The time taken by a block of mass m to slide down from the highest...

Let the length of the inclined plane be $s$.

For the smooth inclined plane, the acceleration of the block is $a_1 = g \sin \theta$.

The time taken to slide down is $t_1 = \sqrt{\dfrac{2s}{a_1}} = \sqrt{\dfrac{2s}{g \sin \theta}}$.

For the rough inclined plane, the acceleration of the block is $a_2 = g \sin \theta - \mu_k g \cos \theta$.

The time taken to slide down is $t_2 = \sqrt{\dfrac{2s}{a_2}} = \sqrt{\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)}}$.

Given that $t_2$ is $50\%$ more than $t_1$, we have $t_2 = t_1 + 0.5 t_1 = \dfrac{3}{2} t_1$.

Squaring both sides gives $t_2^2 = \dfrac{9}{4} t_1^2$.

Substituting the expressions for $t_1^2$ and $t_2^2$:

$\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)} = \dfrac{9}{4} \left( \dfrac{2s}{g \sin \theta} \right)$

$\dfrac{1}{\sin \theta - \mu_k \cos \theta} = \dfrac{9}{4 \sin \theta}$

$4 \sin \theta = 9 \sin \theta - 9 \mu_k \cos \theta$

$9 \mu_k \cos \theta = 5 \sin \theta$

$\mu_k = \dfrac{5}{9} \tan \theta$

Given $\theta = 45^\circ$, we have $\tan 45^\circ = 1$.

$\mu_k = \dfrac{5}{9} \times 1 = \dfrac{5}{9}$

Answer: $5/9$