A block of mass 1.5 kg kept on a rough horiz | vedclass

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Question

(B) Given: Initial velocity $u = 10 \ ms^{-1}$,Final velocity $v = 0 \ ms^{-1}$,Distance $S = 12.5 \ m$,Acceleration due to gravity $g = 10 \ ms^{-2}$

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(B) Given: Initial velocity $u = 10 \ ms^{-1}$,Final velocity $v = 0 \ ms^{-1}$,Distance $S = 12.5 \ m$,Acceleration due to gravity $g = 10 \ ms^{-2}$.The frictional force $f$ acting on the block is $f = \mu N = \mu mg$,where $\mu$ is the coefficient of kinetic friction.The retardation $a$ produced by this frictional force is $a = \frac{f}{m} = \frac{\mu mg}{m} = \mu g$.Using the equation of motion $v^2 = u^2 - 2aS$,where $v = 0$:$0 = u^2 - 2aS$$u^2 = 2aS$$S = \frac{u^2}{2a} = \frac{u^2}{2 \mu g}$.Substituting the given values:$12.5 = \frac{(10)^2}{2 	imes \mu 	imes 10}$$12.5 = \frac{100}{20 \mu}$$12.5 = \frac{5}{\mu}$$\mu = \frac{5}{12.5} = \frac{50}{125} = 0.4$.Therefore,the coefficient of kinetic friction is $0.4$.

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