| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force up incline, find speed |
| Difficulty | Standard +0.3 Part (a) is a standard 'show that' proof requiring resolution of forces and applying limiting equilibrium (F=μR), which is routine bookwork. Part (b) applies the work-energy principle to find an unknown force—a straightforward application requiring calculation of work done by each force and change in kinetic energy. Both parts are standard textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 3.03r Friction: concept and vector form6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | R=mgcosα,F =mgsinα |
| Answer | Marks |
|---|---|
| So µ=tanα | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.1 |
| 2.2a | Attempt to resolve perpendicular and parallel to the |
| Answer | Marks |
|---|---|
| (b) | 1 |
| Answer | Marks |
|---|---|
| P=23.1 | B1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | 52.5 |
Question 6:
6 | (a) | R=mgcosα,F =mgsinα
Point of slipping ⇒mgsinα=mgµcosα
So µ=tanα | M1
A1
[2] | 2.1
2.2a | Attempt to resolve perpendicular and parallel to the
plane. Allow sin/cos confusion only. Missing g is
M0. Allow W for mg. Stating mgsinα=mgµcosα
can imply this mark
AG – must state or clearly imply that F =µR
(b) | 1
Change in KE: ± (5)(52 −22)
2
Change in GPE: ±5g(10sin15)
Work done by pulling force is 10(Pcos25)
1 1
(5)(5)2 − (5)(2)2 =10(Pcos25)−10(3)−5g(10sin15)
2 2
P=23.1 | B1
B1
B1
M1
A1
[5] | 1.1
1.1
1.1
3.3
1.1 | 52.5
126.8213321…
Correct expression for the work done by P
(9.063077…)P
Use of work-energy principle – correct number of
terms – allow sign errors (and sin/cos confusion) but
must be dimensionally consistent
23.0960535…In this question the box should be modelled as a particle.
A box of mass $m$ kg is placed on a rough slope which makes an angle of $\alpha$ with the horizontal.
\begin{enumerate}[label=(\alph*)]
\item Show that the box is on the point of slipping if $\mu = \tan \alpha$, where $\mu$ is the coefficient of friction between the box and the slope. [2]
\end{enumerate}
A box of mass 5 kg is pulled up a rough slope which makes an angle of 15° with the horizontal. The box is subject to a constant frictional force of magnitude 3 N. The speed of the box increases from 2 m s$^{-1}$ at a point A on the slope to 5 m s$^{-1}$ at a point B on the slope with B higher up the slope than A. The distance AB is 10 m.
\includegraphics{figure_6}
The pulling force has constant magnitude $P$ N and acts at a constant angle of 25° above the slope, as shown in the diagram.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Use the work–energy principle to determine the value of $P$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q6 [7]}}