A-Level Physics May/June 2025 Q1(b)(iii): Between time t = 0 and time t = 5.8s, the work done against resistive forces is 4.7 x 1…

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  The total work done by the engine is the sum of the work done against resistive forces and the gain in kinetic energy of the car.

  From part (b)(ii), the gain in kinetic energy is $1.3 \times 10^5$ J.

  Total work done by engine, $W_{total} = W_{resistive} + \Delta E_k$ $W_{total} = (4.7 \times 10^4 \text{ J}) + (1.3 \times 10^5 \text{ J})$ $W_{total} = 1.77 \times 10^5 \text{ J}$

  Average output power is the total work done divided by the time taken. $P = \frac{W_{total}}{t}$ $P = \frac{1.77 \times 10^5}{5.8}$ $P = 30517.2... \text{ W}$

  $P \approx 3.1 \times 10^4 \text{ W}$ (to 2 s.f.)

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  How the marks are awarded

  • C1 — Correctly stating the formula for average power, $P = W/t$.
  • C1 — Correctly calculating the total work done by summing the work against resistance and the gain in kinetic energy ($4.7 \times 10^4 + 1.3 \times 10^5$) and substituting this total work and the time into the power formula.
  • A1 — Obtaining the final correct answer of $3.1 \times 10^4$ W, correctly rounded to two significant figures.

  Common mistakes

  • Forgetting to include the gain in kinetic energy, and only calculating power using the work done against resistive forces (i.e., $4.7 \times 10^4 / 5.8$).
  • Forgetting to include the work done against resistive forces, and only calculating power using the gain in kinetic energy from the previous part (i.e., $1.3 \times 10^5 / 5.8$).
  • Incorrectly rounding the final answer or giving it to an inappropriate number of significant figures, such as 30500 W or 30517 W.
  • Making an arithmetic error when summing the two work components before dividing by time.

  Examiner tip: Always account for all energy transfers; the total work done by an engine must equal the sum of the gain in kinetic energy and the work done against resistive forces.

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