A body of mass M is released from a height of | vedclass

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(B) According to the work-energy theorem,the loss in potential energy is equal to the work done against friction on the horizontal surface.Let $h = 1.

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(B) According to the work-energy theorem,the loss in potential energy is equal to the work done against friction on the horizontal surface.Let $h = 1.5 \ m$ be the initial height and $\mu = 0.2$ be the coefficient of friction.The potential energy at the start is $PE = Mgh$.The work done by friction over a distance $s$ on the horizontal surface is $W_f = f \cdot s = (\mu Mg) \cdot s$.Equating the two: $Mgh = \mu Mgs$.Solving for $s$: $s = \frac{h}{\mu} = \frac{1.5}{0.2} = 7.5 \ m$.Since $7.5 \ m < 15 \ m$,the body comes to rest at a distance of $7.5 \ m$ from point $P$.

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