## College Physics (7th Edition)

$\frac{W}{4}$
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 =$\frac{W}{4}$