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Step-by-Step Solution
Step 1: Define the time taken on a smooth incline
For a smooth ${45^\circ}$ incline (no friction), the acceleration of the object is
$g \sin 45^\circ = \frac{g}{\sqrt{2}}$.
If the object travels a distance $S$ along the incline, using
$s = \frac{1}{2}at^2$,
we have
$S = \frac{1}{2}\left(\frac{g}{\sqrt{2}}\right){t_1}^2.$
Solving for $t_1$ gives
$$
t_1 = \sqrt{\frac{2S\sqrt{2}}{g}}.
$$
Step 2: Define the time taken on a rough incline
For the same ${45^\circ}$ incline but with friction, the net acceleration becomes
$g\sin 45^\circ - \mu_k \, g\cos 45^\circ$
(because friction opposes motion).
Hence the net acceleration of the object is
$\frac{g}{\sqrt{2}} - \mu_k \frac{g}{\sqrt{2}} = \frac{g}{\sqrt{2}}\bigl(1 - \mu_k\bigr).$
Again using
$S = \frac{1}{2}at^2$,
we have
$S = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\bigl(1 - \mu_k\bigr)\right){t_2}^2.$
Solving for $t_2$ gives
$$
t_2 = \sqrt{\frac{2S\sqrt{2}}{g\,(1 - \mu_k)}}.
$$
Step 3: Relate the times using the given ratio
According to the problem, the time taken on the rough incline is $n$ times the time on the smooth incline:
$$
t_2 = n\, t_1.
$$
Substitute the expressions for $t_2$ and $t_1$:
$$
\sqrt{\frac{2S\sqrt{2}}{g\,(1 - \mu_k)}} \;=\; n \,\sqrt{\frac{2S\sqrt{2}}{g}}.
$$
Squaring both sides and simplifying, we get
$$
\frac{2S\sqrt{2}}{g\,(1 - \mu_k)} \;=\; n^2 \,\frac{2S\sqrt{2}}{g}.
$$
Step 4: Solve for the coefficient of kinetic friction $\mu_k$
Canceling common terms and rearranging:
$$
1 - \mu_k = \frac{1}{n^2},
$$
$$
\mu_k = 1 - \frac{1}{n^2}.
$$
Final Answer
Therefore, the coefficient of kinetic friction between the object and the ${45^\circ}$ incline is
$$1 - \frac{1}{n^2}.$$
