# Springs -lost

• Jul 9th 2013, 01:55 AM
baldysm
Springs -lost
Suppose that 2 J of work is needed to stretch a spring from it's natural length of 30cm to a length of 42cm. How much work is needed to stretch the spring from 35cm to 40cm?

The answer is $\displaystyle \frac{25}{24}$.

The difference is 12cm, but Hooke's law is in meters, so I need to use .12. Accordingly I get 2=.12k just plugging into Hooke's law. k is $\displaystyle \frac{50}{3}$.

For the setup, I get:

$\displaystyle \int(\frac{50}{3})x dx$ The interval is supposed to be .05 to .10, but I can't get it to work for some reason. I get 1/16 as the answer, which is wrong.

If I don't convert to meters, I get $\displaystyle k=\frac{1}{6}$ with an interval of 5 to 10, and that yields an answer of $\displaystyle \frac{25}{4}$, also wrong.

What am I doing wrong?
• Jul 9th 2013, 05:14 AM
ebaines
Re: Springs -lost
First, your value for k is incorrect. The work required to stretch the spring from rest to 42 cm is:

$\displaystyle W =2J = \int F dx = \int kx dx = \int_0 ^ {0.12} x dx = \frac 1 2 k (0.12^2 - 0^2)$.

Hence $\displaystyle k = \frac {2J}{0.12m^2} = 277.8\ N/m$

Then use this value of k to evaluate the work to stretch from 35 to 40 cm:

W = $\displaystyle \int _{0.05} ^{0.10} kx dx$.

Remember rhat the deflections are always measured relative to the rest position of the spring.