A block of mass m slides down the plane inclined at angle $30^{o}$ with an acceleration $\frac{g}{4}$ . The value of coefficient of kinetic friction will be:

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$\frac{2 \sqrt{3} - 1}{\sqrt{2}}$

$\frac{3}{\sqrt{2}}$

$\frac{1}{2 \sqrt{3}}$

$\frac{2 \sqrt{3} + 1}{\sqrt{2}}$

answer is A.

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Detailed Solution

Here,
$\begin{array}{l} mg \sin 30^{\circ} - μmg \cos 30^{\circ} = ma \\ \Rightarrow a = g (\frac{1}{2} - \frac{\sqrt{3} μ}{2}) \\ \Rightarrow \frac{g}{4} = \frac{g}{2} (1 - \sqrt{3} μ) \\ \Rightarrow \frac{1}{2} = 1 - \sqrt{3} μ \\ \Rightarrow μ = \frac{1}{2 \sqrt{3}} \end{array}$