# Work Integration

• Feb 6th 2010, 06:32 PM
Kasper
Work Integration
Hey folks, I've got a problem here that I can't figure out how to start. Just a pointer in the right direction would be great.

Quote:

What is the natural length of a spring (in cm) which needs 6 joules of work to stretch it to a length of 10 cm to 12 cm, and another 10 joules to stretch it from 12 cm to 14 cm
Havn't yet hit a problem where I need to work backwards to find this. I'm not sure how to start, because even if I do find the spring constant and the force, I can only find the displacement from the natural length, not the natural length itself. What am I missing?

Thanks!
• Feb 6th 2010, 07:35 PM
Black
Let $\displaystyle \ell$ be the natural length of the spring (in cm) and $\displaystyle k$ be the spring constant. Then we have the following equations:

$\displaystyle \int_{(10-\ell)\times 10^{-2}}^{(12-\ell)\times 10^{-2}}kxdx=\frac{k}{2\cdot10^4}[(12-\ell)^2-(10-\ell)^2 ]=6$

$\displaystyle \int_{(12-\ell)\times 10^{-2}}^{(14-\ell)\times 10^{-2}}kxdx=\frac{k}{2\cdot10^4}[(14-\ell)^2-(12-\ell)^2 ]=10$.

Divide the two equations to get

$\displaystyle \frac{(12-\ell)^2-(10-\ell)^2}{(14-\ell)^2-(12-\ell)^2}=\frac{3}{5}.$

Solve for $\displaystyle \ell$.
• Feb 7th 2010, 09:39 AM
Kasper
Was just looking for a push, thank you very much. Makes sense to include the natural length in the limits rather than trying to integrate and solve for it after. Thanks!