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Hooke's Law / Potential Energy 
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#1
Oct2607, 08:58 PM

P: 263

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_fmgh_i=.5kx^2 mgh_fmgh_i = .5(mg/x)x^2 h_fh_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 


#2
Oct2607, 10:00 PM

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P: 21,865

The balloon is at 65 m, so the length of the cord must not be 110 m or the jumper goes splat.
OK, think about the problem. Falling from 65 m to stop at 10 m, the jumper plans to fall 55 m. The jumper will 'freefall' a distance x, the length of the cord, before it starts stretching, so the jumper has kinetic energy = mgx at this point. Then the jumper must decelerate to stop at 10 m from the ground or total distance of 55 m. The jumper then recoils, subject to a force F = kd, where d is the extension of the cord from equilibrium, and k is the spring constant. 


#3
Oct2607, 10:12 PM

P: 263

I know that F = kx, and U = 1/2kx^2.. but... 


#4
Oct2607, 10:23 PM

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P: 21,865

Hooke's Law / Potential Energy



#5
Oct2607, 10:31 PM

P: 263

mgh_f = mgh_i = .5kx^2 k = 2mgh/x^2...? 


#6
Oct2607, 10:39 PM

P: 263

the weight should just be.. mg?
so mg = kx? k = mg/x 


#7
Oct2607, 10:44 PM

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P: 21,865

The 5 m length stetches 1.5 m, so k(1.5m) = mg. 


#8
Oct2607, 10:49 PM

P: 263

using k = mg/1.5 then...
mg(hfhi) = .5kx^2 mg(hf_hi) = .5(mg/1.5)x^2 h = .5x^2/1.5 3(55) = x^2 x = 12.84 m.. which still doesn't make sense, because the length shold be much greater 


#9
Oct2707, 11:42 AM

Admin
P: 21,865

Try writing the energy balance equation.
Assuming no dissipative forces, the energy in the spring must equal the GPE of the free fall (which equals the kinetic energy of the jumper before the cord stretches) and the work done by the cord when it stretches. Now the cord has length L which one is trying to find, so the jumper freefalls this length (height). Then the cord stretches, some length d, and decelerates the jumper to a stop (v=0). BUT, since the stretched cord must stop at 10 m, and the jumper started at 65 m, the stretched core must have length 55 m, so the displacement d of the core must be 55L. See where that takes one. Think about an elevator slowing down. Think of g+a. 


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