| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find power at constant speed |
| Difficulty | Easy -1.2 This is a straightforward application of the power formula P = Fv at maximum speed where driving force equals total resistance. Part (a) is trivial unit conversion, and part (b) requires only summing resistance forces (3000 + 5×400 = 5000 N) and multiplying by speed. No problem-solving or conceptual insight needed—purely mechanical calculation following a standard template. |
| Spec | 6.02l Power and velocity: P = Fv |
5 A train consists of an engine and five carriages. A constant resistance force of 3000 N acts on the engine, and a constant resistance force of 400 N acts on each of the five carriages.
The maximum speed of the train on a horizontal track is $90 \mathrm {~km} \mathrm {~h} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that this speed is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Hence find the maximum power output of the engine. Give your answer in kilowatts.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2011 Q5 [4]}}