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Secondary 4 Pure Physics Energy Power Quiz

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Questions

Secondary 4 Pure Physics Quiz - Energy Power

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

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working for calculation questions.
Use $g = 10 \text{ N/kg}$ unless otherwise stated.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Multiple Choice Questions (10 marks)

Answer all questions. Choose the correct option and write the letter (A, B, C, or D) in the box provided.

1. [1 mark]

A student lifts a 2.0 kg book from the floor to a shelf 1.5 m high. The work done by the student against gravity is:

A. 3.0 J
B. 15 J
C. 30 J
D. 20 J

Answer: □

2. [1 mark]

A car of mass 1200 kg accelerates uniformly from rest to 25 m/s in 10 s. The average power developed by the engine (ignoring resistive forces) is:

A. 37.5 kW
B. 75 kW
C. 375 kW
D. 750 kW

Answer: □

3. [1 mark]

A block slides down a frictionless inclined plane from height $h$ . At the bottom, its speed is $v$ . If the block starts from a height $4h$ , its speed at the bottom will be:

A. $2v$
B. $4v$
C. $v/2$
D. $v/4$

Answer: □

4. [1 mark]

Which of the following statements about power is correct?

A. Power is the rate of change of momentum.
B. Power is the rate of doing work.
C. Power is the product of force and displacement.
D. Power is the energy stored in a system.

Answer: □

5. [1 mark]

A motor lifts a 500 N load through a vertical height of 12 m in 20 s. The efficiency of the motor is 80%. The electrical energy input to the motor is:

A. 7.5 kJ
B. 9.6 kJ
C. 12 kJ
D. 15 kJ

Answer: □

6. [1 mark]

A pendulum bob is released from rest at position A, which is 0.30 m above the lowest point B. Assuming no air resistance, the speed of the bob at B is:

A. 1.7 m/s
B. 2.4 m/s
C. 3.0 m/s
D. 6.0 m/s

Answer: □

7. [1 mark]

A force of 20 N acts on a box, moving it 5.0 m in the direction of the force. The work done by the force is:

A. 4.0 J
B. 25 J
C. 100 J
D. 400 J

Answer: □

8. [1 mark]

A hydroelectric power station has an efficiency of 90%. Water falls through a vertical height of 80 m at a rate of 500 kg/s. The electrical power output is: (Take $g = 10 \text{ N/kg}$ )

A. 360 kW
B. 400 kW
C. 3.6 MW
D. 4.0 MW

Answer: □

9. [1 mark]

A spring is compressed by a force that increases uniformly from 0 to 40 N over a distance of 0.10 m. The work done in compressing the spring is:

A. 2.0 J
B. 4.0 J
C. 20 J
D. 40 J

Answer: □

10. [1 mark]

A 60 kg student runs up a flight of stairs of vertical height 4.0 m in 5.0 s. The average power developed by the student against gravity is:

A. 48 W
B. 240 W
C. 480 W
D. 2400 W

Answer: □

Section B: Structured Questions (18 marks)

Answer all questions in the spaces provided.

11. [3 marks]

A crate of mass 25 kg is pulled horizontally across a rough floor by a constant force of 120 N. The crate moves a distance of 8.0 m. The frictional force acting on the crate is 40 N.

(a) Calculate the work done by the applied force.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(b) Calculate the work done against friction.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(c) Calculate the net work done on the crate and state the change in its kinetic energy.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

12. [4 marks]

A roller coaster car of mass 500 kg starts from rest at point A, which is 40 m above ground level. It travels down a frictionless track to point B at ground level, then up to point C which is 25 m above ground level.

<image_placeholder> id: Q12-fig1 type: diagram linked_question: Q12 description: Roller coaster track profile showing points A, B, and C with heights labelled. Point A at 40 m, point B at 0 m, point C at 25 m. Car shown at point A. labels: Point A (40 m), Point B (0 m), Point C (25 m), mass = 500 kg values: h_A = 40 m, h_B = 0 m, h_C = 25 m, m = 500 kg, g = 10 N/kg must_show: Track profile with three labelled points, heights indicated, car at starting position </image_placeholder>

(a) Calculate the speed of the car at point B.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

(b) Calculate the speed of the car at point C.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

13. [3 marks]

An electric motor is used to lift a load of 800 N through a vertical height of 15 m in 30 s. The motor draws a current of 5.0 A from a 240 V supply.

(a) Calculate the useful power output of the motor.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(b) Calculate the electrical power input to the motor.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(c) Calculate the efficiency of the motor.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

14. [4 marks]

A ball of mass 0.50 kg is thrown vertically upwards with an initial speed of 20 m/s. Air resistance is negligible.

(a) Calculate the maximum height reached by the ball.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

(b) Calculate the kinetic energy of the ball when it is at half the maximum height.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

15. [4 marks]

A pump lifts water from a well 20 m deep and delivers it through a pipe at a speed of 5.0 m/s. The mass flow rate of water is 10 kg/s. The pump efficiency is 70%.

(a) Calculate the rate at which gravitational potential energy is given to the water.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(b) Calculate the rate at which kinetic energy is given to the water.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(c) Calculate the power input required by the pump.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

Section C: Longer Structured Questions (12 marks)

Answer all questions in the spaces provided.

16. [6 marks]

A block of mass 2.0 kg is pushed up a rough inclined plane at an angle of 30° to the horizontal. The block moves a distance of 5.0 m along the plane at constant speed. The coefficient of kinetic friction between the block and the plane is 0.25.

<image_placeholder> id: Q16-fig1 type: diagram linked_question: Q16 description: Free-body diagram of block on inclined plane showing weight mg, normal reaction N, friction f, and applied force F parallel to plane. labels: m = 2.0 kg, θ = 30°, μ_k = 0.25, distance = 5.0 m, F (applied force up plane), f (friction down plane), N (normal), mg (weight) values: m = 2.0 kg, θ = 30°, μ_k = 0.25, s = 5.0 m, g = 10 N/kg must_show: Inclined plane at 30°, block with four labelled forces, angle marked </image_placeholder>

(a) Draw and label all forces acting on the block in the diagram above.
Answer: (Diagram completed above) [1]

(b) Calculate the magnitude of the frictional force acting on the block.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

(c) Calculate the work done by the applied force $F$ .
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

(d) Explain why the kinetic energy of the block does not change, using the work-energy theorem.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [1]

17. [6 marks]

A wind turbine generates electrical power from kinetic energy of moving air. The turbine has blades that sweep a circular area of radius 25 m. The density of air is 1.2 kg/m³. The wind speed is 12 m/s. The turbine efficiency is 40%.

(a) Calculate the mass of air passing through the swept area per second.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

(b) Calculate the kinetic energy of this mass of air per second (i.e., the power available in the wind).
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

(c) Calculate the electrical power output of the turbine.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(d) State one reason why the turbine cannot extract 100% of the kinetic energy from the wind.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

Section D: Additional Questions (10 marks)

Answer all questions in the spaces provided.

18. [3 marks]

A 1200 kg car is travelling at 20 m/s on a horizontal road. The driver applies the brakes and the car comes to rest in a distance of 50 m. Assume the braking force is constant.

(a) Calculate the initial kinetic energy of the car.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(b) Calculate the average braking force acting on the car.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

19. [3 marks]

A spring with spring constant $k = 200 \text{ N/m}$ is compressed by 0.15 m from its natural length. A block of mass 0.50 kg is placed against the spring on a smooth horizontal surface. The spring is released, propelling the block forward.

(a) Calculate the elastic potential energy stored in the spring when compressed.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(b) Calculate the speed of the block as it leaves the spring.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

20. [4 marks]

A 50 kg girl slides down a water slide from a vertical height of 8.0 m. She starts from rest and reaches the bottom with a speed of 10 m/s. The length of the slide is 20 m.

(a) Calculate the loss in gravitational potential energy of the girl.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(b) Calculate the gain in kinetic energy of the girl.
Answer: _______________________________________________________________________________
_______________________________________________________________________________ [1]

(c) Calculate the average resistive force (friction + air resistance) acting on the girl during the slide.
Answer: _______________________________________________________________________________

_______________________________________________________________________________ [2]

End of Quiz

Answers

Secondary 4 Pure Physics Quiz - Energy Power (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. [1 mark] Answer: C

Working:
Work done against gravity = $mgh = 2.0 \times 10 \times 1.5 = 30 \text{ J}$
Key concept: Work done against gravity = gain in gravitational potential energy = $mgh$ .

2. [1 mark] Answer: A

Working:
Kinetic energy gained = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 25^2 = 375\,000 \text{ J}$
Average power = $\frac{\text{Work done}}{\text{time}} = \frac{375\,000}{10} = 37\,500 \text{ W} = 37.5 \text{ kW}$
Key concept: Work-energy theorem: net work done = change in kinetic energy. Power = work/time.

3. [1 mark] Answer: A

Working:
Conservation of energy: $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$
If $h$ becomes $4h$ , new speed $v' = \sqrt{2g(4h)} = 2\sqrt{2gh} = 2v$
Key concept: Speed is proportional to square root of height ( $v \propto \sqrt{h}$ ).

4. [1 mark] Answer: B

Explanation: Power is defined as the rate of doing work or the rate of energy transfer: $P = \frac{W}{t} = \frac{E}{t}$ .

A describes rate of change of momentum (force).
C describes work (for constant force in direction of displacement).
D describes energy, not power.

5. [1 mark] Answer: A

Working:
Useful work output = $Fh = 500 \times 12 = 6000 \text{ J}$
Efficiency = $\frac{\text{useful output}}{\text{input}} \Rightarrow 0.80 = \frac{6000}{\text{input}}$
Electrical energy input = $\frac{6000}{0.80} = 7500 \text{ J} = 7.5 \text{ kJ}$
Marking note: Common error: using $P = VI$ without current given, or confusing power/energy.

6. [1 mark] Answer: B

Working:
Loss in GPE = Gain in KE: $mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.30} = \sqrt{6} \approx 2.45 \text{ m/s} \approx 2.4 \text{ m/s}$
Key concept: Conservation of mechanical energy for pendulum (no air resistance).

7. [1 mark] Answer: C

Working:
Work done = Force $\times$ distance in direction of force = $20 \times 5.0 = 100 \text{ J}$
Key concept: $W = Fs$ for constant force parallel to displacement.

8. [1 mark] Answer: A

Working:
Mass flow rate = $500 \text{ kg/s}$
GPE loss per second (power available) = $\dot{m}gh = 500 \times 10 \times 80 = 400\,000 \text{ W} = 400 \text{ kW}$
Electrical power output = efficiency $\times$ power available = $0.90 \times 400 \text{ kW} = 360 \text{ kW}$
Note: 360 kW = 0.36 MW, not 3.6 MW.

9. [1 mark] Answer: A

Working:
Force increases uniformly from 0 to 40 N → average force = $\frac{0 + 40}{2} = 20 \text{ N}$
Work done = average force $\times$ distance = $20 \times 0.10 = 2.0 \text{ J}$
Alternative: Area under force-extension graph = $\frac{1}{2} \times 40 \times 0.10 = 2.0 \text{ J}$
Key concept: Work done by variable force = area under $F$ - $x$ graph.

10. [1 mark] Answer: C

Working:
Work done against gravity = $mgh = 60 \times 10 \times 4.0 = 2400 \text{ J}$
Average power = $\frac{2400}{5.0} = 480 \text{ W}$
Key concept: Power = work/time = rate of energy transfer.

Section B: Structured Questions (18 marks)

11. [3 marks]

(a) [1 mark]
Work done by applied force = $F \times s = 120 \times 8.0 = 960 \text{ J}$
Answer: 960 J

(b) [1 mark]
Work done against friction = $f \times s = 40 \times 8.0 = 320 \text{ J}$
Answer: 320 J
Note: Work done by friction is $-320 \text{ J}$ ; work done against friction is $+320 \text{ J}$ .

(c) [1 mark]
Net work done = Work by applied force + Work by friction = $960 + (-320) = 640 \text{ J}$
Change in kinetic energy = Net work done = $640 \text{ J}$ (increase)
Answer: Net work = 640 J; $\Delta KE = 640 \text{ J}$ (increase)
Key concept: Work-energy theorem: $W_{\text{net}} = \Delta KE$ .

12. [4 marks]

(a) [2 marks]
At A: $E_{\text{total}} = mgh_A = 500 \times 10 \times 40 = 200\,000 \text{ J}$
At B: $h = 0$ , so all energy is kinetic: $\frac{1}{2}mv_B^2 = 200\,000$
$v_B^2 = \frac{2 \times 200\,000}{500} = 800$
$v_B = \sqrt{800} = 20\sqrt{2} \approx 28.3 \text{ m/s}$
Answer: $28.3 \text{ m/s}$ (or $20\sqrt{2} \text{ m/s}$ )
Mark breakdown: 1 mark for energy conservation equation, 1 mark for correct calculation.

(b) [2 marks]
At C: $E_{\text{total}} = KE_C + GPE_C = 200\,000$
$GPE_C = mgh_C = 500 \times 10 \times 25 = 125\,000 \text{ J}$
$KE_C = 200\,000 - 125\,000 = 75\,000 \text{ J}$
$\frac{1}{2} \times 500 \times v_C^2 = 75\,000$
$v_C^2 = \frac{150\,000}{500} = 300$
$v_C = \sqrt{300} = 10\sqrt{3} \approx 17.3 \text{ m/s}$
Answer: $17.3 \text{ m/s}$ (or $10\sqrt{3} \text{ m/s}$ )
Mark breakdown: 1 mark for finding KE at C, 1 mark for correct speed calculation.

13. [3 marks]

(a) [1 mark]
Useful work output = $Fh = 800 \times 15 = 12\,000 \text{ J}$
Useful power output = $\frac{12\,000}{30} = 400 \text{ W}$
Answer: 400 W

(b) [1 mark]
Electrical power input = $VI = 240 \times 5.0 = 1200 \text{ W}$
Answer: 1200 W

(c) [1 mark]
Efficiency = $\frac{\text{useful power output}}{\text{power input}} \times 100\% = \frac{400}{1200} \times 100\% = 33.3\%$
Answer: 33.3% (or $\frac{1}{3}$ or 0.333)
Key concept: Efficiency = useful output / total input.

14. [4 marks]

(a) [2 marks]
At max height, $v = 0$ . Conservation of energy:
$\frac{1}{2}mu^2 = mgh_{\text{max}}$
$h_{\text{max}} = \frac{u^2}{2g} = \frac{20^2}{2 \times 10} = \frac{400}{20} = 20 \text{ m}$
Answer: 20 m
Mark breakdown: 1 mark for correct equation/principle, 1 mark for correct answer with unit.

(b) [2 marks]
At half max height ( $h = 10 \text{ m}$ ):
$GPE = mgh = 0.50 \times 10 \times 10 = 50 \text{ J}$
Total energy = initial KE = $\frac{1}{2} \times 0.50 \times 20^2 = 100 \text{ J}$
$KE = \text{Total} - GPE = 100 - 50 = 50 \text{ J}$
Answer: 50 J
Alternative: $v^2 = u^2 - 2gh = 400 - 200 = 200$ , $KE = \frac{1}{2} \times 0.50 \times 200 = 50 \text{ J}$
Mark breakdown: 1 mark for finding total energy or GPE at half height, 1 mark for correct KE.

15. [4 marks]

(a) [1 mark]
Rate of GPE gain = $\dot{m}gh = 10 \times 10 \times 20 = 2000 \text{ W} = 2.0 \text{ kW}$
Answer: 2000 W (or 2.0 kW)

(b) [1 mark]
Rate of KE gain = $\frac{1}{2}\dot{m}v^2 = \frac{1}{2} \times 10 \times 5.0^2 = 5 \times 25 = 125 \text{ W}$
Answer: 125 W

(c) [2 marks]
Total useful power output = Rate of GPE gain + Rate of KE gain = $2000 + 125 = 2125 \text{ W}$
Efficiency = $\frac{\text{useful output}}{\text{input}} \Rightarrow 0.70 = \frac{2125}{P_{\text{input}}}$
$P_{\text{input}} = \frac{2125}{0.70} = 3035.7 \text{ W} \approx 3040 \text{ W}$ (or 3.04 kW)
Answer: 3040 W (or 3.04 kW)
Mark breakdown: 1 mark for total useful power, 1 mark for correct input power calculation.

Section C: Longer Structured Questions (12 marks)

16. [6 marks]

(a) [1 mark]
Diagram should show:

Weight $mg$ vertically downwards
Normal reaction $N$ perpendicular to plane (upwards)
Friction $f$ down the plane (opposing motion)
Applied force $F$ up the plane
Marking note: All four forces correctly labelled and directions shown. Arrow lengths need not be to scale but directions must be correct.

(b) [2 marks]
Normal reaction $N = mg\cos\theta = 2.0 \times 10 \times \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3 \text{ N}$
Frictional force $f = \mu_k N = 0.25 \times 10\sqrt{3} = 2.5\sqrt{3} \approx 4.33 \text{ N}$
Answer: $4.33 \text{ N}$ (or $2.5\sqrt{3} \text{ N}$ )
Mark breakdown: 1 mark for correct normal reaction, 1 mark for correct friction calculation.

(c) [2 marks]
Since constant speed, net force parallel to plane = 0:
$F = mg\sin\theta + f = (2.0 \times 10 \times \sin 30^\circ) + 2.5\sqrt{3} = (20 \times 0.5) + 2.5\sqrt{3} = 10 + 2.5\sqrt{3} \approx 14.33 \text{ N}$
Work done by $F = F \times s = (10 + 2.5\sqrt{3}) \times 5.0 = 50 + 12.5\sqrt{3} \approx 71.65 \text{ J}$
Answer: $71.7 \text{ J}$ (or $50 + 12.5\sqrt{3} \text{ J}$ )
Mark breakdown: 1 mark for finding $F$ (equilibrium condition), 1 mark for work done calculation.

(d) [1 mark]
By the work-energy theorem, net work done on the block equals its change in kinetic energy. Since the block moves at constant speed, $\Delta KE = 0$ . Therefore, the net work done on the block is zero. The positive work done by the applied force is exactly balanced by the negative work done by friction and the negative work done by the component of weight parallel to the plane.
Answer: Net work = 0, so $\Delta KE = 0$ .
Key concept: Work-energy theorem $W_{\text{net}} = \Delta KE$ ; constant speed $\Rightarrow \Delta KE = 0 \Rightarrow W_{\text{net}} = 0$ .

17. [6 marks]

(a) [2 marks]
Swept area $A = \pi r^2 = \pi \times 25^2 = 625\pi \text{ m}^2$
Volume of air per second = $A \times v = 625\pi \times 12 = 7500\pi \text{ m}^3/\text{s}$
Mass of air per second = density $\times$ volume rate = $1.2 \times 7500\pi = 9000\pi \approx 28\,274 \text{ kg/s}$
Answer: $28\,300 \text{ kg/s}$ (or $9000\pi \text{ kg/s}$ )
Mark breakdown: 1 mark for swept area, 1 mark for mass flow rate calculation.

(b) [2 marks]
Kinetic energy per second (power in wind) = $\frac{1}{2} \dot{m} v^2 = \frac{1}{2} \times 9000\pi \times 12^2 = \frac{1}{2} \times 9000\pi \times 144 = 648\,000\pi \approx 2.036 \times 10^6 \text{ W} = 2.04 \text{ MW}$
Answer: $2.04 \text{ MW}$ (or $648\,000\pi \text{ W}$ )
Mark breakdown: 1 mark for correct formula $\frac{1}{2}\dot{m}v^2$ , 1 mark for correct calculation.

(c) [1 mark]
Electrical power output = efficiency $\times$ power in wind = $0.40 \times 2.036 \times 10^6 = 8.14 \times 10^5 \text{ W} = 814 \text{ kW}$
Answer: $814 \text{ kW}$ (or $0.814 \text{ MW}$ )
Marking note: Accept answers consistent with (b) rounded appropriately.

(d) [1 mark]
Reason: If 100% of kinetic energy were extracted, the air would have zero speed after passing through the turbine, meaning no air could flow through (mass flow would stop). Air must retain some kinetic energy to move away from the turbine (Betz's law limit is 59.3%).
Answer: Air must retain some kinetic energy to continue flowing; otherwise mass flow would cease.
Key concept: Betz limit / conservation of mass flow.

Section D: Additional Questions (10 marks)

18. [3 marks]

(a) [1 mark]
Initial kinetic energy = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 20^2 = 600 \times 400 = 240\,000 \text{ J} = 240 \text{ kJ}$
Answer: 240 000 J (or 240 kJ)

(b) [2 marks]
Work done by braking force = change in kinetic energy = $-240\,000 \text{ J}$ (negative because force opposes motion)
Work done = $F \times s = F \times 50 = -240\,000$
$F = \frac{-240\,000}{50} = -4800 \text{ N}$
Magnitude of average braking force = $4800 \text{ N}$
Answer: 4800 N
Mark breakdown: 1 mark for work-energy principle, 1 mark for correct force magnitude.

19. [3 marks]

(a) [1 mark]
Elastic potential energy = $\frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.15)^2 = 100 \times 0.0225 = 2.25 \text{ J}$
Answer: 2.25 J

(b) [2 marks]
On smooth surface, elastic PE converts entirely to KE:
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
$v^2 = \frac{kx^2}{m} = \frac{200 \times 0.0225}{0.50} = \frac{4.5}{0.50} = 9$
$v = 3.0 \text{ m/s}$
Answer: 3.0 m/s
Mark breakdown: 1 mark for energy conservation equation, 1 mark for correct speed.

20. [4 marks]

(a) [1 mark]
Loss in GPE = $mgh = 50 \times 10 \times 8.0 = 4000 \text{ J}$
Answer: 4000 J

(b) [1 mark]
Gain in KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 50 \times 10^2 = 25 \times 100 = 2500 \text{ J}$
Answer: 2500 J

(c) [2 marks]
Work done against resistive forces = Loss in GPE - Gain in KE = $4000 - 2500 = 1500 \text{ J}$
Work done = resistive force $\times$ distance = $F_{\text{res}} \times 20 = 1500$
$F_{\text{res}} = \frac{1500}{20} = 75 \text{ N}$
Answer: 75 N
Mark breakdown: 1 mark for energy difference (work done against resistance), 1 mark for correct force calculation.

End of Answer Key