Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 212: 120c

Answer

$\Delta U = 1.12~J$

Work Step by Step

We can find the elastic potential energy when the spring is compressed $4.0~cm$: $U = \frac{1}{2}kx^2$ $U = \frac{1}{2}(3200~N/m)(0.040~m)^2$ $U = 2.56~J$ We can find the change in elastic potential energy: $\Delta U = (2.56~J)-(1.44~J)$ $\Delta U = 1.12~J$
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