Coefficient of Kinetic Friction on Inclined Plane

To find the coefficient of kinetic friction ($\mu$) between the block and the rough inclined plane, we analyze the motion on both the smooth and rough inclined planes.

1. Motion on the Smooth Inclined Plane

For a block sliding down a smooth inclined plane of angle $\theta = 45^\circ$, the acceleration down the incline is: $a_1 = g \sin\theta$

Let $s$ be the length of the incline. Using the second equation of motion for an initial velocity of zero: $s = \frac{1}{2} a_1 t_1^2 \implies t_1 = \sqrt{\frac{2s}{g \sin\theta}}$

2. Motion on the Rough Inclined Plane

For a block sliding down a rough inclined plane with a coefficient of kinetic friction $\mu$, the opposing frictional force is $f_k = \mu N = \mu m g \cos\theta$.

The net acceleration down the incline is: $a_2 = g \sin\theta - \mu g \cos\theta = g(\sin\theta - \mu \cos\theta)$

The time taken $t_2$ to cover the same distance $s$ is: $t_2 = \sqrt{\frac{2s}{g(\sin\theta - \mu \cos\theta)}}$

3. Calculating the Coefficient of Kinetic Friction

We are given that the time taken on the rough incline is $50\%$ more than that on the smooth incline: $t_2 = t_1 + 0.5 t_1 = 1.5 t_1 = \frac{3}{2} t_1$

Taking the ratio of $t_2$ to $t_1$: $\frac{t_2}{t_1} = \sqrt{\frac{g \sin\theta}{g(\sin\theta - \mu \cos\theta)}} = \frac{3}{2}$

Squaring both sides: $\frac{\sin\theta}{\sin\theta - \mu \cos\theta} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$

Since the inclination angle is $\theta = 45^\circ$, we have $\sin 45^\circ = \cos 45^\circ$. Substituting these values simplifies the expression to: $\frac{1}{1 - \mu} = \frac{9}{4}$

Solving for $\mu$: $1 - \mu = \frac{4}{9}$ $\mu = 1 - \frac{4}{9} = \frac{5}{9}$

Correct Answer: Option C ($\frac{5}{9}$)