| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | March |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Lifting objects vertically |
| Difficulty | Moderate -0.8 This is a straightforward application of basic work-energy principles requiring only W=mgh for gravitational work, adding resistive work, and using P=W/t. All steps are direct substitutions with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 6.02a Work done: concept and definition6.02k Power: rate of doing work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(600g \times 15 = 90000\) | M1 | Attempt potential energy |
| Total work done by crane \(= 90000 + 10000 = 100000\) J | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(100000 = 12500 \times t\) | M1 | Use of work done \(=\) power \(\times\) time to set up an equation from which \(t\) can be found |
| Time \(= 8\) s | A1 FT | FT on *their* work done \(= 100000\) |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Average force \(F = \dfrac{\text{Total WD}}{15}\), Average velocity \(v = \dfrac{s}{t} = \dfrac{15}{t}\), \(P = Fv \rightarrow 12500 = \dfrac{\text{Total WD}}{15} \times \dfrac{15}{t}\) | M1 | A complete method, using \(P = Fv\), for setting up an equation from which \(t\) can be found |
| Time \(= 8\) s | A1 FT | |
| Total: 2 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $600g \times 15 = 90000$ | M1 | Attempt potential energy |
| Total work done by crane $= 90000 + 10000 = 100000$ J | A1 | |
| | **Total: 2** | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $100000 = 12500 \times t$ | M1 | Use of work done $=$ power $\times$ time to set up an equation from which $t$ can be found |
| Time $= 8$ s | A1 FT | FT on *their* work done $= 100000$ |
| | **Total: 2** | |
**Alternative scheme for 1(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Average force $F = \dfrac{\text{Total WD}}{15}$, Average velocity $v = \dfrac{s}{t} = \dfrac{15}{t}$, $P = Fv \rightarrow 12500 = \dfrac{\text{Total WD}}{15} \times \dfrac{15}{t}$ | M1 | A complete method, using $P = Fv$, for setting up an equation from which $t$ can be found |
| Time $= 8$ s | A1 FT | |
| | **Total: 2** | |1 A crane is used to raise a block of mass 600 kg vertically upwards at a constant speed through a height of 15 m . There is a resistance to the motion of the block, which the crane does 10000 J of work to overcome.
\begin{enumerate}[label=(\alph*)]
\item Find the total work done by the crane.
\item Given that the average power exerted by the crane is 12.5 kW , find the total time for which the block is in motion.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q1 [4]}}