## College Physics (7th Edition)

$\sqrt 2v$
From the problem, work done = change in kinetic energy =$\frac{1}{2}mv^{2}-0=W$ Let the required speed =$v^{'}$ So, $\frac{1}{2}m(v^{'})^{2}-0=2W$ From above 2 equations, we get $v^{'}=\sqrt 2v$