A Problem with Hooke's Law

1. Nov 21, 2005

Jon1NL8796

Hey there everyone. First post, so... yeah.

Anywho, I got a few questions on Hooke's law as a homework assignment. But the last one is giving me some trouble. I'll tell what I've done so far that leads up to the problem.

The first question asks to calculate the work done given a Graphical Analysis graph that has force on the y-axis and the elongation of a spring on the x-axis (N/m). I figured out that this would be equal to the area of under the given best fit line, so it was a simple case of using the equation K=(.5)bh, where b is th longest elongation (.084) and h is the corresponding force (5.886). So the equation looks like:

Work = K = (.5)(.084)(5.886)
= .247 J

The next question is to find the maximum speed of the spring when a 150 gram mass is attatched to the end, the spring is stretched to its maximum distance, and then released. I know this dealt with PE elastic and KE. so the equation was:

PE elastic(i) + KE(i) = PE elastic(f) + KE(f)

And since there was no KE initially or PE elastic in the final effect:

PE elastic = KE

So then:

PE elastic = (0.5)kx^2
= (0.5)(69.51)(0.084)^2
= 0.245 J
PE elastic = KE = (0.5)mV^2
0.245 = (0.5)(0.15)V^2
V^2 = 0.245 / ([0.5][0.15])
= 3.27
V = 1.81 m/s

Now here's the final question, which is giving me so much trouble. If the setup from the previous question is rigged to be released vertically at the instant that all of the PE in the spring is transfered to the mass, what will the maximum height the mass rises to be?

I thought that since it involved PEg, PE elastic, and KE, and because there would initially be no KE or PEg or PE elastic in the final result, the equation would look like:

(.5)kx^2 = (.5)mV^2 + mgh

Yet no matter how many times I try to get it to work, I always get h to equal a negative number in the ten-thousanths place. I need help with this desperately, and any assistance given would be greatly appreciated.

Peace out y'all!

2. Nov 21, 2005

Chi Meson

Initial (at bottom, call h=0):
KE = 0 (not moving)
PEg =0 (h=0)
PEelast=max (x is maximum and equal to h)

Final (at top):
KE=0 (not moving)
PEg =maximum
PEelast= 0 (you can set the "x" to an arbitrary 0, just like "h")

NOw you assume that the total energy of the system remaons constant, and is still equal to what it was before. How much energy was there?

Last edited: Nov 21, 2005
3. Nov 21, 2005

Jon1NL8796

Thanks for that, I see how I went wrong and fixed the problem.

The questions are supposed to take place in a frictionless enviorment, so there would be no energy loss between the initial and the final. And seeing as the new equation is PE elastic = PEg, and I had already found PE elastic to be .245 J, PEg would be equal to .245 J.

So the steps for the final question looked like this:

.245 = PEg = mgh
= (.15)(9.81)h
h = .245 / ([.15][9.81])
= 1.98 m

Thanks again Chi!