (1 mark for F=ma or Work-Energy link, 1 mark for mass decrease effect, 1 mark for conclusion on acceleration)

14.
(a) Conservation of Energy for bob: $mgh = \frac{1}{2}mv^2$ .
$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.20} = \sqrt{3.924} \approx 1.98 \text{ m s}^{-1}$ .
[2] (1 mark for formula, 1 mark for answer)

(b) Conservation of Momentum for collision: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v_{final}$ .
$(0.50)(1.98) + 0 = (0.50 + 1.50)v_{final}$ .
$0.99 = 2.0 v_{final} \Rightarrow v_{final} = 0.495 \text{ m s}^{-1}$ .
$KE_{final} = \frac{1}{2}(m_1+m_2)v_{final}^2 = \frac{1}{2}(2.0)(0.495)^2 \approx 0.245 \text{ J}$ .
[3] (1 mark for momentum conservation to find v, 1 mark for v value, 1 mark for final KE)

15.
(a) Total Solar Power Incident $= 800 \times 20 = 16,000 \text{ W}$ .
Electrical Power $= 0.15 \times 16,000 = 2,400 \text{ W} = 2.4 \text{ kW}$ .
[2] (1 mark for incident power, 1 mark for efficiency calc)

(b) Daily Energy Demand $= 10 \text{ kWh} = 10 \times 1000 \text{ Wh} = 10,000 \text{ Wh}$ .
Power generated $= 2.4 \text{ kW}$ .
Time $t = \frac{\text{Energy}}{\text{Power}} = \frac{10 \text{ kWh}}{2.4 \text{ kW}} \approx 4.17 \text{ hours}$ .
Answer: $4.2 \text{ hours}$ .
[3] (1 mark for unit consistency/conversion, 1 mark for formula, 1 mark for answer)

16.
(a) Normal reaction $N = mg = 10 \times 9.81 = 98.1 \text{ N}$ .
Frictional force $F_f = \mu N = 0.30 \times 98.1 = 29.43 \text{ N} \approx 29 \text{ N}$ .
[2] (1 mark for normal force, 1 mark for friction calc)

(b) Work done by student $W = F d = 50 \times 5.0 = 250 \text{ J}$ .
[1]

(c) Net force $F_{net} = F_{applied} - F_f = 50 - 29.43 = 20.57 \text{ N}$ .
Net Work $W_{net} = F_{net} d = 20.57 \times 5.0 = 102.85 \text{ J} \approx 100 \text{ J}$ (2 s.f.).
Alternative: $W_{net} = W_{applied} - W_{friction} = 250 - (29.43 \times 5) = 250 - 147.15 = 102.85 \text{ J}$ .
[2] (1 mark for net force or work diff, 1 mark for answer)

17.
(a) Total mass $M = 800 + 200 = 1000 \text{ kg}$ .
Weight $W = Mg = 1000 \times 9.81 = 9810 \text{ N}$ .
Since speed is constant, Tension $T = W + F_{friction} = 9810 + 1500 = 11,310 \text{ N} \approx 11.3 \text{ kN}$ .
[2] (1 mark for weight + friction logic, 1 mark for answer)

(b) Power $P = T v = 11,310 \times 3.0 = 33,930 \text{ W} \approx 34 \text{ kW}$ .
[2] (1 mark for formula, 1 mark for answer)

18.
(a) $E_p = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.15)^2 = 100 \times 0.0225 = 2.25 \text{ J}$ .
[2] (1 mark for formula, 1 mark for answer)

(b) $E_k = E_p \Rightarrow \frac{1}{2}mv^2 = 2.25$ .
$v^2 = \frac{2 \times 2.25}{0.05} = \frac{4.5}{0.05} = 90$ .
$v = \sqrt{90} \approx 9.49 \text{ m s}^{-1} \approx 9.5 \text{ m s}^{-1}$ .
[2] (1 mark for equating energies, 1 mark for answer)

19.
(a) Work done $W = mgh = 500 \times 9.81 \times 10 = 49,050 \text{ J}$ .
Power $P = \frac{W}{t} = \frac{49,050}{20} = 2,452.5 \text{ W} \approx 2.45 \text{ kW}$ .
[2] (1 mark for work/energy, 1 mark for power)

(b) Efficiency $= \frac{P_{useful}}{P_{input}} \times 100\% = \frac{2452.5}{5000} \times 100\% = 49.05\% \approx 49\%$ .
[2] (1 mark for ratio, 1 mark for answer)

20.
(a) $P = Fv \Rightarrow F = \frac{P}{v} = \frac{100,000}{20} = 5,000 \text{ N}$ .
$F = ma \Rightarrow a = \frac{F}{m} = \frac{5,000}{1,200} = 4.166... \text{ m s}^{-2} \approx 4.2 \text{ m s}^{-2}$ .
[2] (1 mark for force from power, 1 mark for acceleration)

(b) Resistive forces (air resistance, friction) act against the motion, reducing the net force available for acceleration.
[1] (1 mark for mention of resistive forces)

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