Conservation of energy with a bungee jumper

[lordbolton]

1. Starting from rest, a 64kg person bungee jumps from a tethered balloon 65m high. The cord has unstretched length 25.8m. The cord is modeled as a spring that obeys hooke's law with spring constant 81N/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 as a function of the person's variable height y above the ground. Express the elastic potential energy of the cord as a function of y. Assume air resistance is negligible, determine the minimum height of the person above the ground during his plunge.

Homework Equations

Hooke's law f = -kx
Work done by a spring = (1/2)kx^2
Gravitational potential energy U =mgy


3. I have no idea where to start here. I used mgy for the first one. Second one I put (1/2)(81)(25.8-y)^2. For the final one I used mgy to find the person's potential energy at the very top (40,768J)before the jump and used that as total energy. Then, I equated that to mgy + 1/2(81)(25.8-y)^2 and solved which gave me y as 1.16m. I don't think that's correct though since a bungee jumper would risk going that close to the ground before the cord pulls him back up.

 

[Chestermiller]

You got the elastic potential energy of the spring incorrectly. If the unextended length of the spring is 25.8 m, then the altitude of the bottom of the spring at which the spring is unextended is 65.0 - 25.8 = 40.2 m. The amount of spring extension when the person is at altitude y is then ??

Chet

 

[lordbolton]

So I used your correction and recalculated and got y = 19.597 which seems correct. Is that the right answer though? I have no way to check.

 

[Chestermiller]

So I used your correction and recalculated and got y = 19.597 which seems correct. Is that the right answer though? I have no way to check.

I got a different answer. Show us what you did.

Chet

 

[lordbolton]

mgy at 39.2 is 25,213 J.
So I subtract that from the total energy 40,768J which leaves 15,554.56J
Then I equate that with (.5)(81)(y^2) and solve for y which gives 19.61

 

[lordbolton]

Also tried using the equation mgy + (1/2)(k)(39.2-y)^2 = 40,768 and solving using the quadratic formula which gave y as 7.47

 

[Chestermiller]

Also tried using the equation mgy + (1/2)(k)(39.2-y)^2 = 40,768 and solving using the quadratic formula which gave y as 7.47

This is the correct equation to use, but the solution is wrong. I plugged 7.47 into your equation, but the equation was not satisfied. I get ~ 10 m.

Chet

 

[lordbolton]

Tried it again, got x = 10.02m. So thanks for the help. Are my a) and b) correct though?

 

[Chestermiller]

Tried it again, got x = 10.02m. So thanks for the help. Are my a) and b) correct though?

Yes. Those are the terms in your equation, and they are correct.

Chet