Looking for classes? Ksquare Career Institute, Bengaluru →
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:
A
0.25
B
0.40
C
0.5
D
0.75
Explanation
Using $t_r = 2 t_s \implies a_r = a_s / 4$, we get $\mu_k = \left(1 - \frac{1}{n^2}\right) \tan\theta = \left(1 - \frac{1}{4}\right) \tan 45^\circ = 0.75$.
Detailed Solution
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, $g \sin 45^\circ - \mu_k g \cos 45^\circ = \frac{g \sin 45^\circ}{4} \implies \mu_k \cos 45^\circ = \frac{3}{4} \sin 45^\circ \implies \mu_k = \frac{3}{4} \tan 45^\circ = 0.75$.