Spring constant varies with maximum velocity

AI Thread Summary
The discussion centers on a physics problem involving a 75kg circus performer jumping onto a trampoline that follows Hooke's law. The initial calculations suggest a spring constant of 2450 N/m based on a 0.30m stretch from a 5.0m jump, but there is confusion regarding how to apply this when the jump height increases to 8.0m. Participants emphasize the need to use the work-energy theorem to determine the maximum stretch of the trampoline when the performer jumps from different heights, as the kinetic energy at impact must be considered. It is clarified that while the spring constant itself does not vary in real life, the effective energy dynamics change with different jump heights. The discussion highlights the importance of correctly applying energy conservation principles to solve the problem.
endeavor
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"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?"
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
v_{max} = \sqrt{\frac{k}{m}} (A)
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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endeavor said:
"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?"
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 not seem right to me. You are treating the system as if the 0.3 m was the new equilibrium position. It is not.

It seems that you have to use the work energy theorem: find the speed of the performer just before he touches the trampoline. That's the initial kinetic energy (treat now this as the initial position of the problem). When he reaches the lowest point, he will have no kinetic energy left, he will have lost some gravitational potential energy (mg times 0.3 meter) and energy will have been converted into potential energy stored in the spring (1/2 kx^2). That will allow you to find k.
does the spring constant vary when the performer jumps from 5.0m and from 8.0m?
Yes..(well, in real life no, but I am sure they had a constant k in mind).


Patrick
 
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