Starting from rest, the time taken by a body sliding down on a rough inclined plane at $45^{\circ}$ with the horizontal is, twice the time taken to travel on a smooth plane of same inclination and same distance. Then the coefficient of kinetic friction is

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Aicharya · Accepted Answer

$ \mu=	an 	heta\left[1-\frac{1}{n^2}ight] $ Here, $\quad 	heta=45^{\circ}$ and $n=2$ $\begin{aligned} 	herefore \quad \mu & =	an 45^{\circ}\left[1-\frac{1}{2^2}ight] \ & =1-\frac{1}{4}=\frac{3}{4}=0.75\end{aligned}$

Starting from rest, the time taken by a body sliding down on a rough inclined plane at $45^{\circ}$ with the…

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