If 6J work is needed to stretch a spring from 10cm to 12cm and another 10J is needed to stretch it from 12 cm to 14 cm, what is the natural length?
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However,
$\displaystyle U = \frac{1}{2} k (x-x_0)^2 $
$\displaystyle U_1 = \frac{1}{2} k (10-x_0)^2 $
$\displaystyle U_2 = \frac{1}{2} k (12-x_0)^2 $
$\displaystyle U_2 - U_1 = 6 = \frac{1}{2} k [(12-x_0)^2 - (10-x_0)^2]$
Solve for k
Replace k by what you found in
$\displaystyle U_4 - U_3 = 10 = \frac{1}{2} k [(14-x_0)^2 - (12-x_0)^2]$
and solve for $\displaystyle x_0 $
Answer is 8.