| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on horizontal road |
| Difficulty | Standard +0.3 This is a multi-part mechanics question requiring friction calculations, work-energy principles, and kinematic reasoning. While it involves several steps and concepts (resolving forces, friction at limiting equilibrium, work done, power, and energy conservation), each part uses standard M2 techniques without requiring novel insight. The 'show that' parts guide students through the solution, making it slightly easier than average for an M2 question. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let friction be \(F\) N and normal reaction \(R\) N | ||
| \(F_{\max} = 58\cos 35\) | B1 | Need not be explicit |
| \(R = 16g + 58\sin 35\) | M1 | Both terms required |
| A1 | ||
| \(F_{\max} = \mu R\) | M1 | |
| so \(\mu = 0.249968\ldots\) about \(0.25\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| WD is \(70\cos 35 \times 3 = 210\cos 35\) | M1 | Use of \(WD = Fd\). Accept \(\cos 35\) omitted. |
| so \(172.0219\ldots = 172\) J (3 s.f.) | A1 | |
| Average power is WD/time | M1 | |
| so \(34.4043\ldots = 34.4\) W (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using constant acceleration result \(s = \frac{1}{2}(u+v)t\) with \(s=3\), \(u=0\), \(v=1.5\) and \(t=5\) we see that \(3 \neq \frac{1}{2}(0+1.5)\times 5 = 3.75\) | M1 | Attempt to substitute in suvat (sequence) |
| E1 | Conclusion clear |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(172.0219\ldots\) | M1 | Using W-E equation, allow 1 missing term |
| \(= \frac{1}{2} \times 16 \times 1.5^2\) | M1 | KE term attempted |
| A1 | correct | |
| \(+ 0.25 \times (16g + 70\sin 35) \times 3\) | M1 | Attempt at using new \(F\) in \(F_{\max} = \mu R\) |
| \(+\) WD | A1 | All correct |
| so WD by \(S\) is \(6.30916\ldots\) so \(6.31\) J (3 s.f.) | A1 | cao |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let friction be $F$ N and normal reaction $R$ N | | |
| $F_{\max} = 58\cos 35$ | B1 | Need not be explicit |
| $R = 16g + 58\sin 35$ | M1 | Both terms required |
| | A1 | |
| $F_{\max} = \mu R$ | M1 | |
| so $\mu = 0.249968\ldots$ about $0.25$ | E1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| WD is $70\cos 35 \times 3 = 210\cos 35$ | M1 | Use of $WD = Fd$. Accept $\cos 35$ omitted. |
| so $172.0219\ldots = 172$ J (3 s.f.) | A1 | |
| Average power is WD/time | M1 | |
| so $34.4043\ldots = 34.4$ W (3 s.f.) | A1 | cao |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using constant acceleration result $s = \frac{1}{2}(u+v)t$ with $s=3$, $u=0$, $v=1.5$ and $t=5$ we see that $3 \neq \frac{1}{2}(0+1.5)\times 5 = 3.75$ | M1 | Attempt to substitute in suvat (sequence) |
| | E1 | Conclusion clear |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $172.0219\ldots$ | M1 | Using W-E equation, allow 1 missing term |
| $= \frac{1}{2} \times 16 \times 1.5^2$ | M1 | KE term attempted |
| | A1 | correct |
| $+ 0.25 \times (16g + 70\sin 35) \times 3$ | M1 | Attempt at using new $F$ in $F_{\max} = \mu R$ |
| $+$ WD | A1 | All correct |
| so WD by $S$ is $6.30916\ldots$ so $6.31$ J (3 s.f.) | A1 | cao |4 A box of mass 16 kg is on a uniformly rough horizontal floor with an applied force of fixed direction but varying magnitude $P$ N acting as shown in Fig. 4. You may assume that the box does not tip for any value of $P$. The coefficient of friction between the box and the floor is $\mu$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1a605f0b-f595-4bb9-9624-f816c789ad86-5_348_863_429_643}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
Initially the box is at rest and on the point of slipping with $P = 58$.\\
(i) Show that $\mu$ is about 0.25 .
In the rest of this question take $\mu$ to be exactly 0.25 .\\
The applied force on the box is suddenly increased so that $P = 70$ and the box moves against friction with the floor and another horizontal retarding force, $S$. The box reaches a speed of $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from rest after 5 seconds; during this time the box slides 3 m .\\
(ii) Calculate the work done by the applied force of 70 N and also the average power developed by this force over the 5 seconds.\\
(iii) By considering the values of time, distance and speed, show that an object starting from rest that travels 3 m while reaching a speed of $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 5 seconds cannot be moving with constant acceleration.
The reason that the acceleration is not constant is that the retarding force $S$ is not constant.\\
(iv) Calculate the total work done by the retarding force $S$.
\hfill \mbox{\textit{OCR MEI M2 2010 Q4 [18]}}