| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done by constant force - find work done |
| Difficulty | Moderate -0.8 This is a straightforward mechanics problem requiring basic resolution of forces (vertical equilibrium to find sin θ) and direct application of the work formula W = Fs cos θ. Both parts involve standard one-step calculations with no problem-solving insight needed, making it easier than average but not trivial due to the force resolution component. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([20 + 25\sin\theta = 2.7g]\) | M1 | For resolving forces vertically |
| \(\sin\theta = 0.28\) | A1 | AG |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([25 \times 5 \times \sqrt{(1 - 0.28^2)}]\) | M1 | For using \(WD = Fd\cos\theta\) |
| Work done is 120 J | A1 | |
| Total: 2 marks |
## Question 1:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[20 + 25\sin\theta = 2.7g]$ | M1 | For resolving forces vertically |
| $\sin\theta = 0.28$ | A1 | AG |
| **Total: 2 marks** | | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[25 \times 5 \times \sqrt{(1 - 0.28^2)}]$ | M1 | For using $WD = Fd\cos\theta$ |
| Work done is 120 J | A1 | |
| **Total: 2 marks** | | |
---1 A block $B$ of mass 2.7 kg is pulled at constant speed along a straight line on a rough horizontal floor. The pulling force has magnitude 25 N and acts at an angle of $\theta$ above the horizontal. The normal component of the contact force acting on $B$ has magnitude 20 N .\\
(i) Show that $\sin \theta = 0.28$.\\
(ii) Find the work done by the pulling force in moving the block a distance of 5 m .
\hfill \mbox{\textit{CAIE M1 2015 Q1 [4]}}