Explanation

(b) Let body is dragged with force $P$, making an angle $60{^\circ}$ with the horizontal
$F_{k} =$ Kinetic friction in the motion$= \mu_{k}R$
From the figure $F_{k} = p\cos{60{^\circ}}$ and $R = mg - P\sin{60{^\circ}}$
$\therefore P\cos{60{^\circ}} = \mu_{k}(mg - P\sin{60{^\circ}})$
$\Rightarrow \frac{P}{2} = 0.5\left( 60 \times 10 - \frac{P\sqrt{3}}{2} \right)$
$\Rightarrow P = 315.1N$
$\therefore K_{k} = P\cos{60{^\circ}} = \frac{315.1}{2}N$
Work done $= F_{k} \times s = \frac{315.1}{2} \times 2 = 315\ Joule$