Hooke's Law applied to a bungee cord

1. Nov 5, 2009

sollinton

1. The problem statement, all variables and given/known data

"A Daredevil plans to bungee-jump from a hot-air 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, the daredevil finds his body weight stretches the cord by 1.50 m. He intends to drop from rest at the point where the top end of a longer section of the cord is attached to the stationary hot-air balloon.

(a) What length of cord should he use?
(b) What maximum acceleration will he experience?

2. Relevant equations

Hooke's Law:
$$F=-kx$$

Most likely other equations.

3. The attempt at a solution

For part (a) I know I will be using Hooke's law, but my problem is I was not given the mass of the "daredevil". I know I need to use the fact that the daredevil's weight stretched a 5.0m section of the cord by 1.5m, but I do not know what the correlation is without being given the mass.

I have not yet attempted part (b), but I am guessing that he will experience the maximum acceleration when the cord is fully stretched and is just about to reverse his downward direction. If that is true, then I would probably use Hooke's law again, substituting the distance the cord will stretch for x.

I know the rules state that you guys are not supposed to give me the answer, but I just want to reiterate, I only want help with this, if I cannot yet solve a problem like this on my own then I need to realize that now and not during an exam.

Thank you very much in advance!

Last edited: Nov 5, 2009
2. Nov 5, 2009

Andrew Mason

What is the energy stored in the cord for a displacement x from equilibrium? What is the potential energy of the daredevil in the balloon compared to his energy at 10 m above the ground. Equate those two energies.

Use the fact that his weight stretches the cord by 1.5 m. to set up an expression for k in terms of his weight. Be careful. The k of the cord changes with length. For example, his weight would stretch a 10 m. cord by 3 m. (this is like two 5 m. lengths in series). Then use that expression for k in the energy equation. You will see that the mass term ultimately cancels out.

That should give you an expression for the displacement from equilibrium that the cord must have in order to absorb the change in potential energy of the daredevil. This will be a function of length of the original cord.

Note: The cord does not begin to stretch until he has fallen the length of the cord. It then stretches the amount you determined it needs to stretch until he stops.

[PS: the practical problem that is overlooked here is: how do you know that the end attached to the balloon will stay stationary? It would have to be a very big balloon! ]

AM

Last edited: Nov 5, 2009
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