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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. The time taken for it to reach the bottom is

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4.135 sec

answer is C.

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Detailed Solution

In case of smooth incline

$d = ut + \frac{1}{2} {gt}^{2} \Rightarrow \Rightarrow d = \frac{1}{2} \times gsin 45 ° \times 4 = 2 gsin 45^{o}$

Now for friction path we have acceleration

a=gsinθ−μgcosθ

$2 gsin 45 ° = \frac{1}{2} \times (gsinθ - μgcosθ) t^{2}$

$\Rightarrow t = 2 \sqrt{2} \sec$

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