| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Moderate -0.8 This is a straightforward two-part question on basic mechanics. Part (i) requires direct application of W = Fs cos θ with given values. Part (ii) uses vertical equilibrium (N = mg - F sin θ) with no problem-solving insight needed. Both are standard textbook exercises requiring only routine formula application. |
| Spec | 3.03i Normal reaction force6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| For using the idea of work as a force times a distance \((25 \times 2\cos 15°)\) | M1 | |
| Work done is 48.3 J | A1 | 2 marks |
| For resolving forces vertically (3 terms needed) | M1 | |
| \(N + 25\sin 15° = 3 \times 10\) | A1 √ | |
| (N cos instead of sin following sin instead of cos in (i)) | ||
| Component is 23.5 N | A1 | 3 marks |
For using the idea of work as a force times a distance $(25 \times 2\cos 15°)$ | M1 |
Work done is 48.3 J | A1 | 2 marks
For resolving forces vertically (3 terms needed) | M1 |
$N + 25\sin 15° = 3 \times 10$ | A1 √ |
(N cos instead of sin following sin instead of cos in (i)) | |
Component is 23.5 N | A1 | 3 marks
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{5cba3e17-3979-4c22-a415-2cdd60f09289-2_143_611_1050_769}
A crate of mass 3 kg is pulled at constant speed along a horizontal floor. The pulling force has magnitude 25 N and acts at an angle of $15 ^ { \circ }$ to the horizontal, as shown in the diagram. Find\\
(i) the work done by the pulling force in moving the crate a distance of 2 m ,\\
(ii) the normal component of the contact force on the crate.
\hfill \mbox{\textit{CAIE M1 2003 Q3 [5]}}