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A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms^-1. The coefficient of friction between the surface and bock is 0.5. The distance covered by the block before coming to rest is : [use g = 9.8 ms^-2]

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4.9 m

19.6 m

12.5 m

9.8 m

answer is B.

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Detailed Solution

$\begin{aligned} a = - μ g = - 0.5 \times 9.8 = - 4.9 m / s^{2} \\ d = \frac{v^{2}}{2 a} = \frac{9.8 \times 9.8}{2 (4.9)} \\ = 9.8 m \end{aligned}$

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