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Let L be the natural length of the spring (in inches), and let k be the spring constant (with units lbs/in). The work done stretching the spring from 5 to 10 inches is
${\int}_{5-L}^{10-L}kx\mathit{DifferentialD;}xequals;{\left[\frac{1}{2}k{x}^{2}\right]}_{xequals;5-L}^{xequals;10-L}equals;5k\left(\frac{15}{2}-L\right)equals;25$
The work done stretching the spring from 10 to 14 inches is
${\int}_{10-L}^{14-L}kx\mathit{DifferentialD;}xequals;4k\left(12-L\right)equals;40$
Solving these two resulting equations gives L = 3, and $k\=10\/9$.
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