| Exam Board | WJEC |
|---|---|
| Module | Unit 2 (Unit 2) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Lift with occupant problems |
| Difficulty | Moderate -0.8 This is a straightforward two-body dynamics problem requiring application of Newton's second law. Part (a) involves setting up F=ma with given forces (3 marks suggests routine calculation), and part (b) is trivial constant velocity case where acceleration is zero. Both parts are standard textbook exercises with no problem-solving insight required. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03f Weight: W=mg |
The diagram below shows a forklift truck being used to raise two boxes, $P$ and $Q$, vertically. Box $Q$ rests on horizontal forks and box $P$ rests on top of box $Q$. Box $P$ has mass 25 kg and box $Q$ has mass 55 kg.
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\item When the boxes are moving upwards with uniform acceleration, the reaction of the horizontal forks on box $Q$ is 820 N.
Calculate the magnitude of the acceleration. [3]
\item Calculate the reaction of box $Q$ on box $P$ when they are moving vertically upwards with constant speed. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Unit 2 2024 Q7 [4]}}