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(B) Let the body be dragged with a constant force $P$ at an angle of $60^\circ$ with the horizontal.For the body to move at a constant velocity,the ho

Vedclass · Accepted Answer

$315$

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(B) Let the body be dragged with a constant force $P$ at an angle of $60^\circ$ with the horizontal.For the body to move at a constant velocity,the horizontal component of the applied force must balance the kinetic friction force $F_k$.$F_k = \mu R$,where $R$ is the normal reaction.From the free body diagram,the vertical forces are balanced: $R + P \sin 60^\circ = mg \implies R = mg - P \sin 60^\circ$.The horizontal forces are balanced: $P \cos 60^\circ = F_k = \mu(mg - P \sin 60^\circ)$.Substituting the values: $P \cos 60^\circ = 0.5(60 	imes 9.8 - P \sin 60^\circ)$.$P(0.5) = 0.5(588 - P 	imes 0.866)$.$P = 588 - 0.866P \implies 1.866P = 588 \implies P \approx 315.11 \, N$.The work done against friction is $W = F_k 	imes s = (P \cos 60^\circ) 	imes s$.$W = (315.11 	imes 0.5) 	imes 2 = 315.11 \, J$.Rounding to the nearest integer,the work done is $315 \, J$.

$A$ $60 \, kg$ mass is dragged on a horizontal surface by a rope for a distance of $2 \, m$. If the coefficient of friction is $\mu = 0.5$,the angle of the rope with the surface is $60^\circ$,and $g = 9.8 \, m/s^2$,then the work done against friction is ........ $J$.

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