Physics • Year 11 • Module 2 • Lesson 8
Power
Build HSC Band 5–6 extended-response technique by evaluating power data, designing an investigation and synthesising multi-step quantitative arguments.
1. Multi-step problem — car engine and reserve power (Band 4–5)
8 marks Band 4–5
Scenario. A 1500 kg car has a maximum engine power output of 120 kW. At a constant highway speed of 108 km/h on a flat road, the total resistive force acting on the car is 1000 N. Use g = 9.8 m/s2 throughout.
(a) Convert 108 km/h to m/s and calculate the engine’s power output at this constant speed. (2 marks)
(b) Calculate the reserve power available at this speed and explain what reserve power means physically. (2 marks)
(c) If the driver uses full engine power (120 kW) at 108 km/h, calculate the new driving force and the net force. Using Newton’s Second Law, explain what happens to the car’s motion. (3 marks)
(d) State one assumption made in parts (a)–(c) and explain how removing that assumption would affect your answer. (1 mark)
2. Data + scenario — comparing power outputs of athletes (Band 5–6)
8 marks Band 5–6
Scenario. Two athletes, Aisha and Ben, each have a mass of 70 kg and compete in different sports. The table below summarises their performance data during two different physical tasks.
| Measurement | Aisha (sprint cyclist) | Ben (marathon runner) |
|---|---|---|
| Task | Sprint up a 6 m staircase | Climb same 6 m staircase |
| Time taken (Δt) | 2.4 s | 18 s |
| Mass (m) | 70 kg | 70 kg |
| Sustained cycling power (separate test) | 350 W for 30 min | 180 W for 30 min |
| Peak sprint stair power (calculated) | to be determined | to be determined |
Illustrative data. g = 9.8 m/s2.
Q2. Analyse and evaluate the data above to compare the power outputs of Aisha and Ben. In your response you must:
- Calculate the peak stair-sprint power for both athletes and show your working.
- Compare Aisha’s peak sprint power with Ben’s, using a ratio and a written explanation.
- Distinguish between peak power (stair sprint) and sustained power (cycling test), explaining why a high peak power does not guarantee a high sustained power.
- Explain why both athletes do the same amount of work on the staircase despite having very different power outputs.
- Evaluate the statement: “Aisha is a better all-round athlete than Ben because she is more powerful.” Assess whether power alone is a fair measure of overall athletic ability.
3. Experimental design — measuring the power of a small electric motor (Band 5–6)
7 marks Band 5–6
Research question. A student claims: “My small electric motor can lift a 200 g mass to a height of 0.5 m in about 3 seconds.” Design a scientific investigation to determine the mechanical power output of the motor as accurately as possible.
Constraints: You have access to a ruler (millimetre precision), a digital balance (±0.1 g), a digital stopwatch (±0.01 s), a selection of masses from 50 g to 500 g, a metre ruler, and a string-and-pulley arrangement. You have 40 minutes.
Q3. Design the investigation in the format below.
- State your hypothesis (a testable prediction linking mass, height and time to power output).
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least four numbered steps, including how you will improve accuracy of timing.
- State what result would falsify your hypothesis.
- Identify two sources of uncertainty or error in this design and propose one improvement.
Q1 — Car engine sample answers
(a) 108 km/h ÷ 3.6 = 30 m/s [1]. At constant v, Newton 1: Fdrive = Fresist = 1000 N. P = Fdrive × v = 1000 × 30 = 30 000 W = 30 kW [1].
(b) Reserve power = 120 kW − 30 kW = 90 kW [1]. Reserve power is the additional power available beyond what is needed to maintain constant speed; it can be used to accelerate, climb a hill, or carry a payload [1].
(c) New driving force at full power: Fdrive_new = Pmax/v = 120 000/30 = 4000 N [1]. Net force = Fdrive_new − Fresist = 4000 − 1000 = 3000 N forward [1]. By Newton 2: a = Fnet/m = 3000/1500 = 2 m/s2 forward. The car accelerates — it is no longer at constant velocity; it speeds up until the driver reduces power or a higher speed is reached where the new resistance equals 4000 N [1].
(d) Assumption: resistance force remains constant at 1000 N regardless of speed. In reality, air resistance increases with speed (approximately proportional to v2), so at higher speeds the resistance force would be larger, the net force and acceleration would be smaller, and the car would reach a new equilibrium speed at a lower power output than Pmax [1].
Marking criteria (8 marks): 1 = correct unit conversion (108 km/h = 30 m/s); 1 = correct Pcruise = 30 kW using Newton 1; 1 = correct reserve power = 90 kW; 1 = correct physical meaning of reserve power; 1 = correct new driving force (4000 N); 1 = correct net force (3000 N); 1 = correct acceleration with Newton 2 and correct physical interpretation (car accelerates); 1 = valid assumption stated with a physically correct consequence of removing it.
Q2 — Sample Band 6 response (8 marks), annotated
Peak stair power: Both athletes: W = mgh = 70 × 9.8 × 6 = 4116 J. Aisha: P = 4116/2.4 = 1715 W. Ben: P = 4116/18 = 228.7 W [1 — correct calculations with formula shown].
Comparison ratio: PAisha/PBen = 1715/228.7 ≈ 7.5. Aisha’s peak stair power is approximately 7.5 times greater than Ben’s. This is because she completes the same task 7.5 times faster [1].
Peak vs sustained power distinction: Peak power is the maximum rate of energy output a person can achieve over a very short burst (seconds); it relies on fast-twitch muscle fibres and anaerobic metabolism. Sustained power is the rate a person can maintain over a long period (30 minutes), relying on aerobic metabolism. High peak power does not guarantee high sustained power because the energy systems involved are different. Aisha’s sprinting ability does not mean she can deliver 1715 W for 30 minutes [1 — clear distinction with physiological or energy-system reasoning; 1 — applied correctly to the data].
Same work, different power: Both athletes climb 6 m at 70 kg, so W = mgh = 4116 J in both cases [1]. Work depends only on force and displacement, not on time. Power measures how fast the work is done. Aisha does the same work faster, so her power is higher — but neither athlete does more or less work than the other [1].
Evaluation of the claim: The statement is not well supported by the data. Power is only one dimension of athletic ability. Ben’s sustained cycling power (180 W for 30 minutes) is more relevant to endurance performance than Aisha’s peak sprint power. For a marathon, high sustained power over hours matters more than a 2.4 s peak burst. The claim conflates peak power with overall athletic ability, ignoring endurance capacity, technique, flexibility, and sport-specific skills [1 — clear evaluative judgement; 1 — uses evidence from the data to challenge the claim].
Marking criteria (8 marks): 1 = correct PAisha and PBen with formula; 1 = correct ratio (~7.5) and written comparison; 1 = clear distinction between peak and sustained power with reasoning; 1 = applies distinction correctly to the athlete data; 1 = states work = mgh is the same for both; 1 = explains why same work yet different power (time differs); 1 = makes a clear evaluative judgement about the claim; 1 = uses specific data to support the judgement.
Q3 — Sample Band 6 response (7 marks), annotated
Hypothesis: If the motor lifts a mass m a height h in time t, the mechanical power output P = mgh/t; for m = 0.200 kg, h = 0.5 m, t ≈ 3 s, P ≈ 0.2 × 9.8 × 0.5 / 3 ≈ 0.33 W. IV: mass lifted (50–200 g). DV: mechanical power P = mgh/t. Controlled: height (0.5 m), motor (same unit), string length, ambient conditions [1].
Procedure: (1) Attach a 50 g mass to the motor shaft via string and pulley. Measure the lift height h = 0.500 m using a metre ruler, marking start and end positions with tape. (2) Start the motor and simultaneously start the stopwatch; stop both when the mass reaches the end position. Record time t. (3) Calculate P = mgh/t for that mass. (4) Repeat for masses of 100 g, 150 g, 200 g and 250 g, performing 5 trials at each mass to reduce random timing error. Discard outliers and average the remaining values. Record all results in a table [1].
Falsification: If the power output varies systematically and significantly with mass (e.g. doubles as mass doubles, or drops to near zero at higher masses), this would falsify the hypothesis that the motor delivers approximately constant power output; it would indicate the motor is unable to run at constant power under load [1].
Error 1 — Reaction time in manual timing: Starting and stopping the stopwatch manually introduces ±0.2 s reaction-time error per trial, which is significant relative to a ~3 s lift. [1] Error 2 — Friction in pulley: The pulley introduces friction, meaning some of the motor’s power does not go into lifting the mass; the measured P = mgh/t underestimates the true motor power output [1].
Improvement: Replace manual timing with a photogate at the start and end positions to record time electronically, reducing timing uncertainty to <0.001 s; this would allow much more accurate P calculations, especially for short lift times [1].
Marking criteria (7 marks): 1 = testable hypothesis with IV and DV stated; 1 = four clear procedure steps including a repeat/average strategy; 1 = valid falsification condition; 1 = first named source of error with physical explanation; 1 = second distinct source of error; 1 = one specific improvement with justification; 1 = uses precise physics terminology throughout (P = mgh/t, independent/dependent/controlled variable, power, work, watts).