# Chapter 6 - Section 6.3 - Forces That Vary - Example - Page 97: 6.4

15,000 Joules

#### Work Step by Step

If the length doubled, it has been increased by 11 meters. We know that work equals change in Spring energy in this case, so we find: $W = \frac{1}{2}kx^2 = \frac{1}{2}(250)(11)^2 \approx 15,000 J$

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