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: 130c

Answer

$\Delta U_e = 0.25~J$

Work Step by Step

We can find the change in elastic potential energy: $\Delta U_e = \frac{1}{2}ky_2^2-\frac{1}{2}ky_1^2$ $\Delta U_e = \frac{1}{2}ky_2^2-0$ $\Delta U_e = \frac{1}{2}(200~N/m)(-0.050~m)^2$ $\Delta U_e = 0.25~J$
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