Coefficient of Friction on Inclined Plane

Let the length of the inclined plane be $s$ and the angle of inclination be $\theta = 45^\circ$.

When the block slides down from rest on the rough inclined plane, the net force acting along the incline is: $F_{\text{net}} = m g \sin\theta - \mu m g \cos\theta$

Thus, the acceleration down the incline is: $a_{\text{rough}} = g(\sin\theta - \mu\cos\theta)$

Given that $\theta = 45^\circ$, we have $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$. Therefore: $a_{\text{rough}} = \frac{g}{\sqrt{2}}(1 - \mu)$

The time $t$ taken to slide down a distance $s$ starting from rest is given by the kinematic equation: $s = \frac{1}{2} a_{\text{rough}} t^2 \implies t^2 = \frac{2s}{a_{\text{rough}}}$

When the plane is made smooth (frictionless, i.e., $\mu = 0$), the acceleration down the incline becomes: $a_{\text{smooth}} = g \sin 45^\circ = \frac{g}{\sqrt{2}}$

The time taken to slide down the smooth plane is given as $t_{\text{smooth}} = \frac{t}{2}$: $s = \frac{1}{2} a_{\text{smooth}} \left(\frac{t}{2}\right)^2 \implies \left(\frac{t}{2}\right)^2 = \frac{2s}{a_{\text{smooth}}}$

Equating the distance $s$ covered in both cases: $\frac{1}{2} a_{\text{rough}} t^2 = \frac{1}{2} a_{\text{smooth}} \left(\frac{t}{2}\right)^2$

$a_{\text{rough}} \cdot t^2 = a_{\text{smooth}} \cdot \frac{t^2}{4}$

$a_{\text{rough}} = \frac{a_{\text{smooth}}}{4}$

Substituting the expressions for $a_{\text{rough}}$ and $a_{\text{smooth}}$: $\frac{g}{\sqrt{2}}(1 - \mu) = \frac{1}{4} \left(\frac{g}{\sqrt{2}}\right)$

$1 - \mu = \frac{1}{4}$

$\mu = 1 - \frac{1}{4} = \frac{3}{4} = 0.75$

The coefficient of friction is given as $\left(\frac{\alpha}{100}\right)$: $\frac{\alpha}{100} = 0.75 \implies \alpha = 75$