| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Lifting objects vertically |
| Difficulty | Moderate -0.8 This is a straightforward work-energy question requiring direct application of standard formulas (PE = mgh, KE = ½mv², Power = Work/time). All values are given explicitly, requiring only substitution and arithmetic with no problem-solving insight or multi-step reasoning beyond adding energy components. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02k Power: rate of doing work |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| PE gain is \(32\,000\) J | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([KE \text{ gain} = \frac{1}{2} \times 160 \times 1.25^2]\) | M1 | For using KE gain \(= \frac{1}{2}mv^2\) |
| KE gain is 125 J | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| WD by drum \(= 32\,000 + 125 + 20\,000\) | B1ft | |
| \([P = 52\,125 \div 41.7]\) | M1 | For using \(P = \Delta(WD) \div \Delta T\) |
| Power is 1250 W | A1 [3] |
## Question 3(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| PE gain is $32\,000$ J | B1 [1] | |
## Question 3(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[KE \text{ gain} = \frac{1}{2} \times 160 \times 1.25^2]$ | M1 | For using KE gain $= \frac{1}{2}mv^2$ |
| KE gain is 125 J | A1 [2] | |
## Question 3(iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| WD by drum $= 32\,000 + 125 + 20\,000$ | B1ft | |
| $[P = 52\,125 \div 41.7]$ | M1 | For using $P = \Delta(WD) \div \Delta T$ |
| Power is 1250 W | A1 [3] | |
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\includegraphics[max width=\textwidth, alt={}, center]{01e73486-5a95-4e65-bf18-518d1adc7cfb-2_502_661_1219_742}
A load of mass 160 kg is pulled vertically upwards, from rest at a fixed point $O$ on the ground, using a winding drum. The load passes through a point $A , 20 \mathrm {~m}$ above $O$, with a speed of $1.25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram). Find, for the motion from $O$ to $A$,\\
(i) the gain in the potential energy of the load,\\
(ii) the gain in the kinetic energy of the load.
The power output of the winding drum is constant while the load is in motion.\\
(iii) Given that the work done against the resistance to motion from $O$ to $A$ is 20 kJ and that the time taken for the load to travel from $O$ to $A$ is 41.7 s , find the power output of the winding drum.
\hfill \mbox{\textit{CAIE M1 2012 Q3 [6]}}