A $60 \,kg$ weight is dragged on a horizontal surface by a rope upto $ 2$ metres. If coefficient of friction is $\mu = 0.5$, the angle of rope with the surface is $60^°$ and $g = 9.8\,m/{\sec ^2}$, then work done is ........ $joules$
A $60 \,kg$ weight is dragged on a horizontal surface by a rope upto $ 2$ metres. If coefficient of friction is $\mu = 0.5$, the angle of rope with the surface is $60^°$ and $g = 9.8\,m/{\sec ^2}$, then work done is ........ $joules$
$294$
$315$
$588$
$197$
A $60 \,kg$ weight is dragged on a horizontal surface by a rope upto $ 2$ metres. If coefficient of friction is $\mu = 0.5$, the angle of rope with the surface is $60^°$ and $g = 9.8\,m/{\sec ^2}$, then work done is ........ $joules$
Let body is dragged with force $P$, making an angle $60^°$ with the horizontal.
${F_k} = $ Kinetic friction in the motion = ${\mu _k}R$
From the figure ${F_k} = P\cos 60^\circ \;$ and $R = mg - P\sin 60^\circ $
$\therefore P\cos 60^\circ = {\mu _k}(mg - P\sin 60^\circ )$
$⇒$ $\frac{P}{2} = 0.5\left( {60 \times 10 - \frac{{P\sqrt 3 }}{2}} \right)$ $⇒$ $P = 315.1\;N$
$\therefore {F_k} = P\cos 60^\circ = \frac{{315.1}}{2}N$ Work done $ = {F_k} \times s = \frac{{315.1}}{2} \times 2 = 315\;Joule$
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