College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 5 - Work and Energy - Learning Path Questions and Exercises - Conceptual Questions - Page 173: 10


$\sqrt 2v$

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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