| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Lifting objects vertically |
| Difficulty | Standard +0.3 This is a straightforward multi-part work-energy question requiring standard applications of W=Fd, P=Fv, and energy conservation. All parts follow directly from basic principles with clear numerical values given. The 'show that' in part (i) guides students to the correct approach, and no novel problem-solving insight is required—just systematic application of mechanics formulae. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| WD is \(800 \times 9.8 \times 6 + 400 \times 6\) J | M1 | WD as \(Fd\), used in TWO terms |
| \(= 49440\) | E1 | |
| Power is \(49440 \div 12 = 4120 \text{ W}\) | M1, A1 | Power is \(\text{WD}/\Delta t\); cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Power is \((800 \times 9.8 + 400) \times 0.55\) | M1 | Power as \(Fv\) in one term |
| \(= 4532 \text{ W}\) | A1, A1 | All correct; cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 800v^2 = 800 \times 9.8 \times 3 - 400 \times 3\) | M1 | Use of W-E equation. Must include KE and at least one WD term |
| A1 | Allow only sign errors | |
| A1 | All correct | |
| \(v^2 = 55.8\) so \(v = 7.4699\ldots\) | ||
| Speed is \(7.47 \text{ m s}^{-1}\) | A1 | SC: Use of N2L and suvat: M1 complete method, A1 \(7.47\) cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 800 \times \dfrac{v^2}{4} - \frac{1}{2} \times 800 \times v^2 = (800 \times 9.8 - 400) \times 0.8 - \text{WD}\) | M1 | Use of W-E equation. Must include 2 KE terms and a WD term |
| Final KE term correct (FT their \(v\)) | B1 | |
| One correct WD term | B1 | |
| All terms present, allow sign errors and FT their \(v\) | A1 | cao |
| WD is \(22692\) so \(22700 \text{ J}\) | A1 | SC Use of N2L and suvat: Award max B1 for 'Average force \((28365) \times 0.8\)' |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| WD is $800 \times 9.8 \times 6 + 400 \times 6$ J | M1 | WD as $Fd$, used in TWO terms |
| $= 49440$ | E1 | |
| Power is $49440 \div 12 = 4120 \text{ W}$ | M1, A1 | Power is $\text{WD}/\Delta t$; cao |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Power is $(800 \times 9.8 + 400) \times 0.55$ | M1 | Power as $Fv$ in one term |
| $= 4532 \text{ W}$ | A1, A1 | All correct; cao |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 800v^2 = 800 \times 9.8 \times 3 - 400 \times 3$ | M1 | Use of W-E equation. Must include KE and at least one WD term |
| | A1 | Allow only sign errors |
| | A1 | All correct |
| $v^2 = 55.8$ so $v = 7.4699\ldots$ | | |
| Speed is $7.47 \text{ m s}^{-1}$ | A1 | SC: Use of N2L and suvat: M1 complete method, A1 $7.47$ cao |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 800 \times \dfrac{v^2}{4} - \frac{1}{2} \times 800 \times v^2 = (800 \times 9.8 - 400) \times 0.8 - \text{WD}$ | M1 | Use of W-E equation. Must include 2 KE terms and a WD term |
| Final KE term correct (FT their $v$) | B1 | |
| One correct WD term | B1 | |
| All terms present, allow sign errors and FT their $v$ | A1 | cao |
| WD is $22692$ so $22700 \text{ J}$ | A1 | SC Use of N2L and suvat: Award max B1 for 'Average force $(28365) \times 0.8$' |
---2 A fairground ride consists of raising vertically a bench with people sitting on it, allowing the bench to drop and then bringing it to rest using brakes. Fig. 2 shows the bench and its supporting tower. The tower provides lifting and braking mechanisms. The resistances to motion are modelled as having a constant value of 400 N whenever the bench is moving up or down; the only other resistance to motion comes from the action of the brakes.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c8f26b7e-1be1-4abf-8fea-6847185fad81-3_552_741_479_628}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
On one occasion, the mass of the bench (with its riders) is 800 kg .\\
With the brakes not applied, the bench is lifted a distance of 6 m in 12 seconds. It starts from rest and ends at rest.\\
(i) Show that the work done in lifting the bench in this way is 49440 J and calculate the average power required.
For a short period while the bench is being lifted it has a constant speed of $0.55 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Calculate the power required during this period.
With neither the lifting mechanism nor the brakes applied, the bench is now released from rest and drops 3 m .\\
(iii) Using an energy method, calculate the speed of the bench when it has dropped 3 m .
The brakes are now applied and they halve the speed of the bench while it falls a further 0.8 m .\\
(iv) Using an energy method, calculate the work done by the brakes.
\hfill \mbox{\textit{OCR MEI M2 2013 Q2 [16]}}