CAIE M1 2022 November — Question 2 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeConstant speed up/down incline
DifficultyModerate -0.3 This is a straightforward mechanics problem requiring standard formulas: distance = speed × time, change in PE = mgh (with h = d sin 20°), and work = force × distance. The constant speed simplifies the force calculation (pulling force = friction + component of weight down slope). All steps are routine applications of M1 content with no conceptual challenges.
Spec6.02a Work done: concept and definition6.02e Calculate KE and PE: using formulae

2 A box of mass 5 kg is pulled at a constant speed of \(1.8 \mathrm {~ms} ^ { - 1 }\) for 15 s up a rough plane inclined at an angle of \(20 ^ { \circ }\) to the horizontal. The box moves along a line of greatest slope against a frictional force of 40 N . The force pulling the box is parallel to the line of greatest slope.
  1. Find the change in gravitational potential energy of the box.
  2. Find the work done by the pulling force.

Question 2:
Part (a)
AnswerMarks Guidance
AnswerMarks Guidance
\(PE = 5g \times 15 \times 1.8\sin20\)M1 Attempt to find PE gain.
\(PE = 462\text{ J}\)A1 From \(461.727\ldots\)
Total: 2
Part (b)
AnswerMarks Guidance
AnswerMarks Guidance
\(WD = 5g \times 15 \times 1.8\sin20 + 40 \times 15 \times 1.8\) or \(WD = (5g\sin20 + 40) \times 15 \times 1.8\)M1 Uses WD by pulling force \(=\) PE gain \(+\) WD against friction or \(WD = Fs\).
\(WD = 1540\text{ J}\)A1 FT From \(1541.727\ldots\) FT '\(1080 +\) PE from (a)'.
Total: 2
## Question 2:

**Part (a)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $PE = 5g \times 15 \times 1.8\sin20$ | **M1** | Attempt to find PE gain. |
| $PE = 462\text{ J}$ | **A1** | From $461.727\ldots$ |
| | **Total: 2** | |

**Part (b)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $WD = 5g \times 15 \times 1.8\sin20 + 40 \times 15 \times 1.8$ **or** $WD = (5g\sin20 + 40) \times 15 \times 1.8$ | **M1** | Uses WD by pulling force $=$ PE gain $+$ WD against friction or $WD = Fs$. |
| $WD = 1540\text{ J}$ | **A1 FT** | From $1541.727\ldots$ FT '$1080 +$ PE from **(a)**'. |
| | **Total: 2** | |
2 A box of mass 5 kg is pulled at a constant speed of $1.8 \mathrm {~ms} ^ { - 1 }$ for 15 s up a rough plane inclined at an angle of $20 ^ { \circ }$ to the horizontal. The box moves along a line of greatest slope against a frictional force of 40 N . The force pulling the box is parallel to the line of greatest slope.
\begin{enumerate}[label=(\alph*)]
\item Find the change in gravitational potential energy of the box.
\item Find the work done by the pulling force.
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q2 [4]}}