Explanation

(a) When friction absent
$a_{1} = g\sin\theta$
$\therefore\ s_{1} = \frac{1}{2}a_{2}t_{2}^{2}\ \ \ldots.(i)$
When friction in present
$a_{2} = g\sin{\theta - \mu_{k}g\cos\theta}$
$\therefore\ s_{2} = \frac{1}{2}a_{2}t_{2}^{2}\ \ \ \ldots.(ii)$
From Eqs. (i) and (ii), we have
$\frac{1}{2}a_{1}t_{1}^{2} = \frac{1}{2}a_{2}t_{2}^{2}$
$or\ \ a_{1}t_{1}^{2} = a_{2}({nt}_{1})^{2}\ \ \ \ (\because\ t_{2} = {nt}_{2})$
$or\ \ a_{1} = n^{2}a_{2}$
$or\ \ \ \frac{a_{2}}{a_{1}} = \frac{g\sin{\theta - \mu_{k}g\cos\theta}}{g\sin\theta} = \frac{1}{n^{2}}$
$or\ \ \ \frac{g\sin{45{^\circ} - \mu_{k}g\cos{45{^\circ}}}}{g\sin{45{^\circ}}} = \frac{1}{n^{2}}$
$or\ \ 1 - \mu_{k} = \frac{1}{n^{2}}$
$or\ \ \ \mu_{k} = 1 - \frac{1}{n^{2}}$