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(C) The net work done on the ball is the sum of the work done by the gravitational force and the work done by the spring force.$1$. Work done by gravi

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$mg(h + d) - \frac{1}{2}Kd^2$

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(C) The net work done on the ball is the sum of the work done by the gravitational force and the work done by the spring force.$1$. Work done by gravity $(W_g)$: The ball falls through a total vertical distance of $(h + d)$. Since the force acts in the direction of displacement,$W_g = mg(h + d)$.$2$. Work done by the spring $(W_s)$: The spring is compressed by a distance $d$. The work done by the spring force is given by $W_s = -\frac{1}{2}Kd^2$ because the spring force acts opposite to the displacement.$3$. Net work done $(W_{net})$: $W_{net} = W_g + W_s = mg(h + d) - \frac{1}{2}Kd^2$.

$A$ vertical spring with force constant $K$ is fixed on a table. $A$ ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is

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