Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 7 - Exercises and Problems - Page 133: 68

Answer

$U_{total} = \frac{1}{2}kx_1^2 + \frac{1}{2}(k+ka)(3x_1^2)$

Work Step by Step

We know that at $x_1$, the spring energy is equal to: $=\frac{1}{2}kx_1^2 $ Since spring constants are additive when springs are in parallel, it follows: $U_{total} = \frac{1}{2}kx_1^2 + \frac{1}{2}(k+ka)((2x_1)^2-x_1^2)$ $U_{total} = \frac{1}{2}kx_1^2 + \frac{1}{2}(k+ka)(3x_1^2)$
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