OCR MEI M1 — Question 7 4 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeHorizontal force on slope
DifficultyModerate -0.3 This is a straightforward equilibrium problem requiring resolution of forces parallel and perpendicular to the slope. Students must resolve the weight, horizontal force, and friction into components, then apply equilibrium conditions. While it involves multiple force components and careful angle work, it follows a standard method taught in M1 with no novel problem-solving required, making it slightly easier than average.
Spec3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form

7 A block of weight 100 N is on a rough plane that is inclined at \(35 ^ { \circ }\) to the horizontal. The block is in equilibrium with a horizontal force of 40 N acting on it, as shown in Fig. 5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{82f933a6-c17e-4b41-ae2b-3cc9d0ba975c-5_490_880_316_623} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure} Calculate the frictional force acting on the block.

Question 7:
AnswerMarks Guidance
AnswerMark Guidance
Take \(F\) +ve up the plane: \(F + 40\cos35 = 100\sin35\)M1 Resolve \(\parallel\) plane (or horiz or vert). All forces present. At least one resolved. Allow \(\sin \leftrightarrow \cos\) and sign errors. Allow \(100g\) used
B1Either \(\pm40\cos35\) or \(\pm100\sin35\) or equivalent seen
\(F = 24.5915\ldots\) so \(24.6\ \text{N}\) (3 s.f.)A1 Accept \(\pm24.5915\ldots\) or \(\pm90.1237\ldots\) even if inconsistent or wrong signs used
up the planeA1 \(24.6\ \text{N}\) up the plane (specified or from diagram) or equiv all obtained from consistent and correct working
Total: 4
## Question 7:

| Answer | Mark | Guidance |
|--------|------|----------|
| Take $F$ +ve up the plane: $F + 40\cos35 = 100\sin35$ | M1 | Resolve $\parallel$ plane (or horiz or vert). All forces present. At least one resolved. Allow $\sin \leftrightarrow \cos$ and sign errors. Allow $100g$ used |
| | B1 | Either $\pm40\cos35$ or $\pm100\sin35$ or equivalent seen |
| $F = 24.5915\ldots$ so $24.6\ \text{N}$ (3 s.f.) | A1 | Accept $\pm24.5915\ldots$ or $\pm90.1237\ldots$ even if inconsistent or wrong signs used |
| up the plane | A1 | $24.6\ \text{N}$ up the plane (specified or from diagram) or equiv all obtained from consistent and correct working |
| **Total: 4** | | |
7 A block of weight 100 N is on a rough plane that is inclined at $35 ^ { \circ }$ to the horizontal. The block is in equilibrium with a horizontal force of 40 N acting on it, as shown in Fig. 5.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{82f933a6-c17e-4b41-ae2b-3cc9d0ba975c-5_490_880_316_623}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}

Calculate the frictional force acting on the block.

\hfill \mbox{\textit{OCR MEI M1  Q7 [4]}}