Hey there everyone. First post, so... yeah.(adsbygoogle = window.adsbygoogle || []).push({});

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!

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# A Problem with Hooke's Law

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