Moderate -0.3 This is a straightforward application of P=Fv at maximum speed where driving force equals total resistance. Students must convert units (km/h to m/s), sum resistances (2400 + 8×300 = 4800N), then calculate power. It's slightly below average difficulty as it's a standard textbook exercise requiring only one key insight (equilibrium at max speed) with routine calculations.
5 A train consisting of an engine and eight carriages moves on a straight horizontal track.
A constant resistive force of 2400 N acts on the engine.
A constant resistive force of 300 N acts on each of the eight carriages.
The maximum speed of the train on the track is \(120 \mathrm {~km} \mathrm {~h} ^ { - 1 }\)
Find the maximum power output of the engine.
Fully justify your answer.
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| For maximum power, train travels at maximum speed when driving force equals resistance $R = F$ | E1 | Explains that maximum power occurs at maximum speed when driving force equals resistance |
| $F = 2400 + 8(300) = 4800$ N | B1 | Correct driving force or resistive force of 4800 N |
| $P = Fv$ | M1 | Translates problem into equation modelling power as $Fv$ |
| $v = \frac{120 \times 1000}{60 \times 60} = \frac{100}{3}$ ms$^{-1}$ | B1 | Converts km h$^{-1}$ to m s$^{-1}$ correctly |
| $P = 4800 \times \frac{100}{3} = 160000$, Max $P = 160$ kW | A1 | Must state units |
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5 A train consisting of an engine and eight carriages moves on a straight horizontal track.
A constant resistive force of 2400 N acts on the engine.\\
A constant resistive force of 300 N acts on each of the eight carriages.\\
The maximum speed of the train on the track is $120 \mathrm {~km} \mathrm {~h} ^ { - 1 }$\\
Find the maximum power output of the engine.\\
Fully justify your answer.\\
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2020 Q5 [5]}}