#### Answer

$\sqrt 2v$

#### Work Step by Step

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 $v^{'}=\sqrt 2v$

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Published by
Pearson

ISBN 10:
0-32160-183-1

ISBN 13:
978-0-32160-183-4

$\sqrt 2v$

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