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Newton's Second Law

A body is mov...

Question

A body is moving with a velocity of $72 k m / h$ on a rough horizontal surface of coefficient of friction $0.5$ . If the acceleration due to gravity is $10 m / s^{2}$ . Find the distance covered it comes to rest.

A

10 m

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B

20 m

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C

30 m

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D

40 m

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Solution

The correct option is D $40 m$
$40 m$

$s = \frac{v^{2} - u^{2}}{2 a}$

$= \frac{0^{2} - (20)^{2}}{2 (- 0.5 \times 10)}$

$= \frac{400}{10} = 40 m$

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