| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Standard +0.3 This is a standard M2 work-energy question with two straightforward parts: (a) calculating work done against gravity and friction up an incline using W = (mg sin θ + F)d, and (b) applying work-energy principle for motion down the slope. Both parts use routine mechanics formulas with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Work done against friction \(= 50\times\mu R = 50\times\frac{1}{4}\times 30\cos 20°\times 9.8\) | M1, A1 | |
| Gain in GPE \(= 30\times 9.8\times 50\sin 20°\) | M1 A1 | |
| Total work done \(=\) WD against friction \(+\) gain in GPE \(= 8480\text{ (J)},\ 8500\text{ (J)}\) | DM1, A1 | |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Loss in GPE \(=\) WD against friction \(+\) gain in KE | M1 | 3 terms |
| \(30\times9.8\times50\sin20°= 50\times\frac{1}{4}\times30\times9.8\times\cos20°+\frac{1}{2}\times30\times v^2\) | A2,1,0 | \(-1\) ee |
| \(\frac{1}{2}v^2 = 50\times9.8\times(\sin20°-\frac{1}{4}\cos20°)\) | DM1 | |
| \(v = 10.2 \text{ m s}^{-1}\) | A1 | |
| (5) [11] |
## Question 4:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Work done against friction $= 50\times\mu R = 50\times\frac{1}{4}\times 30\cos 20°\times 9.8$ | M1, A1 | |
| Gain in GPE $= 30\times 9.8\times 50\sin 20°$ | M1 A1 | |
| Total work done $=$ WD against friction $+$ gain in GPE $= 8480\text{ (J)},\ 8500\text{ (J)}$ | DM1, A1 | |
| | **(6)** | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Loss in GPE $=$ WD against friction $+$ gain in KE | M1 | 3 terms |
| $30\times9.8\times50\sin20°= 50\times\frac{1}{4}\times30\times9.8\times\cos20°+\frac{1}{2}\times30\times v^2$ | A2,1,0 | $-1$ ee |
| $\frac{1}{2}v^2 = 50\times9.8\times(\sin20°-\frac{1}{4}\cos20°)$ | DM1 | |
| $v = 10.2 \text{ m s}^{-1}$ | A1 | |
| | **(5) [11]** | |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c5760fa5-3c7f-4e29-87a2-b3b4145b9361-06_365_776_264_584}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A box of mass 30 kg is held at rest at point $A$ on a rough inclined plane. The plane is inclined at $20 ^ { \circ }$ to the horizontal. Point $B$ is 50 m from $A$ up a line of greatest slope of the plane, as shown in Figure 1. The box is dragged from $A$ to $B$ by a force acting parallel to $A B$ and then held at rest at $B$. The coefficient of friction between the box and the plane is $\frac { 1 } { 4 }$. Friction is the only non-gravitational resistive force acting on the box. Modelling the box as a particle,
\begin{enumerate}[label=(\alph*)]
\item find the work done in dragging the box from $A$ to $B$.
The box is released from rest at the point $B$ and slides down the slope. Using the workenergy principle, or otherwise,
\item find the speed of the box as it reaches $A$.\\
January 2011
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2011 Q4 [11]}}