Standard +0.3 This is a straightforward mechanics problem requiring resolution of forces on a slope, application of F=μR, and use of constant acceleration equations. While it involves multiple steps (resolving perpendicular and parallel to slope, finding net force, calculating deceleration, then using kinematics), these are all standard techniques with no novel insight required. The calculation is slightly above average difficulty due to the multi-step nature and need to handle friction correctly, but remains a typical textbook exercise.
12 A box of mass \(m \mathrm {~kg}\) slides down a rough slope inclined at \(15 ^ { \circ }\) to the horizontal. The coefficient of friction between the box and the slope is 0.4 . The box has an initial velocity of \(1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) down the slope.
Calculate the distance the box travels before coming to rest.
All terms present; allow sign errors, sin/cos interchange for weight and their \(F\). Correct equation (\(a\) need not be explicitly evaluated here)
Using \(v^2 = u^2 + 2as\): \(0^2 = 1.2^2 + 2 \times (-1.25)s\) giving \(s = 0.576\) m
M1, A1 [7]
Use of suvat equation(s) leading to a value for \(s\) using \(v = 0\). FT their negative \(a\)
## Question 12:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Normal reaction $mg\cos 15°$ | B1 | Correct normal reaction |
| Max friction $\mu N = 0.4mg\cos 15°$ | M1 | Attempt to evaluate friction FT their normal reaction. Only allow $0.4mg$ if it is clear that $mg$ is their normal reaction and not just weight |
| Resolve down the slope: $mg\sin 15° - F = ma$ | B1 | Correct component of weight seen ($2.536m$) |
| $mg\sin 15° - 0.4mg\cos 15° = ma$ giving $a = -1.25 \text{ m s}^{-2}$ | M1, A1 | All terms present; allow sign errors, sin/cos interchange for weight and their $F$. Correct equation ($a$ need not be explicitly evaluated here) |
| Using $v^2 = u^2 + 2as$: $0^2 = 1.2^2 + 2 \times (-1.25)s$ giving $s = 0.576$ m | M1, A1 [7] | Use of suvat equation(s) leading to a value for $s$ using $v = 0$. FT their negative $a$ |
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12 A box of mass $m \mathrm {~kg}$ slides down a rough slope inclined at $15 ^ { \circ }$ to the horizontal. The coefficient of friction between the box and the slope is 0.4 . The box has an initial velocity of $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ down the slope.
Calculate the distance the box travels before coming to rest.
\hfill \mbox{\textit{OCR MEI Paper 1 2021 Q12 [7]}}