Spring constant varies with maximum velocity

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SUMMARY

The discussion focuses on the application of Hooke's Law and the work-energy theorem to analyze the behavior of a trampoline when a 75kg circus performer jumps from varying heights. The spring constant, calculated as 2450 N/m, is used to determine the trampoline's stretch under different conditions. The key equations include F = -kx and the work-energy theorem, which relates gravitational potential energy to the potential energy stored in the spring. The spring constant does not vary in a theoretical context, but practical applications may show differences based on maximum velocity and jump height.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Familiarity with the work-energy theorem
  • Basic knowledge of gravitational potential energy (PE = mgh)
  • Ability to calculate kinetic energy (KE = 1/2 mv^2)
NEXT STEPS
  • Explore the work-energy theorem in-depth
  • Learn about energy conservation in elastic systems
  • Investigate the effects of varying jump heights on trampoline dynamics
  • Study real-world applications of Hooke's Law in engineering
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Physics students, mechanical engineers, and anyone interested in the dynamics of elastic materials and energy transfer in jumping scenarios.

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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