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A block of mass 2.5 kg is kept on a rough horizontal surface. It is found that the block does not slide if a horizontal force less than 15 N is applied to it. Also it is found that it takes 5 seconds to slide through the first 10 m if a horizontal force of 15 N is applied and the block is gently pushed to start the motion. Taking $g = 10.0 m / s^{2}$ , calculate the coefficient of kinetic friction between the block and the surface.

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0.45

0.32

0.52

0.8

answer is C.

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Detailed Solution

$From s = u t + \frac{1}{2} a t^{2}$

$10 = 0 + \frac{1}{2} \times a (5)^{2}, a = \frac{20}{25} = 0.8 m s^{2}$

$Using Newton s 2nd law F - μ_{k} m g = m a$

$μ_{k} = \frac{F - m a}{m g} = \frac{15 - 2.5 \times 0.8}{2.5 \times 10} = \frac{13}{25} = 0.52$

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