Work Problem

**sfgiants13** · October 22nd 2008, 04:27 PM

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?

**vincisonfire** · October 22nd 2008, 04:55 PM

This is a physic problem I suggest you this forum Physics Help and Math Help - Physics Forums
However,
$U = \frac{1}{2} k (x-x_0)^2$
$U_1 = \frac{1}{2} k (10-x_0)^2$
$U_2 = \frac{1}{2} k (12-x_0)^2$
$U_2 - U_1 = 6 = \frac{1}{2} k [(12-x_0)^2 - (10-x_0)^2]$
Solve for k
Replace k by what you found in
$U_4 - U_3 = 10 = \frac{1}{2} k [(14-x_0)^2 - (12-x_0)^2]$
and solve for $x_0$
Answer is 8.

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