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 203: 19b


62.7 J

Work Step by Step

In Problem 19a, we found that the spring constant was k = 784 N/m. To find the elastic potential energy of the compressed spring, we use $U = \frac{1}{2} kx^{2}$ The spring is compressed 10.0 cm and then an additional 30.0 cm, so x = 10.0 cm + 30.0 cm = 40.0 cm = 0.40 m. Plugging in these values: $U = \frac{1}{2} kx^{2}$ $ = \frac{1}{2} (784 N/m)(0.40 m)^{2}$ = 62.7 J
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