A block sliding down a rough 45° inclined plane has half the velocity it would have had, the inclined plane been smooth. The coefficient of sliding friction between block and the inclined plane is

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$\frac{1}{4}$

$\frac{3}{4}$

$\frac{1}{2 \sqrt{2}}$

$\frac{1}{\sqrt{2}}$

answer is B.

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Detailed Solution

$\begin{array}{l} V_{rough} = \frac{V_{smooth}}{2} \\ \sqrt{2 gl (\sin θ - μ_{k} \cos θ)} = \frac{\sqrt{2 glsin θ}}{2} \\ \sin θ - μ_{k} \cos θ = \frac{\sin θ}{4} \\ 3 \sin θ = 4 μ_{k} \cos θ \Rightarrow μ_{k} = \frac{3}{4} \tan θ \\ μ_{k} = \frac{3}{4} \tan 45^{\circ} = \frac{3}{4} \end{array}$

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