"A 75kg circus performer jumps from a height of 5.0m onto a trampoline and stretches it a depth of 0.30m. Assume that the trampoline obeys Hooke's law. (a) How far will it stretch if the performer jumps from a height of 8.0m? (b) How far will it be stretched when the performer stands still on it while taking a bow?"(adsbygoogle = window.adsbygoogle || []).push({});

I'm not sure how to solve this problem, but here's some ideas I have:

Since Hooke's law is F = -kx,

-mg = -kx, where x = 0.3

k = 2450 N/m

is that right?

does the spring constant vary when the performer jumps from 5.0m and from 8.0m?

by only using Hooke's law, I can't seem to determine part (a), because hte equation doesn't use the height...

I thought about using

[tex] v_{max} = \sqrt{\frac{k}{m}} (A)[/tex]

to find the spring constant, but then it seems like the spring constant varies with maximum velocity...

is it correct to use max velocity = square root of (2gh), where h = 5.0m or 8.0m?

I'm not sure what to do...

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# Homework Help: Shm 2

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