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(A) Let $\ell$ be the length of the inclined plane and $	heta = 45^{\circ}$.Case-$1$: Smooth inclined plane (no friction).The acceleration is $a_1 =

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$1-\frac{1}{n^2}$

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(A) Let $\ell$ be the length of the inclined plane and $	heta = 45^{\circ}$.Case-$1$: Smooth inclined plane (no friction).The acceleration is $a_1 = g \sin 	heta$.Using the equation of motion $s = ut + \frac{1}{2}at^2$,with $u=0$,we have $\ell = \frac{1}{2}(g \sin 	heta) t_1^2$.So,$t_1 = \sqrt{\frac{2 \ell}{g \sin 	heta}}$.Case-$2$: Rough inclined plane (with friction).The acceleration is $a_2 = g \sin 	heta - \mu g \cos 	heta$.Similarly,$\ell = \frac{1}{2}(g \sin 	heta - \mu g \cos 	heta) t_2^2$.So,$t_2 = \sqrt{\frac{2 \ell}{g(\sin 	heta - \mu \cos 	heta)}}$.Given that $t_2 = n t_1$,we have:$\sqrt{\frac{2 \ell}{g(\sin 	heta - \mu \cos 	heta)}} = n \sqrt{\frac{2 \ell}{g \sin 	heta}}$.Squaring both sides:$\frac{1}{\sin 	heta - \mu \cos 	heta} = \frac{n^2}{\sin 	heta}$.$\sin 	heta = n^2 \sin 	heta - n^2 \mu \cos 	heta$.$n^2 \mu \cos 	heta = (n^2 - 1) \sin 	heta$.$\mu = \frac{n^2 - 1}{n^2} 	an 	heta$.Since $	heta = 45^{\circ}$,$	an 45^{\circ} = 1$.Therefore,$\mu = \frac{n^2 - 1}{n^2} = 1 - \frac{1}{n^2}$.

$A$ given object takes $n$ times the time to slide down a $45^{\circ}$ rough inclined plane as it takes to slide down an identical perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the surface of the inclined plane is:

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