# Chapter 7 - Section 7.3 - Conservation of Mechanical Energy - Example - Page 117: 7.5

1.7 meters

#### Work Step by Step

We know from the situation that: $mgh = \frac{1}{2} kx^2$ Thus, we find: $h = \frac{kx^2}{2mg}$ $h = \frac{140(.11)^2}{2(.05)(9.81)}=1.7m$

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