A professional AFL midfielder produces peak power outputs exceeding 1,000 watts during sprints — roughly the same power as a microwave oven. In physics, work has a precise meaning: it is the transfer of energy when a force moves an object through a distance. Understanding work and power lets us quantify who is working hardest, which machines are most effective, and where energy is actually going.
Imagine two students climbing the stairs to the top of the Sydney Tower Eye, 250 metres above street level. Student A walks slowly, taking 10 minutes. Student B runs up, taking 5 minutes. Both students have the same mass of 60 kg.
Estimate and predict: Use g ≈ 10 N/kg to estimate the gravitational force on each student. Then calculate the work done against gravity (W = F × d). Predict: who does more work, who has greater power output, and does the faster student use more total energy or just use it faster? Write your estimates and reasoning — you will compare them to the physics at the end of the lesson.
📚 Core Content
Wrong: "If I hold a heavy box still, I am doing work because I am using effort."
Right: In physics, work is only done when the object moves in the direction of the force. Holding a box still requires muscle effort, but no work is done on the box because the distance moved is zero. Your muscles are converting chemical energy into thermal energy (you get tired and warm), but zero work is done on the box.
Wrong: "A more powerful machine always does more work than a less powerful one."
Right: Power measures how fast work is done, not how much work is done. A 2,000 W kettle boils water faster than a 1,000 W kettle, but both do the same amount of work (transfer the same energy) to boil the same amount of water. The 2,000 W kettle just does it in half the time.
In everyday language, "work" means any activity that requires effort. In physics, work has a much more specific meaning: work is done when a force causes an object to move in the direction of that force. If the object does not move, no work is done — no matter how much effort is applied.
The formula is beautifully simple:
W = F × d
Where W is work in joules (J), F is force in newtons (N), and d is distance in metres (m). One joule equals one newton-metre — the work done when a force of one newton moves an object one metre.
For example, lifting a 10 kg bag of cement requires a force of about 100 N (since gravity pulls with roughly 10 N per kilogram). If you lift it 2 metres onto a scaffold, the work done is:
W = 100 N × 2 m = 200 J
This 200 J is the amount of energy transferred from your muscles to the bag. That energy is now stored as gravitational potential energy in the bag. When the bag is lowered, that gravitational potential energy can be returned — though in practice, some becomes thermal energy through friction and air resistance.
A worker on a Sydney Metro construction site lifts a 50 kg steel beam 3 metres onto a platform. Calculate the work done. (Use g = 10 N/kg for simplicity.)
Calculate the force: F = m × g = 50 kg × 10 N/kg = 500 N
Apply the formula: W = F × d = 500 N × 3 m = 1,500 J
Answer: 1,500 J of work is done. This energy is transferred from the worker's muscles to the beam, where it is stored as gravitational potential energy.
Two workers can lift the same beam the same distance, doing the same amount of work. But if one takes 10 seconds and the other takes 30 seconds, the first worker is working more powerfully. Power measures the rate of energy transfer — how many joules per second.
P = W ÷ t or P = E ÷ t
Where P is power in watts (W), W is work in joules (J), E is energy in joules (J), and t is time in seconds (s). One watt equals one joule per second. A 60-watt light bulb converts 60 joules of electrical energy into light and thermal energy every single second.
Returning to our two workers who both did 1,500 J of work:
Worker A (10 s): P = 1,500 J ÷ 10 s = 150 W
Worker B (30 s): P = 1,500 J ÷ 30 s = 50 W
Worker A is three times more powerful. They did not do more work — they just did the same work faster. This is why power is such a useful concept: it tells us about performance, not just output.
| Source | Power Output |
|---|---|
| Human brain | ~20 W |
| Human at rest | ~100 W |
| Human walking | ~250 W |
| Professional cyclist (sustained) | ~400 W |
| AFL midfielder (peak) | ~1,000 W |
| Family car engine | ~75,000 W (75 kW) |
| Hornsdale Power Reserve (Tesla battery) | ~150,000,000 W (150 MW) |
| Loy Yang Power Station | ~2,200,000,000 W (2.2 GW) |
A professional cyclist in the Tour Down Under produces 400 W of useful mechanical power while climbing a hill for 20 minutes. Calculate the total work done.
Convert time: 20 minutes = 20 × 60 = 1,200 seconds
Rearrange: W = P × t = 400 W × 1,200 s = 480,000 J (480 kJ)
Answer: 480,000 J. But remember — the human body is only about 25% efficient at converting food energy into mechanical work. The cyclist actually metabolised about 480,000 × 4 = 1,920,000 J of chemical energy from food. The other 1,440,000 J became thermal energy.
Work and power are not abstract classroom concepts. They govern every machine, every worker, and every power station in Australia. Understanding them helps engineers design better equipment, helps employers set realistic workloads, and helps us appreciate the staggering scale of Australia's energy infrastructure.
The World's Longest Fence, the Dingo Fence, stretches 5,614 kilometres across Queensland, South Australia and New South Wales — longer than the Great Wall of China. Building it in the 1880s required enormous work: workers had to dig post holes, carry timber, and stretch wire across some of Australia's most arid terrain. The work done against gravity alone — lifting posts, carrying supplies, climbing dunes — would total billions of joules. Today, maintaining the fence still requires significant work and power, with vehicles, machinery and helicopters used to patrol and repair sections.
During the third quarter of an AFL match at the MCG, a midfielder produces peak power outputs of over 1,000 watts during sprints — comparable to a toaster or microwave oven. Over the full game, their average power output is about 250 watts. This means they do approximately 250 J of work every second for 120 minutes: total work = 250 × 7,200 = 1,800,000 J (1.8 MJ). But because the human body is only 25% efficient, the player must actually burn food energy of about 7.2 MJ — equivalent to about four slices of pizza or two energy gels. This is why AFL players consume 4,000–5,000 kilojoules during a game just to maintain performance.
1 A miner pushes a loaded cart with a horizontal force of 200 N for 15 metres along a tunnel. Calculate the work done.
2 A student climbs a rope in a gym, lifting their 55 kg body 4 metres off the ground. Calculate the work done against gravity. (Use F = m × g, where g = 10 N/kg.)
3 A crane lifts a 2,000 kg steel beam 12 metres in 8 seconds. Calculate the work done and the power output of the crane. (Use g = 10 N/kg.)
4 A 60 W light bulb is left on for 5 hours. Calculate the total electrical energy transferred. Give your answer in both joules and kilowatt-hours.
1. A person holds a 20 kg box at waist height for 30 seconds without moving it. How much work is done on the box?
2. A crane lifts a 500 kg load 8 metres in 4 seconds. What is the power output of the crane? (Use g = 10 N/kg.)
3. Two students of equal mass climb the same staircase. Student A takes 20 seconds. Student B takes 40 seconds. Which statement is correct?
4. A 2,000 W electric kettle boils water in 3 minutes. A 1,000 W kettle boils the same amount of water in 6 minutes. Which statement about the energy used is correct?
5. A Pilbara haul truck does 200 MJ of useful work climbing a ramp. The engine has a power output of 2,500 kW and is 35% efficient. Approximately how much chemical energy in diesel is required, and how long does the climb take?
6. A Sydney Tower Eye elevator lifts a total mass of 1,200 kg (passengers plus cabin) from ground level to the observation deck 250 metres above. Calculate the work done by the elevator motor. If the lift takes 45 seconds, calculate the power output. (Use g = 10 N/kg.) 1 mark for correct force calculation. 1 mark for correct work. 1 mark for correct power.
7. A family is choosing between two air conditioners for their home in Darwin. Unit A has a cooling power of 2,500 W and costs $800. Unit B has a cooling power of 5,000 W and costs $1,200. Both units run on electricity that costs $0.30 per kWh. The family needs to remove 18,000 kJ of thermal energy from their house each hour on a hot day.
Calculate how long each unit would take to remove 18,000 kJ. Calculate the electrical energy used by each unit in one hour of operation. Explain which unit is the better choice, considering both upfront cost and operating cost. 1 mark for calculating time for each unit to remove 18,000 kJ. 1 mark for calculating energy used by each unit in one hour. 1 mark for explaining that Unit B removes heat faster but uses more electricity per hour. 1 mark for a justified recommendation considering Darwin's climate (long hot season makes operating cost significant).
8. A mining company claims their new electric haul truck is "better than diesel because it has a more powerful motor." The diesel truck has a 2,500 kW engine at 35% efficiency. The electric truck has a 2,000 kW motor at 85% efficiency. Both trucks do the same 200 MJ of useful work climbing a ramp.
Analyse this claim by calculating: (a) the energy input required by each truck, (b) the time each truck takes to climb the ramp, and (c) whether "more powerful" is the right way to compare them. Use the concepts of work, power and efficiency in your answer. 1 mark for calculating diesel energy input (≈571 MJ). 1 mark for calculating electric energy input (≈235 MJ). 1 mark for calculating time for each truck (diesel 80 s, electric 100 s). 1 mark for explaining that the electric truck uses less than half the energy despite being less powerful. 1 mark for a balanced conclusion about what "better" means in this context.
1. Miner pushing cart: W = F × d = 200 N × 15 m = 3,000 J.
2. Student climbing rope: F = m × g = 55 × 10 = 550 N. W = 550 N × 4 m = 2,200 J.
3. Crane lifting beam: F = 2,000 × 10 = 20,000 N. W = 20,000 × 12 = 240,000 J (240 kJ). P = 240,000 ÷ 8 = 30,000 W (30 kW).
4. Light bulb energy: Time = 5 hours = 5 × 3,600 = 18,000 s. Energy in joules = P × t = 60 × 18,000 = 1,080,000 J (1.08 MJ). Energy in kWh = 0.060 kW × 5 h = 0.3 kWh.
Experienced shearer: time = 2 min = 120 s. Energy = P × t = 200 × 120 = 24,000 J [0.5]. Power = 200 W (given) [0.5].
Beginner shearer: time = 6 min = 360 s. Energy = 200 × 360 = 72,000 J [0.5]. Power = 200 W (same machine) [0.5].
Both use the same power (200 W) because it is the same machine. But the beginner takes longer, so uses more total energy [1 mark]. The experienced shearer is more "productive" — they do the same job (shear one sheep) using less time and therefore less total energy. The machine's power is fixed; the human's efficiency varies [1 mark].
Cost: Energy in kWh = 24,000 ÷ 3,600,000 = 0.0067 kWh. Cost = 0.0067 × $0.30 = $0.002 (experienced) [0.5]. Beginner: 72,000 ÷ 3,600,000 = 0.02 kWh. Cost = 0.02 × $0.30 = $0.006 [0.5]. The cost difference is small per sheep but significant over a day shearing hundreds of sheep.
1. C — No work is done because the box does not move. W = F × d = F × 0 = 0 J.
2. B — F = 500 × 10 = 5,000 N. W = 5,000 × 8 = 40,000 J. P = 40,000 ÷ 4 = 10,000 W.
3. A — Same work (same force, same distance), but Student A takes half the time so has twice the power. Option B confuses time with work. Option C confuses speed with work. Option D ignores time.
4. D — Energy = P × t. 2,000 W × 180 s = 360,000 J. 1,000 W × 360 s = 360,000 J. Same energy, different rates. Option A doubles incorrectly. Option B doubles incorrectly. Option C introduces irrelevant efficiency.
5. B — Energy input = 200 MJ ÷ 0.35 = 571 MJ (approx). Time = 200,000,000 ÷ 2,500,000 = 80 s. Option A ignores efficiency. Option C uses wrong time calculation. Option D miscalculates energy.
Q6 (3 marks): F = 1,200 × 10 = 12,000 N [1 mark]. W = 12,000 × 250 = 3,000,000 J (3 MJ) [1 mark]. P = 3,000,000 ÷ 45 = 66,667 W (≈ 67 kW) [1 mark].
Q7 (4 marks): Unit A time = 18,000,000 J ÷ 2,500 W = 7,200 s = 2 hours [0.5]. Unit B time = 18,000,000 ÷ 5,000 = 3,600 s = 1 hour [0.5]. Unit A energy per hour = 2.5 kW × 1 h = 2.5 kWh [0.5]. Unit B energy per hour = 5 kW × 1 h = 5 kWh [0.5]. Unit B removes heat faster (1 hour vs 2 hours) but uses twice as much electricity per hour [1 mark]. In Darwin's climate with a long hot season, operating costs matter. If the house needs cooling 8 hours per day for 200 days: Unit A costs 2.5 × 8 × 200 × $0.30 = $1,200/year. Unit B costs 5 × 8 × 200 × $0.30 = $2,400/year. The $400 upfront saving of Unit A is recovered in less than 4 months [1 mark]. Recommendation depends on priorities: Unit B for faster cooling comfort, Unit A for long-term cost savings.
Q8 (5 marks): Diesel energy input = 200 MJ ÷ 0.35 = 571 MJ (≈ 571,000,000 J) [1 mark]. Electric energy input = 200 MJ ÷ 0.85 = 235 MJ (≈ 235,000,000 J) [1 mark]. Diesel time = 200,000,000 ÷ 2,500,000 = 80 s. Electric time = 200,000,000 ÷ 2,000,000 = 100 s [1 mark]. The claim is flawed because "more powerful" only means faster work, not better overall performance. The electric truck uses less than half the energy (235 MJ vs 571 MJ) to do the same job [1 mark]. Conclusion: "Better" depends on priorities. If speed matters most, the diesel truck is better (80 s vs 100 s). If energy cost and environmental impact matter most, the electric truck is dramatically better. The mining company's claim oversimplifies by focusing only on power, ignoring efficiency and total energy use [1 mark].
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Tick when you can calculate work, power and energy, and compare efficiency across different systems.