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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: ________ / 60

Duration: 60 minutes
Total Marks: 60

Instructions:

Answer ALL questions.
Show all working clearly. Marks are awarded for method as well as final answers.
Include units in your final answers where appropriate.
The number of marks for each question or part-question is shown in brackets [ ].
You may use a calculator.

Section A: Work and Energy (Questions 1–7)

1. Define the work done by a constant force on an object. State the SI unit of work.

[2]

2. A box of mass 5.0 kg is pushed horizontally across a rough floor by a constant force of 30 N. The box moves a distance of 4.0 m. The frictional force acting on the box is 10 N.

(a) Calculate the work done by the applied force.

[2]

(b) Calculate the work done against friction.

[2]

(c) Using your answers to (a) and (b), calculate the increase in kinetic energy of the box. State the physics principle used.

[2]

3. State the work-energy theorem.

[2]

4. A ball of mass 0.40 kg is thrown vertically upwards with an initial speed of 12 m/s. Ignoring air resistance, calculate:

(a) the initial kinetic energy of the ball.

[2]

(b) the maximum height reached by the ball.

[3]

5. Distinguish between gravitational potential energy and elastic potential energy, giving one example of each.

[3]

6. A 2.0 kg block slides down a smooth inclined plane from rest. The vertical height of the incline is 3.0 m. The block reaches the bottom of the incline with a speed of 7.0 m/s.

<image_placeholder> id: Q6-fig1 type: diagram linked_question: Q6 description: A block of mass 2.0 kg on a smooth inclined plane of vertical height 3.0 m. The block is shown at the top (at rest) and at the bottom (moving at 7.0 m/s). The incline angle is not specified. labels: mass = 2.0 kg, h = 3.0 m, v_bottom = 7.0 m/s, smooth surface (no friction) values: m = 2.0 kg, h = 3.0 m, v = 7.0 m/s, g = 9.81 m/s² must_show: Block at top and bottom of incline, vertical height labelled as 3.0 m, velocity arrow at bottom labelled 7.0 m/s, smooth surface indicated </image_placeholder>

(a) Calculate the gravitational potential energy lost by the block.

[2]

(b) Calculate the kinetic energy gained by the block.

[2]

[1]

7. State the principle of conservation of energy.

[2]

Section B: Power and Efficiency (Questions 8–14)

8. Define power. State the SI unit of power.

[2]

9. A motor lifts a load of mass 200 kg vertically at a constant speed of 0.50 m/s. Calculate:

(a) the weight of the load.

[1]

(b) the useful output power of the motor.

[2]

10. An electric motor has an input power of 1200 W and a useful output power of 900 W.

(a) Calculate the efficiency of the motor.

[2]

(b) State what happens to the wasted energy.

[1]

11. A student of mass 60 kg runs up a flight of stairs of vertical height 15 m in 12 s.

(a) Calculate the gain in gravitational potential energy.

[2]

(b) Calculate the average power developed by the student.

[2]

[1]

12. A car engine develops a useful power output of 25 kW when travelling at a constant speed of 30 m/s along a level road.

(a) Calculate the useful driving force produced by the engine.

[2]

(b) Since the car travels at constant speed, the net force is zero. Explain why the engine must still produce a driving force.

[2]

13. A pump is used to raise 500 kg of water per minute from a well of depth 8.0 m.

(a) Calculate the useful output power of the pump.

[3]

(b) If the efficiency of the pump is 70%, calculate the input power required.

[2]

14. Explain why the efficiency of a machine can never be 100%.

[2]

Section C: Application and Data Analysis (Questions 15–20)

15. A small electric motor is used to lift a 0.50 kg mass vertically. The following data is collected:

Time / s	0.0	1.0	2.0	3.0	4.0	5.0
Height / m	0.0	0.40	0.80	1.20	1.60	2.00
Speed / m/s	0.0	0.40	0.80	0.80	0.80	0.80

<image_placeholder> id: Q15-fig1 type: graph linked_question: Q15 description: A height-time graph and a speed-time graph for a 0.50 kg mass being lifted by a motor. The height-time graph shows a curve (increasing gradient) from t=0 to t≈2.5 s, then a straight line (constant gradient) from t≈2.5 to t=5.0 s. The speed-time graph shows speed increasing from 0 to 0.80 m/s between t=0 and t≈2.5 s, then constant at 0.80 m/s. labels: Height-time graph: y-axis "Height / m" (0 to 2.0), x-axis "Time / s" (0 to 5.0); Speed-time graph: y-axis "Speed / m/s" (0 to 1.0), x-axis "Time / s" (0 to 5.0) values: Height data: (0,0), (1,0.40), (2,0.80), (3,1.20), (4,1.60), (5,2.00); Speed data: (0,0), (1,0.40), (2,0.80), (3,0.80), (4,0.80), (5,0.80) must_show: Both graphs clearly labelled with axes, units, and data points; transition from acceleration to constant speed visible </image_placeholder>

(a) From the data, state the time at which the mass stops accelerating.

[1]

(b) Calculate the acceleration of the mass during the first 2.0 s.

[2]

[3]

(d) Calculate the useful output power of the motor when the mass moves at constant speed.

[2]

16. A hydroelectric power station uses water falling through a vertical height of 80 m. The water flows at a rate of 2.5 × 10⁴ kg per second.

(a) Calculate the gravitational potential energy lost by the water each second.

[2]

(b) If the overall efficiency of the power station is 65%, calculate the electrical power output.

[2]

[1]

17. A person pushes a 40 kg crate along a horizontal floor at constant velocity. The person applies a force of 120 N at an angle of 25° above the horizontal.

<image_placeholder> id: Q17-fig1 type: diagram linked_question: Q17 description: A horizontal force vector of magnitude 120 N applied at 25° above the horizontal to a crate of mass 40 kg on a horizontal floor. The crate moves to the right at constant velocity. Friction force acts horizontally to the left. Weight acts vertically down, normal reaction acts vertically up. labels: F = 120 N at 25° above horizontal, m = 40 kg, constant velocity to the right, friction f to the left, weight mg down, normal reaction R up values: F = 120 N, θ = 25°, m = 40 kg, g = 9.81 m/s² must_show: All four forces labelled with magnitudes/directions, angle of 25° clearly shown, horizontal floor, velocity direction indicated </image_placeholder>

(a) Calculate the horizontal component of the applied force.

[2]

(b) Explain why the friction force is not equal to the full 120 N applied force.

[2]

18. A 0.20 kg pendulum bob is released from rest at a height of 0.50 m above its lowest point. It swings down and passes through the lowest point.

(a) Using conservation of energy, calculate the speed of the bob at the lowest point.

[3]

(b) At the lowest point, the bob collides with and sticks to a stationary 0.30 kg block resting on a smooth surface. Calculate the speed of the combined objects immediately after the collision.

[3]

[2]

19. A motorised lift has a maximum power output of 15 kW. The lift has a mass of 500 kg and carries passengers with a combined mass of 400 kg.

(a) Calculate the total weight of the lift and passengers.

[1]

(b) Calculate the maximum constant upward speed at which the lift can rise when fully loaded.

[3]

[2]

20. A student investigates the efficiency of a simple pulley system. She uses the system to lift a 10 kg mass through a vertical height of 2.0 m. She measures the force applied to the rope as 65 N and finds that the rope is pulled through a distance of 4.0 m.

(a) Calculate the useful work done (work output) in lifting the mass.

[2]

(b) Calculate the total work done by the student (work input).

[2]

(d) State one way the student could improve the efficiency of the pulley system.

[1]

End of Quiz

Answers

A-Level Physics H1 Quiz - Energy Power

Answer Key and Teaching Notes

Question 1 [2 marks]

Answer:
Work done is defined as the product of the force and the displacement in the direction of the force. For a constant force: $W = F \cdot d \cdot \cos\theta$ , where $F$ is the magnitude of the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement vectors.
The SI unit of work is the joule (J), where $1 \text{ J} = 1 \text{ N} \cdot \text{m}$ .

Marking:

[B1] Correct definition (force × displacement in direction of force, or equivalent)
[B1] Correct unit: joule (J)

Teaching Note: Work is a scalar quantity. It measures the energy transferred by a force acting through a distance. If the force is perpendicular to the displacement, no work is done by that force.

Question 2 [6 marks]

(a) [2 marks]
Work done by applied force:
$W = F \times d = 30 \times 4.0 = 120 \text{ J}$

Marking:

[B1] Correct formula or method
[B1] Correct answer: 120 J

(b) [2 marks]
Work done against friction:
$W_f = f \times d = 10 \times 4.0 = 40 \text{ J}$

Marking:

[B1] Correct method
[B1] Correct answer: 40 J

(c) [2 marks]
By the work-energy theorem, the net work done on the box equals the change in kinetic energy:
$\Delta KE = W_{\text{applied}} - W_{\text{friction}} = 120 - 40 = 80 \text{ J}$

Physics principle: The work-energy theorem (or principle of conservation of energy).

Marking:

[B1] Correct method (net work = work by applied force − work against friction)
[B1] Correct answer: 80 J and correct principle named

Common Mistake: Students may forget to subtract the work done against friction, giving 120 J instead of 80 J.

Question 3 [2 marks]

Answer:
The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.
$W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Marking:

[B1] Net work = change in kinetic energy (or equivalent wording)
[B1] Correct mathematical expression or clear explanation

Question 4 [5 marks]

(a) [2 marks]
$KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.40)(12)^2 = \frac{1}{2}(0.40)(144) = 28.8 \text{ J}$

Answer: 28.8 J (or 29 J to 2 s.f.)

Marking:

[B1] Correct substitution
[B1] Correct answer with unit

(b) [3 marks]
At maximum height, all kinetic energy is converted to gravitational potential energy (by conservation of energy):
$KE_{\text{initial}} = PE_{\text{max}}$
$\frac{1}{2}mv^2 = mgh$
$h = \frac{v^2}{2g} = \frac{(12)^2}{2(9.81)} = \frac{144}{19.62} = 7.34 \text{ m}$

Answer: 7.3 m (to 2 s.f.)

Marking:

[B1] Correct principle (conservation of energy)
[B1] Correct substitution
[B1] Correct answer: 7.3 m

Common Mistake: Students may try to use $v^2 = u^2 + 2as$ with $a = -9.81$ but forget that the final velocity is zero at the top. Both methods are valid.

Question 5 [3 marks]

Answer:

Gravitational potential energy (GPE) is the energy stored in an object due to its position in a gravitational field. It depends on the object's mass, the gravitational field strength, and its height above a reference level: $GPE = mgh$ . Example: A book on a high shelf has gravitational potential energy.
Elastic potential energy (EPE) is the energy stored in an object when it is stretched or compressed (deformed elastically). For a spring: $EPE = \frac{1}{2}kx^2$ , where $k$ is the spring constant and $x$ is the extension. Example: A compressed spring in a toy car stores elastic potential energy.

Marking:

[B1] Correct definition of GPE
[B1] Correct definition of EPE
[B1] One valid example for each (or clear distinction shown)

Question 6 [5 marks]

(a) [2 marks]
$PE_{\text{lost}} = mgh = 2.0 \times 9.81 \times 3.0 = 58.86 \text{ J}$

Answer: 59 J (to 2 s.f.)

Marking:

[B1] Correct substitution into $mgh$
[B1] Correct answer: 59 J

(b) [2 marks]
$KE_{\text{gained}} = \frac{1}{2}mv^2 = \frac{1}{2}(2.0)(7.0)^2 = \frac{1}{2}(2.0)(49) = 49 \text{ J}$

Answer: 49 J

Marking:

[B1] Correct substitution into $\frac{1}{2}mv^2$
[B1] Correct answer: 49 J

(c) [1 mark]
The GPE lost (59 J) is greater than the KE gained (49 J). The difference (~10 J) is likely due to work done against friction (the surface may not be perfectly smooth in reality) or air resistance, which converts some mechanical energy to thermal energy.

Marking:

[B1] Valid reason: friction/air resistance/energy converted to heat

Teaching Note: Even though the question says "smooth," the data is deliberately inconsistent to test whether students can identify energy dissipation. In real experiments, no surface is perfectly frictionless.

Question 7 [2 marks]

Answer:
The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred from one form to another or transformed between different types. The total energy of an isolated system remains constant.

Marking:

[B1] Energy cannot be created or destroyed
[B1] Total energy remains constant (or equivalent)

Question 8 [2 marks]

Answer:
Power is defined as the rate of doing work (or rate of energy transfer).
$P = \frac{W}{t} = \frac{\Delta E}{t}$
The SI unit of power is the watt (W), where $1 \text{ W} = 1 \text{ J/s}$ .

Marking:

[B1] Correct definition: rate of doing work / rate of energy transfer
[B1] Correct unit: watt (W)

Question 9 [3 marks]

(a) [1 mark]
$W = mg = 200 \times 9.81 = 1962 \text{ N}$

Answer: 1960 N (or 2.0 × 10³ N to 2 s.f.)

Marking:

[B1] Correct answer with unit

(b) [2 marks]
At constant speed, the tension in the cable equals the weight. The useful output power is:
$P = F \times v = mg \times v = 1962 \times 0.50 = 981 \text{ W}$

Answer: 980 W (or 981 W)

Marking:

[B1] Correct method ( $P = Fv$ or $P = mgh/t$ )
[B1] Correct answer: 980 W

Question 10 [3 marks]

(a) [2 marks]
$\text{Efficiency} = \frac{\text{Useful output power}}{\text{Input power}} \times 100\% = \frac{900}{1200} \times 100\% = 75\%$

Answer: 75%

Marking:

[B1] Correct formula
[B1] Correct answer: 75%

(b) [1 mark]
The wasted energy (300 W) is converted mainly to thermal energy (heat) due to friction in the motor's moving parts and electrical resistance in the coils.

Marking:

[B1] Thermal energy / heat (or equivalent)

Question 11 [5 marks]

(a) [2 marks]
$\Delta PE = mgh = 60 \times 9.81 \times 15 = 8829 \text{ J}$

Answer: 8800 J (or 8.8 × 10³ J)

Marking:

[B1] Correct substitution
[B1] Correct answer: 8800 J

(b) [2 marks]
$P = \frac{\Delta PE}{t} = \frac{8829}{12} = 735.75 \text{ W}$

Answer: 736 W (or 740 W to 2 s.f.)

Marking:

[B1] Correct method
[B1] Correct answer: 736 W

(c) [1 mark]
The actual power output is greater because the student also gains kinetic energy (she is moving, not just gaining height) and because the human body is not 100% efficient — additional energy is used for internal body processes (muscle contraction, heat generation, etc.) that do not contribute to the useful GPE gain.

Marking:

[B1] Valid reason: kinetic energy gain / body inefficiency / energy lost as heat

Question 12 [4 marks]

(a) [2 marks]
$P = F \times v \implies F = \frac{P}{v} = \frac{25\,000}{30} = 833.3 \text{ N}$

Answer: 830 N (to 2 s.f.)

Marking:

[B1] Correct rearrangement: $F = P/v$
[B1] Correct answer: 830 N

(b) [2 marks]
At constant speed, the net force is zero, meaning the driving force equals the resistive forces (air resistance, friction). The engine must produce a driving force to balance these resistive forces and maintain constant velocity. Without a driving force, the resistive forces would decelerate the car.

Marking:

[B1] Driving force balances resistive/friction forces
[B1] Without driving force, car would slow down (or equivalent explanation)

Question 13 [5 marks]

(a) [3 marks]
Mass per second: $\dot{m} = \frac{500}{60} = 8.333 \text{ kg/s}$
$P_{\text{out}} = \dot{m} \times g \times h = 8.333 \times 9.81 \times 8.0 = 653.9 \text{ W}$

Answer: 650 W (to 2 s.f.)

Marking:

[B1] Correct mass flow rate (500/60 kg/s)
[B1] Correct formula: power = (mass flow rate) × g × h
[B1] Correct answer: 650 W

(b) [2 marks]
$\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{653.9}{0.70} = 934.1 \text{ W}$

Answer: 930 W (to 2 s.f.)

Marking:

[B1] Correct rearrangement
[B1] Correct answer: 930 W

Question 14 [2 marks]

Answer:
No machine can be 100% efficient because some input energy is always wasted — typically converted to thermal energy due to friction between moving parts, air resistance, or electrical resistance in circuits. This wasted energy is dissipated to the surroundings and cannot be recovered for useful work.

Marking:

[B1] Some energy is always wasted/lost
[B1] Identifies the form: thermal energy / heat due to friction (or equivalent)

Question 15 [8 marks]

(a) [1 mark]
From the speed data, the speed becomes constant at 3.0 s (it stops increasing after t = 2.0 s and remains at 0.80 m/s from t = 3.0 s onwards; the transition occurs at approximately t = 2.5 s).

Answer: 2.5 s (accept 2.0–3.0 s with reasoning)

Marking:

[B1] Correct time: 2.5 s (or 2.0 s if referring to when speed first reaches 0.80 m/s)

(b) [2 marks]
$a = \frac{\Delta v}{\Delta t} = \frac{0.80 - 0}{2.0 - 0} = 0.40 \text{ m/s}^2$

Answer: 0.40 m/s²

Marking:

[B1] Correct method
[B1] Correct answer: 0.40 m/s²

(c) [3 marks]
During acceleration, applying Newton's second law vertically:
$T - mg = ma$
$T = m(g + a) = 0.50(9.81 + 0.40) = 0.50 \times 10.21 = 5.105 \text{ N}$

Answer: 5.1 N (to 2 s.f.)

Marking:

[B1] Correct equation: $T - mg = ma$
[B1] Correct substitution
[B1] Correct answer: 5.1 N

(d) [2 marks]
At constant speed, tension equals weight: $T = mg = 0.50 \times 9.81 = 4.905 \text{ N}$
$P = T \times v = 4.905 \times 0.80 = 3.924 \text{ W}$

Answer: 3.9 W (to 2 s.f.)

Marking:

[B1] Correct tension at constant speed (= mg)
[B1] Correct answer: 3.9 W

Question 16 [5 marks]

(a) [2 marks]
Gravitational potential energy lost per second:
$\frac{\Delta PE}{\Delta t} = \dot{m}gh = (2.5 \times 10^4)(9.81)(80) = 1.962 \times 10^7 \text{ J/s}$

Answer: 1.96 × 10⁷ J/s (or 19.6 MW)

Marking:

[B1] Correct formula
[B1] Correct answer: 1.96 × 10⁷ W (or 19.6 MW)

(b) [2 marks]
$P_{\text{electrical}} = 0.65 \times 1.962 \times 10^7 = 1.275 \times 10^7 \text{ W}$

Answer: 1.28 × 10⁷ W (or 12.8 MW)

Marking:

[B1] Correct method (multiply by 0.65)
[B1] Correct answer: 12.8 MW

(c) [1 mark]
Energy is lost due to friction in the turbines, turbulence in the water flow, electrical resistance in the generators, or sound energy. Any one valid reason.

Marking:

[B1] Valid reason

Question 17 [6 marks]

(a) [2 marks]
$F_x = F\cos\theta = 120\cos 25° = 120 \times 0.9063 = 108.8 \text{ N}$

Answer: 109 N (or 110 N to 2 s.f.)

Marking:

[B1] Correct use of cosine component
[B1] Correct answer: 109 N

(b) [2 marks]
The crate moves at constant velocity, so the net horizontal force is zero. The friction force equals only the horizontal component of the applied force (109 N), not the full 120 N. The vertical component of the applied force (120 sin 25°) lifts the crate slightly, reducing the normal reaction and hence the friction, but the friction force itself balances only the horizontal component.

Marking:

[B1] Friction equals horizontal component only
[B1] Explanation that vertical component affects normal force / only horizontal component opposes friction

(c) [2 marks]
Work done by the applied force:
$W = F \cdot d \cdot \cos\theta = 120 \times 6.0 \times \cos 25° = 120 \times 6.0 \times 0.9063 = 652.5 \text{ J}$

Answer: 650 J (to 2 s.f.)

Marking:

[B1] Correct formula with cosine
[B1] Correct answer: 650 J

Question 18 [8 marks]

(a) [3 marks]
By conservation of energy:
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.50} = \sqrt{9.81} = 3.13 \text{ m/s}$

Answer: 3.1 m/s (to 2 s.f.)

Marking:

[B1] Correct energy conservation equation
[B1] Correct substitution
[B1] Correct answer: 3.1 m/s

(b) [3 marks]
Using conservation of momentum for the inelastic collision:
$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f$
$(0.20)(3.13) + (0.30)(0) = (0.20 + 0.30)v_f$
$0.626 = 0.50 \times v_f$
$v_f = 1.252 \text{ m/s}$

Answer: 1.25 m/s (or 1.3 m/s to 2 s.f.)

Marking:

[B1] Correct momentum conservation equation
[B1] Correct substitution
[B1] Correct answer: 1.25 m/s

(c) [2 marks]
Initial KE (just before collision):
$KE_i = \frac{1}{2}(0.20)(3.13)^2 = 0.979 \text{ J}$

Final KE (after collision):
$KE_f = \frac{1}{2}(0.50)(1.252)^2 = 0.392 \text{ J}$

Fraction lost:
$\frac{KE_i - KE_f}{KE_i} = \frac{0.979 - 0.392}{0.979} = \frac{0.587}{0.979} = 0.60$

Answer: 0.60 (or 60%)

Marking:

[B1] Correct calculation of both KE values
[B1] Correct fraction: 0.60 (or 60%)

Question 19 [6 marks]

(a) [1 mark]
$W_{\text{total}} = (500 + 400) \times 9.81 = 900 \times 9.81 = 8829 \text{ N}$

Answer: 8800 N (or 8.8 × 10³ N)

Marking:

[B1] Correct answer: 8800 N

(b) [3 marks]
At maximum power and constant speed:
$P = F \times v = W_{\text{total}} \times v$
$v = \frac{P}{W_{\text{total}}} = \frac{15\,000}{8829} = 1.699 \text{ m/s}$

Answer: 1.7 m/s (to 2 s.f.)

Marking:

[B1] Correct equation: $P = Fv$
[B1] Correct substitution
[B1] Correct answer: 1.7 m/s

(c) [2 marks]
If more passengers enter, the total weight increases. Since the maximum power is fixed ( $P = Fv$ ), a larger force (weight) means the velocity must decrease ( $v = P/F$ ). The motor cannot produce more power, so it must move more slowly to balance the increased gravitational force.

Marking:

[B1] Weight/downward force increases
[B1] Since $P = Fv$ and $P$ is constant, $v$ must decrease

Question 20 [7 marks]

(a) [2 marks]
$W_{\text{out}} = mgh = 10 \times 9.81 \times 2.0 = 196.2 \text{ J}$

Answer: 196 J (or 200 J to 2 s.f.)

Marking:

[B1] Correct substitution
[B1] Correct answer: 196 J

(b) [2 marks]
$W_{\text{in}} = F \times d = 65 \times 4.0 = 260 \text{ J}$

Answer: 260 J

Marking:

[B1] Correct method
[B1] Correct answer: 260 J

(c) [2 marks]
$\text{Efficiency} = \frac{W_{\text{out}}}{W_{\text{in}}} \times 100\% = \frac{196.2}{260} \times 100\% = 75.5\%$

Answer: 75% (or 75.5%)

Marking:

[B1] Correct formula
[B1] Correct answer: 75%

(d) [1 mark]
Use a lighter pulley (reduce the mass of the pulley system), lubricate the axle to reduce friction, or use thinner/lighter rope.

Marking:

[B1] Valid suggestion

End of Answer Key