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(C) Work done is defined as $W = F S \cos 	heta$,where $F$ is the force,$S$ is the displacement,and $	heta$ is the angle between the force and displ

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$100$

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(C) Work done is defined as $W = F S \cos 	heta$,where $F$ is the force,$S$ is the displacement,and $	heta$ is the angle between the force and displacement vectors.$1$. For the horizontal motion: The force applied is vertical ($10 \,N$ upwards) and the displacement is horizontal $(5 \,m)$. The angle $	heta = 90^{\circ}$.$W_{1} = F S \cos 90^{\circ} = 10 	imes 5 	imes 0 = 0 \,J$.$2$. For the vertical motion: The force applied is vertical ($10 \,N$ upwards) and the displacement is vertical ($10 \,m$ upwards). The angle $	heta = 0^{\circ}$.$W_{2} = F S \cos 0^{\circ} = 10 	imes 10 	imes 1 = 100 \,J$.Total work done = $W_{1} + W_{2} = 0 + 100 = 100 \,J$.

$A$ person is holding a bucket by applying a vertical force of $10 \,N$. He moves a horizontal distance of $5 \,m$ and then climbs up a vertical distance of $10 \,m$. Calculate the total work done by him. (in $J$)

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