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**jfinkbei** What is the natural length of a spring (in cm) which needs 6 Joules of work to stretch the length from 10cm to 12cm and another **10 Joules to stretch it from 12cm to 14cm?**

I believe I solved it but a friend said the answer was wrong but I'm pretty sure mine is right. He says my technique is wrong but I don't know why it would be, it makes sense to me. Anyways, here my work:

$\displaystyle 6 = \int^{12-y}_{10-y} kx\ dx$

$\displaystyle \textcolor{red}{12} = \int^{14-y}_{12-y} kx\ dx$

Where k is the spring constant, x is force, and y is natural length.

I solve for both and since k is a constant make each equation equal which results in $\displaystyle y=9$. My friend claims the answer is 8cm and he said he needed to use linear algebra or something silly like that. Hope you can clear things up for me!