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A body of mass 'm' slides down a smooth inclined plane having an inclination of $45^{0}$ with the horizontal. It takes 4sec to reach the bottom. If the body is placed on a similar plane having coefficient of friction 0.5. what is the time taken for it to reach the bottom?

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$4 \sqrt{2} \sec$

4 sec

$3 \sqrt{2} \sec$

2 sec

answer is A.

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Detailed Solution

$4 = \sqrt{\frac{2 l}{g \sin 45^{0}}} \Rightarrow 4 = \sqrt{\frac{2 \sqrt{2} l}{g}}$

$t = \sqrt{\frac{2 l}{g [\frac{1}{\sqrt{2}} - \frac{1}{2} . \frac{1}{\sqrt{2}}]}} = \sqrt{\frac{2 \sqrt{2} 2 l}{g}}$

$= 4 \sqrt{2} \sec$

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