AQA M1 2014 June — Question 8 12 marks

Exam BoardAQA
ModuleM1 (Mechanics 1)
Year2014
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeHorizontal force on slope
DifficultyStandard +0.3 This is a standard M1 mechanics question involving resolving forces on a slope with friction. Part (a) is routine limiting equilibrium, part (b) requires resolving a horizontal force into components perpendicular and parallel to the slope, then applying F=ma. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

8 A crate, of mass 40 kg , is initially at rest on a rough slope inclined at \(30 ^ { \circ }\) to the horizontal, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{788534a5-abbb-4d6a-87b2-c54e859a128a-18_355_882_411_587} The coefficient of friction between the crate and the slope is \(\mu\).
  1. Given that the crate is on the point of slipping down the slope, find \(\mu\).
  2. A horizontal force of magnitude \(X\) newtons is now applied to the crate, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{788534a5-abbb-4d6a-87b2-c54e859a128a-18_357_881_1208_575}
    1. Find the normal reaction on the crate in terms of \(X\).
    2. Given that the crate accelerates up the slope at \(0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\), find \(X\).
      [0pt] [5 marks]
      \includegraphics[max width=\textwidth, alt={}]{788534a5-abbb-4d6a-87b2-c54e859a128a-19_2484_1707_221_153}

Question 8(a)
Resolve perpendicular to slope: \(R = 40g\cos30°\)
Resolve parallel to slope (on point of slipping down, friction acts up):
\(\mu R = 40g\sin30°\)
\[\mu = \frac{40g\sin30°}{40g\cos30°} = \tan30° = \frac{1}{\sqrt{3}} \approx 0.577\]
Question 8(b)(i)
Resolve perpendicular to slope:
\[R = 40g\cos30° + X\sin30°\]
Question 8(b)(ii)
Using \(\mu = \frac{1}{\sqrt{3}}\), applying Newton's second law up the slope:
\[X\cos30° - 40g\sin30° - \frac{1}{\sqrt{3}}(40g\cos30° + X\sin30°) = 40 \times 0.2\]
Solving gives \(X \approx 277\) N
## Question 8(a)

**Resolve perpendicular to slope:** $R = 40g\cos30°$

**Resolve parallel to slope (on point of slipping down, friction acts up):**
$\mu R = 40g\sin30°$

$$\mu = \frac{40g\sin30°}{40g\cos30°} = \tan30° = \frac{1}{\sqrt{3}} \approx 0.577$$

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## Question 8(b)(i)

**Resolve perpendicular to slope:**
$$R = 40g\cos30° + X\sin30°$$

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## Question 8(b)(ii)

Using $\mu = \frac{1}{\sqrt{3}}$, applying Newton's second law up the slope:

$$X\cos30° - 40g\sin30° - \frac{1}{\sqrt{3}}(40g\cos30° + X\sin30°) = 40 \times 0.2$$

Solving gives **$X \approx 277$ N**
8 A crate, of mass 40 kg , is initially at rest on a rough slope inclined at $30 ^ { \circ }$ to the horizontal, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{788534a5-abbb-4d6a-87b2-c54e859a128a-18_355_882_411_587}

The coefficient of friction between the crate and the slope is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item Given that the crate is on the point of slipping down the slope, find $\mu$.
\item A horizontal force of magnitude $X$ newtons is now applied to the crate, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{788534a5-abbb-4d6a-87b2-c54e859a128a-18_357_881_1208_575}
\begin{enumerate}[label=(\roman*)]
\item Find the normal reaction on the crate in terms of $X$.
\item Given that the crate accelerates up the slope at $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, find $X$.\\[0pt]
[5 marks]

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{788534a5-abbb-4d6a-87b2-c54e859a128a-19_2484_1707_221_153}
\end{center}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA M1 2014 Q8 [12]}}