| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a straightforward M2 work-energy-power question requiring standard application of P=Fv and F=ma. Part (a) uses Newton's second law with power to find resistance (routine calculation). Part (c) applies equilibrium on an incline with power. Both parts follow standard textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
A car of mass 1000 kg is moving along a straight horizontal road. The resistance to motion is modelled as a constant force of magnitude $R$ newtons. The engine of the car is working at a rate of 12 kW. When the car is moving with speed 15 m s$^{-1}$, the acceleration of the car is 0.2 m s$^{-2}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $R = 600$. [4]
\end{enumerate}
The car now moves with constant speed $U$ m s$^{-1}$ downhill on a straight road inclined at $\theta$ to the horizontal, where $\sin \theta = \frac{1}{30}$. The engine of the car is now working at a rate of 7 kW. The resistance to motion from non-gravitational forces remains of magnitude $R$ newtons.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Calculate the value of $U$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2006 Q3 [9]}}