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(C) When a block moves up an inclined plane,the retardation $a$ is given by $a = g(\sin 	heta + \mu \cos 	heta)$.Using the equation of motion $v = u

Vedclass · Accepted Answer

$0.6$

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(C) When a block moves up an inclined plane,the retardation $a$ is given by $a = g(\sin 	heta + \mu \cos 	heta)$.Using the equation of motion $v = u - at$,where $v = 0$ (final velocity at rest),$u = 5 \, m/s$,and $t = 0.5 \, s$:$0 = u - at \implies a = \frac{u}{t} = \frac{5}{0.5} = 10 \, m/s^2$.Now,substitute $a = 10 \, m/s^2$,$g = 10 \, m/s^2$,and $	heta = 30^o$ into the retardation formula:$10 = 10(\sin 30^o + \mu \cos 30^o)$.$1 = 0.5 + \mu (\frac{\sqrt{3}}{2})$.$0.5 = \mu (\frac{1.732}{2}) = \mu (0.866)$.$\mu = \frac{0.5}{0.866} \approx 0.577$.Rounding to the nearest provided option,the value is approximately $0.6$.

$A$ block is projected up an inclined plane of angle $\theta = 30^o$ with an initial velocity of $5 \, m/s$. It comes to rest in $0.5 \, s$. What is the coefficient of friction?

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