(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A daredevil plans to bungee jump from a balloon 65.0 m above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test, hanging at rest from a 5.00 m length of the cord, he finds that his body weight stretches it by 1.50 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.

(a) What length of cord should he use?

(b) What maximum acceleration will he experience?

2. Relevant equations

F_cord = -kx

F_gravity = mg

U_cord = .5kx^2

U_body = mgh

3. The attempt at a solution

Uh.. well, I searched the forum and someone had asked a strikingly similar question. From what I read, I should find k by setting the F_gravity to the F_cord, so mg = -kx

In that case.. k would be -mg/x

Then, the potential energy lost from traveling from 65 m to 10 m should equal the potential energy of the cord at 10 m. Thus,

mgh_f-mgh_i=.5kx^2

mgh_f-mgh_i = .5(-mg/x)x^2

h_f-h_i = -.5x

-55 = -.5x

110 = x

... but that would be the length the cord stretched... right? And that would be... the farthest distance it could cover.. Then, the length of the cord would be 110 m?

Then.. to find the max acceleration, ... I'm quite confused. I know the only forces acting on the person is due to gravity and that in the spring.. but what acceleration is there?

Thanks for any help

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# Homework Help: Hooke's Law / Potential Energy

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