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A body of mass m is launched up on a rough inclined plane making an angle 45^o with horizontal. If the time of ascent is half of the time of descent, the frictional coefficient between plane and body is

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a

$\frac{2}{5}$

b

$\frac{3}{5}$

c

$\frac{3}{4}$

d

$\frac{4}{5}$

answer is B.

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Detailed Solution

$\begin{array}{l} t_{A} = \sqrt{\frac{2 s}{a_{A}}} \Rightarrow t_{D} = \sqrt{\frac{2 s}{a_{D}}}, t_{A} = \frac{1}{2} t_{D} \\ \sqrt{\frac{2 s}{g \sin θ + μg \cos θ}} = \frac{1}{2} \sqrt{\frac{2 s}{g \sin θ - μg \cos θ}} \\ 4 g \sin θ - 4 μg \cos θ = g \sin θ + μg \cos θ \\ 3 g \sin θ = 5 μg \cos θ or μ = \frac{3}{5} \tan θ = \frac{3}{5} \end{array}$

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