The time taken by a block of mass \(m\) to slide down from the highest point to the lowest point on a rough inclined plane is \(50\%\) more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at \(45^\circ\) with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is _____.

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: \(5/9\)