| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find power at constant speed |
| Difficulty | Moderate -0.8 This is a straightforward application of the power formula P=Fv at constant speed on an incline. Students resolve forces parallel to the slope (weight component minus resistance equals driving force since speed is constant), then multiply by velocity. It requires only standard mechanics techniques with no problem-solving insight, making it easier than average. |
| Spec | 3.03f Weight: W=mg3.03g Gravitational acceleration6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Resolve \(\curvearrowright\): \(T_r + 2000g \times \sin \alpha = 1600\) | M1 A1 A1 | (\(T_r = 816\)) |
| \(P = 816 \times 14\) (W) \(\approx 11\) (kW) | M1 A1 ft | accept 11.4 |
| A1 cso | [6] |
**Part (a):**
Resolve $\curvearrowright$: $T_r + 2000g \times \sin \alpha = 1600$ | M1 A1 A1 | ($T_r = 816$)
$P = 816 \times 14$ (W) $\approx 11$ (kW) | M1 A1 ft | accept 11.4
| A1 cso | [6]\begin{enumerate}
\item A lorry of mass 2000 kg is moving down a straight road inclined at angle $\alpha$ to the horizontal, where $\sin \alpha = \frac { 1 } { 25 }$. The resistance to motion is modelled as a constant force of magnitude 1600 N . The lorry is moving at a constant speed of $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
Find, in kW , the rate at which the lorry's engine is working.\\
\hfill \mbox{\textit{Edexcel M2 2008 Q1 [6]}}