| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Maximum speed on incline vs horizontal |
| Difficulty | Standard +0.3 This is a straightforward M2 power question requiring application of P=Fv and resolving forces at constant speed. Part (a) uses standard inclined plane resolution with sin(10°), and part (b) is a direct application of the same power on horizontal ground. The multi-step nature and need to work with power equations elevates it slightly above average, but it follows a standard template with no conceptual surprises. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P = 15(35000 + 20000g\sin 10°) = 1035525.6\) W \(\approx 1040\) kW | M1 M1 A1 A1 | |
| (b) \(1035525.6 = v \times 35000\) | M1 A1 | |
| \(v = 29.6\) m/s\(^{-2}\) | M1 A1 | Total: 7 marks |
**(a)** $P = 15(35000 + 20000g\sin 10°) = 1035525.6$ W $\approx 1040$ kW | M1 M1 A1 A1 |
**(b)** $1035525.6 = v \times 35000$ | M1 A1 |
$v = 29.6$ m/s$^{-2}$ | M1 A1 | **Total: 7 marks**An engine of mass $20\,000$ kg climbs a hill inclined at $10°$ to the horizontal. The total non-gravitational resistance to its motion has magnitude $35\,000$ N and the maximum speed of the engine on the hill is $15$ ms$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find, in kW, the maximum rate at which the engine can work. [4 marks]
\item Find the maximum speed of the engine when it is travelling on a horizontal track against the same non-gravitational resistance as before. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [7]}}