Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - Problems - Page 165: 27

Answer

1.01 m.

Work Step by Step

Set the potential energy stored in the spring to the given quantity, and solve for x. $$45 J = \frac{1}{2} kx^{2}$$ $$x=\sqrt{\frac{2(45 J)}{88.0 N/m}}=1.01 m$$
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