Secondary 2 Science Practice Paper 3

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TuitionGoWhere Practice Paper - Science Secondary 2

TuitionGoWhere Practice Paper (AI) — Version 3

Subject: Science
Level: Secondary 2 (G3)
Paper: Practice Paper – Physical Sciences
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

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Instructions to Candidates

1. Write your name, class, and date in the spaces provided above.
2. Answer all questions.
3. Write your answers in the spaces provided on the question paper.
4. The number of marks is given in brackets [ ] at the end of each question or part question.
5. The total marks for this paper is 60.
6. You may use a calculator.
7. Where necessary, take the acceleration due to gravity, $g = 10 \text{ m/s}^2$.
8. Show all working for calculation questions.

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Section A: Multiple Choice Questions (10 marks)

Answer all questions. For each question, choose the one correct answer and write the letter (A, B, C, or D) in the box provided.

1. A 2.0 kg block is lifted vertically from the ground to a shelf 1.5 m high. What is the gain in gravitational potential energy of the block?
A. 3.0 J
B. 15 J
C. 30 J
D. 45 J

☐

2. A ball is dropped from a height of 10 m. Ignoring air resistance, which of the following describes the energy conversion as the ball falls?
A. Kinetic energy → Gravitational potential energy
B. Gravitational potential energy → Kinetic energy
C. Chemical energy → Kinetic energy
D. Gravitational potential energy → Chemical energy

☐

3. A force of 25 N is used to push a box horizontally across a floor for a distance of 4 m. How much work is done by the force?
A. 6.25 J
B. 29 J
C. 100 J
D. 400 J

☐

4. A 500 W electric kettle is used to boil water for 3 minutes. How much electrical energy is consumed?
A. 1500 J
B. 90 000 J
C. 150 000 J
D. 900 000 J

☐

5. Which of the following is a non-renewable energy resource?
A. Solar
B. Wind
C. Natural gas
D. Hydroelectric

☐

6. A car of mass 1000 kg accelerates from rest to 20 m/s in 10 s. What is the average power developed by the car's engine? (Assume no energy losses.)
A. 2000 W
B. 20 000 W
C. 40 000 W
D. 200 000 W

☐

7. A pendulum swings from its highest point A to its lowest point B. At which point is the kinetic energy of the pendulum bob maximum?
A. At point A only
B. At point B only
C. Halfway between A and B
D. At both A and B

☐

8. A student does 500 J of work in pulling a sled up a slope. The gain in gravitational potential energy of the sled is 400 J. What is the efficiency of this process?
A. 20%
B. 80%
C. 100%
D. 125%

☐

9. Which energy conversion takes place in a hydroelectric power station?
A. Electrical → Gravitational potential → Kinetic
B. Gravitational potential → Kinetic → Electrical
C. Chemical → Kinetic → Electrical
D. Nuclear → Thermal → Electrical

☐

10. A 0.2 kg toy car moves at a constant speed of 3 m/s. What is its kinetic energy?
A. 0.3 J
B. 0.9 J
C. 1.8 J
D. 3.6 J

☐

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Section B: Structured Questions (30 marks)

Answer all questions in the spaces provided.

11. A roller coaster car of mass 500 kg is at rest at the top of a hill, point P, which is 40 m above the ground. The car then rolls down the track to point Q, which is 10 m above the ground. Ignore friction and air resistance.

Generated diagram for Q11.

(a) State the principle of conservation of energy.
[2]

(b) Calculate the gravitational potential energy of the car at point P.
[2]

(c) Determine the kinetic energy of the car at point Q.
[2]

(d) Calculate the speed of the car at point Q.
[2]

12. A student investigates the efficiency of a small electric motor. The motor lifts a load of mass 0.5 kg through a vertical height of 1.2 m in 4.0 s. The voltage across the motor is 6.0 V and the current through it is 0.8 A.

(a) Calculate the work done (useful energy output) in lifting the load.
[2]

(b) Calculate the electrical energy input to the motor in 4.0 s.
[2]

(c) Determine the efficiency of the motor.
[2]

(d) Suggest one reason why the efficiency is less than 100%.
[1]

13. The diagram below shows a simple pendulum. The bob is pulled aside to position A, held at rest, and then released. It swings through the lowest point B and rises to position C on the other side. Positions A and C are at the same vertical height above B.

Generated diagram for Q13.

(a) Describe the energy conversion that takes place as the bob swings from A to B.
[2]

(b) Explain why the bob does not rise to a height greater than A on the other side.
[2]

(c) In a real pendulum, the bob eventually comes to rest at B. Explain what happens to the initial gravitational potential energy of the bob at A.
[2]

14. A 1200 kg car is travelling at a constant speed of 25 m/s on a level road. The driver applies the brakes and the car comes to rest in 50 m.

(a) Calculate the initial kinetic energy of the car.
[2]

(b) The work done by the braking force equals the loss in kinetic energy. Calculate the average braking force acting on the car.
[2]

(c) If the car's mass were doubled but its initial speed remained the same, how would the braking distance change? Assume the same average braking force. Explain your answer.
[2]

15. A solar panel of area 2.0 m² receives sunlight with an intensity of 800 W/m². The panel converts 18% of the incident solar energy into electrical energy.

(a) Calculate the total solar power incident on the panel.
[2]

(b) Calculate the electrical power output of the panel.
[2]

(c) The panel is used to charge a 12 V battery. If the charging current is 2.0 A, how long will it take to supply 1.0 MJ of electrical energy to the battery?
[2]

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Section C: Longer Structured and Data-Based Questions (20 marks)

Answer all questions in the spaces provided.

16. A hydroelectric power station uses water falling from a reservoir to generate electricity. Water falls through a vertical height of 80 m. The mass flow rate of water is 500 kg/s. The overall efficiency of the power station is 85%.

(a) Calculate the gravitational potential energy lost by the water each second.
[2]

(b) Calculate the electrical power output of the power station.
[2]

(c) The electricity generated is transmitted at 400 kV. Calculate the current in the transmission cables.
[2]

(d) Explain why electrical energy is transmitted at high voltage rather than low voltage.
[2]

17. The table below shows the energy consumption of various household appliances.

Appliance	Power Rating / W	Typical Daily Usage / h
Refrigerator	150	24
Air conditioner	1500	6
Washing machine	500	1
LED lights (total)	60	5
Laptop	50	4

(a) Calculate the total electrical energy consumed by all these appliances in one day, in kWh.
[3]

(b) If electricity costs $0.28 per kWh, calculate the cost of running these appliances for 30 days.
[2]

(c) The refrigerator runs on a thermostat and is only actively cooling for 30% of the time. Recalculate the daily energy consumption of the refrigerator.
[1]

(d) Suggest two ways a household could reduce its electricity consumption without replacing appliances.
[2]

18. A student sets up an experiment to investigate the relationship between the height of a ramp and the speed of a trolley at the bottom. The trolley is released from rest at the top of the ramp. The student measures the time taken for the trolley to travel the last 1.0 m of the ramp using a light gate.

Generated experimental_setup for Q18.

The student obtains the following data:

Height of ramp, $h$ / m	Time for last 1.0 m, $t$ / s
0.10	0.85
0.20	0.60
0.30	0.49
0.40	0.43
0.50	0.38

(a) For $h = 0.30$ m, calculate the average speed of the trolley over the last 1.0 m.
[1]

(b) Assuming the trolley accelerates uniformly, the speed at the midpoint of the last 1.0 m is approximately equal to the average speed. Calculate the kinetic energy of the trolley (mass = 0.5 kg) at the bottom of the ramp for $h = 0.30$ m.
[2]

(c) Plot a graph of kinetic energy at bottom (y-axis) against height of ramp (x-axis) for all five data points. Use the grid below.

Generated graph for Q18.

[3]

(d) State the relationship between the kinetic energy at the bottom of the ramp and the height of the ramp, based on your graph.
[1]

(e) The graph does not pass through the origin. Suggest one reason for this.
[1]

19. Fossil fuels (coal, oil, natural gas) are the primary energy sources for electricity generation in many countries. However, there is a global shift towards renewable energy sources.

(a) State two disadvantages of burning fossil fuels for electricity generation.
[2]

(b) Explain why solar energy is considered a renewable energy source.
[1]

(c) A solar farm has an area of 50 000 m². The average solar irradiance is 200 W/m² (averaged over day and night). The solar panels have an efficiency of 20%. Calculate the average electrical power output of the solar farm in MW.
[3]

(d) State one advantage and one disadvantage of solar farms compared to fossil fuel power stations.
[2]

20. A 60 kg student runs up a flight of stairs. The vertical height of the stairs is 3.5 m. The student takes 4.0 s to run up the stairs.

(a) Calculate the work done by the student against gravity.
[2]

(b) Calculate the power developed by the student.
[2]

(c) The student's muscles convert chemical energy from food into mechanical work. If the efficiency of this conversion is 25%, calculate the chemical energy expended by the student.
[2]

(d) After reaching the top, the student walks down the stairs slowly. Explain why the work done by gravity on the student during the descent is positive, while the work done by the student against gravity is negative.
[2]

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End of Paper

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TuitionGoWhere Practice Paper - Science Secondary 2 (Answer Key)

Subject: Science
Level: Secondary 2 (G3)
Paper: Practice Paper – Physical Sciences (Version 3)
Total Marks: 60

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Section A: Multiple Choice Questions (10 marks)

1. C
Working:
Gravitational potential energy gained = $mgh = 2.0 \times 10 \times 1.5 = 30 \text{ J}$
Marking: 1 mark for correct answer.

2. B
Explanation: As the ball falls, its height decreases so gravitational potential energy decreases, while its speed increases so kinetic energy increases. Energy is converted from gravitational potential energy to kinetic energy.
Marking: 1 mark for correct answer.

3. C
Working:
Work done = Force × Distance = $25 \times 4 = 100 \text{ J}$
Marking: 1 mark for correct answer.

4. B
Working:
Energy = Power × Time = $500 \text{ W} \times (3 \times 60) \text{ s} = 500 \times 180 = 90\,000 \text{ J}$
Marking: 1 mark for correct answer.

5. C
Explanation: Natural gas is a fossil fuel formed over millions of years and cannot be replenished on a human timescale. Solar, wind, and hydroelectric are renewable.
Marking: 1 mark for correct answer.

6. B
Working:
Kinetic energy gained = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 20^2 = 200\,000 \text{ J}$
Average power = $\frac{\text{Work}}{\text{Time}} = \frac{200\,000}{10} = 20\,000 \text{ W}$
Marking: 1 mark for correct answer.

7. B
Explanation: At the lowest point B, gravitational potential energy is minimum and kinetic energy is maximum (by conservation of energy, ignoring air resistance).
Marking: 1 mark for correct answer.

8. B
Working:
Efficiency = $\frac{\text{Useful energy output}}{\text{Energy input}} \times 100\% = \frac{400}{500} \times 100\% = 80\%$
Marking: 1 mark for correct answer.

9. B
Explanation: Water in a reservoir has gravitational potential energy → converts to kinetic energy as it falls → drives turbines to generate electrical energy.
Marking: 1 mark for correct answer.

10. B
Working:
Kinetic energy = $\frac{1}{2}mv^2 = \frac{1}{2} \times 0.2 \times 3^2 = 0.1 \times 9 = 0.9 \text{ J}$
Marking: 1 mark for correct answer.

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Section B: Structured Questions (30 marks)

11. (a) Principle of conservation of energy: Energy cannot be created or destroyed. It can only be converted from one form to another, and the total amount of energy in a closed system remains constant.
[2]
Marking: 1 mark for "cannot be created or destroyed"; 1 mark for "converted from one form to another" or "total energy remains constant".

(b) GPE at P = $mgh = 500 \times 10 \times 40 = 200\,000 \text{ J}$ (or $2.0 \times 10^5 \text{ J}$)
[2]
Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

(c) By conservation of energy:
Loss in GPE from P to Q = Gain in KE at Q
$mg(h_P - h_Q) = 500 \times 10 \times (40 - 10) = 500 \times 10 \times 30 = 150\,000 \text{ J}$
KE at Q = 150 000 J
[2]
Marking: 1 mark for correct concept (loss in GPE = gain in KE); 1 mark for correct calculation and unit.

(d) $\text{KE} = \frac{1}{2}mv^2$
$150\,000 = \frac{1}{2} \times 500 \times v^2$
$v^2 = \frac{150\,000 \times 2}{500} = 600$
$v = \sqrt{600} \approx 24.5 \text{ m/s}$
[2]
Marking: 1 mark for correct rearrangement/substitution; 1 mark for correct answer with unit (accept 24.5 or 24.49 m/s).

Common mistake: Forgetting to take square root, or using $h = 40$ m instead of height difference.

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12. (a) Work done (useful output) = Gain in GPE = $mgh = 0.5 \times 10 \times 1.2 = 6.0 \text{ J}$
[2]
Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

(b) Electrical energy input = $VIt = 6.0 \times 0.8 \times 4.0 = 19.2 \text{ J}$
[2]
Marking: 1 mark for correct formula $E = VIt$; 1 mark for correct answer with unit.

(c) Efficiency = $\frac{\text{Useful output}}{\text{Input}} \times 100\% = \frac{6.0}{19.2} \times 100\% = 31.25\%$
[2]
Marking: 1 mark for correct formula; 1 mark for correct answer (accept 31.3% or 31.25%).

(d) Reason: Energy is lost as heat (thermal energy) in the motor coils due to electrical resistance, and as sound energy from friction in moving parts.
[1]
Marking: 1 mark for any valid reason (heat loss, sound, friction).

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13. (a) As the bob swings from A to B, gravitational potential energy is converted to kinetic energy. At A, the bob has maximum GPE and zero KE. At B, GPE is minimum and KE is maximum.
[2]
Marking: 1 mark for identifying GPE → KE conversion; 1 mark for describing the change (max at A to max at B).

(b) By conservation of energy, the total mechanical energy (GPE + KE) remains constant (ignoring air resistance). At C, the bob momentarily comes to rest so KE = 0. All energy is GPE. Since no energy is added, the GPE at C cannot exceed the initial GPE at A, so the height cannot be greater than A.
[2]
Marking: 1 mark for conservation of energy argument; 1 mark for explaining height at C ≤ height at A.

(c) In a real pendulum, air resistance and friction at the pivot do negative work on the bob. The initial GPE at A is gradually converted to thermal energy (heat) in the surrounding air and at the pivot, and sound energy. Eventually, all mechanical energy is dissipated as thermal/sound energy, and the bob comes to rest at B.
[2]
Marking: 1 mark for identifying air resistance/friction; 1 mark for stating energy is dissipated as heat/sound.

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14. (a) Initial KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 25^2 = 600 \times 625 = 375\,000 \text{ J}$
[2]
Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

(b) Work done by braking force = Loss in KE = $375\,000 \text{ J}$
Work = Force × Distance
$F \times 50 = 375\,000$
$F = \frac{375\,000}{50} = 7\,500 \text{ N}$
[2]
Marking: 1 mark for equating work to KE loss; 1 mark for correct force with unit.

(c) Braking distance would double.
Explanation: Initial KE = $\frac{1}{2}mv^2$. If mass doubles, initial KE doubles. Work done by braking force = $F \times d$. Since $F$ is constant, $d$ must double to do twice the work.
[2]
Marking: 1 mark for correct prediction (doubles); 1 mark for correct explanation linking KE ∝ m and Work = Fd.

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15. (a) Solar power incident = Intensity × Area = $800 \times 2.0 = 1600 \text{ W}$
[2]
Marking: 1 mark for correct formula; 1 mark for correct answer with unit.

(b) Electrical power output = Efficiency × Incident power = $0.18 \times 1600 = 288 \text{ W}$
[2]
Marking: 1 mark for correct use of efficiency; 1 mark for correct answer with unit.

(c) Electrical power to battery = $VI = 12 \times 2.0 = 24 \text{ W}$
Time = $\frac{\text{Energy}}{\text{Power}} = \frac{1.0 \times 10^6}{24} = 41\,666.7 \text{ s}$
Convert to hours: $\frac{41\,666.7}{3600} \approx 11.6 \text{ hours}$ (or 11 h 34 min)
[2]
Marking: 1 mark for correct power calculation; 1 mark for correct time with unit.

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Section C: Longer Structured and Data-Based Questions (20 marks)

16. (a) GPE lost per second = Mass flow rate × $g$ × height = $500 \times 10 \times 80 = 400\,000 \text{ J/s} = 400\,000 \text{ W}$
[2]
Marking: 1 mark for correct formula (power = $\dot{m}gh$); 1 mark for correct answer with unit.

(b) Electrical power output = Efficiency × Input power = $0.85 \times 400\,000 = 340\,000 \text{ W} = 340 \text{ kW}$
[2]
Marking: 1 mark for correct use of efficiency; 1 mark for correct answer with unit (accept 340 kW or 340 000 W).

(c) $P = VI$ → $I = \frac{P}{V} = \frac{340\,000}{400\,000} = 0.85 \text{ A}$
[2]
Marking: 1 mark for correct formula/rearrangement; 1 mark for correct answer with unit.

(d) Explanation: Transmitting at high voltage reduces the current for the same power ($P = VI$). Lower current reduces the power loss in the cables ($P_{\text{loss}} = I^2R$), making transmission more efficient.
[2]
Marking: 1 mark for stating current is reduced; 1 mark for linking to reduced $I^2R$ losses / heating in cables.

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17. (a) Daily energy per appliance (kWh):

• Refrigerator: $\frac{150}{1000} \times 24 = 3.6 \text{ kWh}$
• Air conditioner: $\frac{1500}{1000} \times 6 = 9.0 \text{ kWh}$
• Washing machine: $\frac{500}{1000} \times 1 = 0.5 \text{ kWh}$
• LED lights: $\frac{60}{1000} \times 5 = 0.3 \text{ kWh}$
• Laptop: $\frac{50}{1000} \times 4 = 0.2 \text{ kWh}$
  Total = $3.6 + 9.0 + 0.5 + 0.3 + 0.2 = 13.6 \text{ kWh}$
  [3]
  Marking: 1 mark for correct method (Power in kW × time); 1 mark for correct individual calculations; 1 mark for correct total with unit.

(b) Cost for 30 days = 13.6 \times 30 \times 0.28 = 408 \times 0.28 = \114.24$[2] **Marking:** 1 mark for correct monthly energy (408 kWh); 1 mark for correct cost with$ sign.

(c) Actual refrigerator energy = $3.6 \times 0.30 = 1.08 \text{ kWh}$
[1]
Marking: 1 mark for correct answer with unit.

(d) Two ways:

1. Switch off appliances at the socket (avoid standby power).
2. Set air conditioner to a higher temperature (e.g., 25°C instead of 22°C).
3. Use natural light instead of LED lights during the day.
4. Wash clothes with full loads only.
   (Any two valid suggestions)
   [2]
   Marking: 1 mark each for two valid, practical suggestions.

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18. (a) Average speed = $\frac{\text{Distance}}{\text{Time}} = \frac{1.0}{0.49} \approx 2.04 \text{ m/s}$
[1]
Marking: 1 mark for correct calculation with unit.

(b) Speed at bottom ≈ average speed over last 1.0 m = $2.04 \text{ m/s}$
KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times (2.04)^2 = 0.25 \times 4.1616 \approx 1.04 \text{ J}$
[2]
Marking: 1 mark for using average speed as final speed; 1 mark for correct KE calculation with unit.

(c) Graph plotting:
First, calculate KE for each height using $v_{\text{avg}} = 1.0/t$, $KE = \frac{1}{2} \times 0.5 \times (1.0/t)^2 = \frac{0.25}{t^2}$:

$h$ (m)	$t$ (s)	$v_{\text{avg}}$ (m/s)	KE (J)
0.10	0.85	1.176	0.346
0.20	0.60	1.667	0.694
0.30	0.49	2.041	1.041
0.40	0.43	2.326	1.353
0.50	0.38	2.632	1.732

Graph requirements:

• Axes labelled with units: "Height of ramp / m" (x-axis), "Kinetic Energy / J" (y-axis)
• Appropriate scales covering all points
• All 5 points plotted accurately
• Best-fit straight line through origin (theoretically) or close to points
  [3]
  Marking: 1 mark for correct axes labels and scales; 1 mark for all 5 points plotted correctly; 1 mark for best-fit line.

(d) Relationship: The kinetic energy at the bottom of the ramp is directly proportional to the height of the ramp. (Or: KE increases linearly with height.)
[1]
Marking: 1 mark for "directly proportional" or "linear increase".

(e) Reason: The graph does not pass through the origin because of friction between the trolley and ramp, and air resistance. Some gravitational potential energy is converted to thermal/sound energy instead of kinetic energy, so even at very small heights, there is a threshold below which the trolley does not move, or the KE is less than predicted by $mgh$.
[1]
Marking: 1 mark for identifying friction/air resistance as cause of energy loss.

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19. (a) Two disadvantages:

1. Releases carbon dioxide (CO₂), a greenhouse gas, contributing to global warming and climate change.
2. Releases pollutants (e.g., sulfur dioxide, nitrogen oxides, particulates) causing air pollution, acid rain, and health problems.
   (Also acceptable: finite/non-renewable resource; habitat destruction from mining; oil spills.)
   [2]
   Marking: 1 mark each for two distinct valid disadvantages.

(b) Explanation: Solar energy comes from the Sun, which will continue to radiate energy for billions of years. It is replenished naturally on a human timescale and cannot be depleted by human use.
[1]
Marking: 1 mark for "replenished naturally" or "virtually inexhaustible" or "Sun will shine for billions of years".

(c) Average electrical power output:
Total solar power incident = Irradiance × Area = $200 \times 50\,000 = 10\,000\,000 \text{ W} = 10 \text{ MW}$
Electrical power output = Efficiency × Incident power = $0.20 \times 10 \text{ MW} = 2 \text{ MW}$
[3]
Marking: 1 mark for incident power calculation; 1 mark for applying efficiency; 1 mark for correct answer in MW.

(d) Advantage: No fuel cost / no greenhouse gas emissions during operation / renewable.
Disadvantage: Intermittent (depends on sunlight, weather, day/night) / requires large land area / high initial cost / energy storage needed for continuous supply.
[2]
Marking: 1 mark for valid advantage; 1 mark for valid disadvantage.

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20. (a) Work done against gravity = Gain in GPE = $mgh = 60 \times 10 \times 3.5 = 2100 \text{ J}$
[2]
Marking: 1 mark for correct formula/substitution; 1 mark for correct answer with unit.

(b) Power = $\frac{\text{Work}}{\text{Time}} = \frac{2100}{4.0} = 525 \text{ W}$
[2]
Marking: 1 mark for correct formula; 1 mark for correct answer with unit.

(c) Chemical energy expended:
Efficiency = $\frac{\text{Useful work output}}{\text{Chemical energy input}}$
$0.25 = \frac{2100}{E_{\text{chem}}}$
$E_{\text{chem}} = \frac{2100}{0.25} = 8400 \text{ J}$
[2]
Marking: 1 mark for correct efficiency formula rearrangement; 1 mark for correct answer with unit.

(d) Explanation:

• Work done by gravity during descent: The force of gravity (weight) acts downwards, and the displacement is downwards. Since force and displacement are in the same direction, work done by gravity is positive ($W = Fd \cos 0^\circ = +Fd$). Gravity transfers energy to the student (increases kinetic energy if uncontrolled).
• Work done by the student against gravity: The student must exert an upward force to control the descent, while displacement is downwards. Force and displacement are in opposite directions, so work done by the student is negative ($W = Fd \cos 180^\circ = -Fd$). The student's muscles absorb energy (chemical energy is still expended to maintain tension).
  [2]
  Marking: 1 mark for explaining positive work by gravity (force and displacement same direction); 1 mark for explaining negative work by student (force opposite to displacement).

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End of Answer Key