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(A) Let the length of the inclined plane be $L$. The acceleration of the block on a rough surface is $a_1 = g(\sin 	heta - \mu \cos 	heta)$ and on a

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) Let the length of the inclined plane be $L$. The acceleration of the block on a rough surface is $a_1 = g(\sin 	heta - \mu \cos 	heta)$ and on a smooth surface is $a_2 = g \sin 	heta$.Using the equation of motion $s = ut + \frac{1}{2}at^2$, with initial velocity $u = 0$, we have $L = \frac{1}{2} a_1 t^2$ and $L = \frac{1}{2} a_2 (t/2)^2$.Equating the two expressions for $L$: $\frac{1}{2} a_1 t^2 = \frac{1}{2} a_2 \frac{t^2}{4}$, which simplifies to $a_1 = \frac{a_2}{4}$ or $4a_1 = a_2$.Substituting the expressions for $a_1$ and $a_2$ with $	heta = 45^\circ$ (where $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$):$4g(\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}) = g(\frac{1}{\sqrt{2}})$.Dividing both sides by $\frac{g}{\sqrt{2}}$, we get $4(1 - \mu) = 1$.$4 - 4\mu = 1 \implies 4\mu = 3 \implies \mu = 0.75$.Since $\mu = \frac{\alpha}{100}$, we have $\frac{\alpha}{100} = 0.75$, which gives $\alpha = 75$.

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