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Laws of Motion
Friction on an Inclined Plane
2 questions from 2 papers, each with its answer.
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- years asked, 2017–2026
- 1 of 10
- of Laws of Motion
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Appears in
Concepts tested
- Angle of repose and kinetic friction
- Inclined Plane Kinematics with Friction
All Questions
2025 1 question
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There are two inclined surfaces of equal length ($L$) and same angle of inclination $45^\circ$ with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction ($\mu_k$) between the object and the rough surface is close to:
Using $t_r = 2 t_s \implies a_r = a_s / 4$, we get $\mu_k = \left(1 - \frac{1}{n^2}\right) \tan\theta = \left(1 - \frac{1}{4}\right) \tan 45^\circ = 0.75$.
Acceleration on a smooth incline: $a_s = g \sin 45^\circ$. Acceleration on a rough incline: $a_r = g (\sin 45^\circ - \mu_k \cos 45^\circ)$. Since $L = \frac{1}{2} a t^2$, time taken is $t = \sqrt{2L/a}$. Given $t_r = 2 t_s$, we have $\sqrt{a_s / a_r} = 2 \implies a_r = a_s / 4$. Therefore, $g \sin 45^\circ - \mu_k g \cos 45^\circ = \frac{g \sin 45^\circ}{4} \implies \mu_k \cos 45^\circ = \frac{3}{4} \sin 45^\circ \implies \mu_k = \frac{3}{4} \tan 45^\circ = 0.75$.
2015 AIPMT-II 1 question
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A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches $30^\circ$, the box starts to slip and slides 4.0 m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively:
$\mu_s = \tan 30^\circ = \frac{1}{\sqrt3} \approx 0.6$
$a = g\sin30^\circ - \mu_kg\cos30^\circ$
$s = ut + \frac{1}{2}at^2$: $4 = \frac{1}{2}\left[\frac{g}{2} - \mu_k\frac{\sqrt3 g}{2}\right]\times 16$
$\mu_k \approx 0.5$