A body of mass 'm' slides down a smooth inclined plane having an inclination of 45⁰ with the horizontal. It takes 2s to reach the bottom. If the body is placed on a similar plane having coefficient friction 0.5, what is the time taken for it to reach the bottom?

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3.725 sec

4.135 sec

2.650 sec

2.828 sec

answer is C.

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Detailed Solution

t = \sqrt{\frac{2 l}{g S i n θ}} \Rightarrow 4 = \frac{2 l}{g S i n θ} \Rightarrow \frac{2 l}{g} = 4 S i n 45^{0}

$= 4 \times \frac{1}{\sqrt{2}} = 2 \sqrt{2}$

$t^{1} = \sqrt{\frac{2 l}{g (\sin θ - μ_{k} \cos θ)}}$
$\sqrt{\frac{2 \sqrt{2}}{\frac{1}{\sqrt{2}} - \frac{1}{2} \frac{1}{\sqrt{2}}}} = \sqrt{\frac{2 \sqrt{2}}{\frac{1}{2 \sqrt{2}}}}$ $= 2 \sqrt{2}$ = 2 × 1.414 = 2.828 S

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