Physics • Year 11 • Module 2 • Lesson 8

Power

Apply your understanding of P = ΔE/Δt, P = Fv and P = mgh/Δt to real data, error-spotting, and cause-and-effect reasoning.

Apply · Data & Reasoning

1. Interpret power data — comparing machines and athletes

The table below lists six scenarios. Complete the missing cells by selecting the correct formula, calculating power, and converting to appropriate units. 9 marks (1.5 per row)

# Scenario Given values Formula Power output
1 Athlete sprints up stairs m = 70 kg, h = 5 m, Δt = 4 s
2 Walker climbs same stairs m = 70 kg, h = 5 m, Δt = 40 s
3 Car at constant highway speed v = 30 m/s, Fresist = 800 N
4 Electric motor transfers energy ΔE = 90 000 J, Δt = 45 s
5 Cyclist pedalling at constant speed v = 8 m/s, Fdrag = 120 N
6 Construction crane lifting load m = 500 kg, h = 20 m, Δt = 40 s

1.1 Compare the athlete (row 1) and the walker (row 2). State how their work done and power outputs compare, and explain the physics reason for the difference. 2 marks

1.2 Rows 3 and 5 both use P = Fv. Identify the key reasoning step required before applying this formula in row 3 but not needed in row 5. Explain why. 2 marks

Stuck? Use the Decision Flowchart from the lesson. For rows 3 and 5, Newton’s First Law is the key step when velocity is constant.

2. Interpret graph — engine power versus vehicle speed

The graph below shows the power required to maintain constant speed against a constant resistance force of 800 N for a vehicle. 7 marks

0 10 20 30 40 50 0 10 20 30 40 50 Speed (m/s) Power required (kW) 24 kW at 30 m/s

Figure 2.1 — Power required (kW) to maintain constant speed (m/s) against a constant resistance of 800 N. Relationship: P = Fresist × v. Illustrative data.

2.1 Describe the relationship between power and speed shown in the graph. Is it linear or non-linear? Explain why, using the formula P = Fv. 2 marks

2.2 Use the graph to estimate the power required at 35 m/s. Show your reasoning. 1 mark

2.3 If the same vehicle doubles its highway speed from 25 m/s to 50 m/s while keeping resistance constant at 800 N, by what factor does the required power increase? Use the graph or the formula to justify your answer. 2 marks

2.4 Explain why highway driving consumes more fuel per kilometre than city driving, using power and speed concepts from this graph. 2 marks

Stuck? Revisit Card 2 (P = Fv) and the Power vs Speed table in the lesson. For 2.3, read off values from the graph at 25 m/s and 50 m/s.

3. Cause-and-effect chain — a cyclist crouches to reduce drag

A professional cyclist competes at constant speed but changes posture partway through a race. Trace the chain of physical reasoning. Fill in the empty effect boxes. 5 marks (1 per effect + 1 for final outcome)

Cyclist crouches — frontal area decreases
→
Effect 1:
Air resistance force (Fdrag) decreases
→
Effect 2:
At constant v, Fdrive needed = Fdrag (by Newton 1)
→
Effect 3:
P = Fdrive × v; Fdrive has decreased; v unchanged
→
Effect 4:

Overall outcome (so…):

Stuck? Think about what happens to Fdrag when drag decreases, then use P = Fv to trace what that means for power.

4. Predict and justify — a hydroelectric dam scenario

The Tumut 3 Power Station in Snowy Hydro (NSW) uses water falling from a height of 150 m to generate electricity. In one second, approximately 130 kg of water passes through each turbine. 5 marks

4.1 Calculate the gravitational power input to one turbine from the falling water. State your formula, substitutions and units clearly. 2 marks

4.2 In reality, the electrical power output of each turbine is about 160 MW. Discuss whether your calculated value is consistent with this. What accounts for any difference? 2 marks

4.3 If engineers doubled the height of the water drop while keeping flow rate constant, predict what would happen to the power available. Justify your prediction using the formula. 1 mark

Stuck? For 4.1, use P = mgh/Δt. For 4.2, consider efficiency losses. For 4.3, P ∝ h when m/Δt is constant.
Answers — Do not peek before attempting

Q1 — Power data table

Row 1 (athlete): P = mgh/Δt = 70×9.8×5/4 = 3430/4 = 857.5 W.

Row 2 (walker): P = mgh/Δt = 70×9.8×5/40 = 3430/40 = 85.75 W.

Row 3 (car, constant v): Fdrive = Fresist = 800 N (Newton 1). P = Fv = 800×30 = 24 000 W = 24 kW.

Row 4 (motor): P = ΔE/Δt = 90 000/45 = 2 000 W = 2 kW.

Row 5 (cyclist): Fdrive = Fdrag = 120 N. P = Fv = 120×8 = 960 W.

Row 6 (crane): P = mgh/Δt = 500×9.8×20/40 = 98 000/40 = 2 450 W = 2.45 kW.

Q1.1 — Athlete vs walker

Both do the same work: W = mgh = 70×9.8×5 = 3430 J [1]. The athlete’s power (857.5 W) is 10× greater than the walker’s (85.75 W) because power = work/time, and the athlete completes the same task 10× faster. Work depends on force and displacement only; power depends on how quickly the work is done [1].

Q1.2 — Key step for P = Fv

For row 3 (car), the velocity is described as constant, so Newton’s First Law must be applied first: Fdrive = Fresist = 800 N. The problem gives resistance, not driving force [1]. For row 5 (cyclist), the 120 N is already the drag force, and at constant speed Fdrive = Fdrag; but the problem also states “constant speed” so the same Newton 1 step applies — the difference is that the resistance force given directly equals the driving force in both cases. The key reasoning step is always: constant v → Newton 1 → Fdrive = Fresist before substituting into P = Fv [1].

Q2.1 — Graph trend

The graph shows a linear (straight-line) relationship between power and speed [1]. This is because P = Fv and Fresist is constant (800 N), so P increases in direct proportion to v: doubling the speed exactly doubles the power required [1].

Q2.2 — Power at 35 m/s

Reading the graph between 30 m/s (24 kW) and 40 m/s (32 kW): at 35 m/s, P ≈ 28 kW. By formula: P = 800×35 = 28 000 W = 28 kW [1].

Q2.3 — Doubling speed factor

At 25 m/s: P = 800×25 = 20 kW. At 50 m/s: P = 800×50 = 40 kW. Power doubles [1]. Because P = Fv and F is constant, doubling v doubles P — a factor of 2 increase [1].

Q2.4 — Highway vs city fuel use

At highway speed (e.g. 30 m/s) the engine must supply 24 kW; at city speed (e.g. 15 m/s) it needs only 12 kW [1]. Higher power means energy is being used at a faster rate. Over the same distance, highway travel requires proportionally more energy per second, so more fuel is consumed per kilometre [1].

Q3 — Cause-and-effect chain

Effect 1: Air resistance (drag force Fdrag) decreases, because drag depends on frontal area. Effect 2: At constant velocity, Fdrive required decreases (Newton 1: Fdrive = Fdrag). Effect 3: The driving force the cyclist needs to apply decreases. Effect 4: Power output decreases: P = Fdrive × v is smaller because Fdrive is smaller at the same speed. Overall outcome: By crouching, the cyclist uses less power to maintain the same speed — meaning the same power output can sustain a higher speed, giving a competitive advantage.

Q4.1 — Tumut 3 gravitational power

P = mgh/Δt. m/Δt = 130 kg/s. P = (130 × 9.8 × 150) / 1 = 130 × 1470 = 191 100 W ≈ 191 kW per kilogram-second. With 130 kg/s: P = 191 100 W = 191.1 kW (per turbine at this flow rate). [Award marks for correct formula, correct substitution, correct unit.]

Note to marker: 130 kg/s × 9.8 m/s² × 150 m = 191 100 W ≈ 191 kW. This is gravitational power input per turbine at the given flow rate.

Q4.2 — Comparison with 160 MW

The calculated 191 kW per turbine is far less than 160 MW. The discrepancy arises because real power stations pass far more water per second (thousands of kg/s, not 130 kg/s) through large turbines [1]. The 130 kg/s figure given was a simplification. Additionally, turbines have efficiencies less than 100% — mechanical friction, heat, and electrical conversion losses mean not all gravitational PE becomes electrical energy [1].

Q4.3 — Doubling height

P = mgh/Δt. If h doubles and m/Δt is constant, P doubles [1]. The power available is directly proportional to the height of the water drop.

HSCScience • Physics • Year 11 • Module 2 • Lesson 8 • Worksheet 2 (Apply)