Starting from rest a body slides down a $45^{\circ}$ inclined plane in twice the time it takes to sl

Question

Starting from rest a body slides down a $45^{\circ}$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

Professor PC Thomas Classes · Answer

Let $t$ be time on smooth incline and $2t$ on rough. Distance $S = \frac{1}{2} a_s t^2 = \frac{1}{2} a_r (2t)^2 \implies a_s = 4 a_r$. We know $a_s = g \sin 45^{\circ}$ and $a_r = g(\sin 45^{\circ} - \mu \cos 45^{\circ})$. Thus, $g \sin 45^{\circ} = 4(g \sin 45^{\circ} - \mu g \cos 45^{\circ})$. Since $\sin 45^{\circ} = \cos 45^{\circ}$, $1 = 4(1 - \mu) \implies 1 = 4 - 4\mu \implies 4\mu = 3 \implies \mu = 0.75$.