## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 7 - Exercises and Problems - Page 128: 66

#### 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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