Starting from rest, a 64.0 kg person bungee jumps from a tethered balloon h = 65

Question

Starting from rest, a 64.0 kg person bungee jumps from a tethered balloon h = 65.0 m above the ground. The bungee cord has negligible mass and unstretched length L = 23.4 m. One end is tied to the basket of the balloon and the other end to a harness around the person's body. The cord is modeled as a spring that obeys Hooke's law with a spring constant of 83.0 N/m, and the person's body is modeled as a particle. The balloon does not move.


(a) Express the gravitational potential energy of the person-Earth system as a function of the person's variable height y above the ground. (Use m, g, and y as necessary.)
Ug = .


(b) Express the elastic potential energy of the cord, Us, as a function of y for points below y = h - L = 41.6 m. (Use m, g, h, L, k, and y as necessary.)
Us = .

(c) Express the total potential energy of the person-cord-Earth system as a function of y for points below y = 41.6 m. (Use m, g, h, L, k, and y as necessary.)
Ug + Us = .

(d) Plot a graph of the gravitational, elastic, and total potential energies as functions of y. (Do this on paper. Your instructor may ask you to turn in this work.)

(e) Assume that air resistance is negligible. Determine the minimum height of the person above the ground during his plunge.
. m

(f) Does the potential energy graph show any equilibrium points or positions? If so, at what elevations? Are they stable or unstable?
There is a ---Select--- stable unstable . equilibrium at y = m

(g) Determine the jumper's maximum speed.

Explanation / Answer

2Max acceleration occurs when a spring is stretched its max and max velocity occur when the force of spring is equal to the force of gravity. Knowing this you can determined the total energy of the system and solve for these quanities. First off, we need to find how far the "spring" will with the man just dangling from the end.Using F=kx we get 64.0(9.8)=83.0x giving us x=7.5566.... Knowing that we can say the total energy is equal to the energy at the very bottom and since the man started 64+7.5566m above the equilibrium point, he will go no further than 64+7.5566 below. Meaning the "spring" is fully stretched. Using the equation for potential energy of spring (PE=.5kx2) we find PE=212494J Now we think of the spot where the spring is unstretch and so all the energy is kinetic. so 212494=.5(64)v2 ? v=81.48898662m/s Since the force of gravity is constant the max acceleartion will entirely on the spring. So the max force will be 83.0*(64+7.5566)-9.8(64)=5312N Then simply use F=ma to find the max acceleration to be 83m/s2 (about 8.5g!)

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