OCR MEI M1 2007 June — Question 5 4 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Year2007
SessionJune
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's a standard textbook exercise with no novel problem-solving required—slightly easier than average for A-level mechanics.
Spec3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form

5 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]{3be85526-3872-42ac-8278-1d4a3cf75ff7-4_490_874_379_591} \captionsetup{labelformat=empty} \caption{Fig. 5}
\end{figure} Calculate the frictional force acting on the block.

Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
Resolve parallel to plane: \(F + 40\cos35° - 100\sin35° = 0\)M1 A1 Correct resolution parallel to slope
or resolve perpendicular and use components
\(F = 100\sin35° - 40\cos35°\)A1
\(F = 57.36 - 32.77 = 24.6\) N (up the plane)A1 cao, must indicate direction
# Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve parallel to plane: $F + 40\cos35° - 100\sin35° = 0$ | M1 A1 | Correct resolution parallel to slope |
| or resolve perpendicular and use components | | |
| $F = 100\sin35° - 40\cos35°$ | A1 | |
| $F = 57.36 - 32.77 = 24.6$ N (up the plane) | A1 | cao, must indicate direction |

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5 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]{3be85526-3872-42ac-8278-1d4a3cf75ff7-4_490_874_379_591}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}

Calculate the frictional force acting on the block.

\hfill \mbox{\textit{OCR MEI M1 2007 Q5 [4]}}