Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 10 - Interactions and Potential Energy - Exercises and Problems: 15

Answer

We should stretch the spring a distance of 0.63 m

Work Step by Step

The energy stored in a spring is $\frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the stretch distance. Therefore; $\frac{1}{2}kx^2 = 200~J$ $x^2 = \frac{(2)(200~J)}{k}$ $x = \sqrt{\frac{(2)(200~J)}{k}}$ $x = \sqrt{\frac{(2)(200~J)}{1000~N/m}}$ $x = 0.63~m$ We should stretch the spring a distance of 0.63 m.
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