Answer
$\frac{W}{4}$
Work Step by Step
Work done in first case = change in kinetic energy =$\frac{1}{2}mv^{2}-0=W$
In the second case, work done =$\frac{1}{2}m(\frac{v}{2})^{2}-0=\frac{W}{4}$
Thus, work required =$\frac{W}{4}$
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