Conservation of Energy/Force in Bungee Jumping

1. Oct 28, 2012

dtseng96

1. The problem statement, all variables and given/known data
A person stands on a bridge that is 100 m above a river and attaches a 30-m-long bungee cord to his harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 40 N/m. Assume that your mass is 80 kg. After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation?

2. Relevant equations
Uspring, initial + KEinitial + Ugravity = Uspring, final + KEfinal + Ugravity

kx = mg

3. The attempt at a solution

So my main problem with this question is that if I use the law of conservation of energy, where
Uspring, initial + KEinitial + Ugravity = Uspring, final + KEfinal + Ugravity, I get a different answer for this question than if I set the force of the string equal the force of gravity.

For spring problems such as these, can anyone clarify for me when to set the force of the spring equal to the force of gravity, or when to use the law of conservation of energy? Thanks!

2. Oct 28, 2012

frogjg2003

What was the potential energy of the bungee cord at the top of the jump?

3. Oct 28, 2012

omiver4

well by my understanding, you will get a different answer because you shouldn't use g for your acceleration. Your acceleration is instantly changes when you begin to stretch the bungee cord.

- I hope this helps!

4. Oct 28, 2012

frogjg2003

But gravity still pulls with the same force, no matter what speed he is accelerating at.

5. Oct 28, 2012

omiver4

sorry

Last edited: Oct 28, 2012
6. Oct 28, 2012

frogjg2003

1. Please read the forum rules. We do not give away the answer. The poster has to show an attempt at a solution first. Second, we still can't do all the work for them. We can point out mistakes, give them hints, etc. But we do not do the problem for them.
2. That's still wrong. The bungee cord was not stretched from the time he jumped off the cliff. Until the jumper fell 30m, there was no potential energy change in the cord. There are a few ways to account for this, but I won't say anything more until the original poster gives a response.

7. Oct 28, 2012

omiver4

sorry I'm new and still haven't read all the rules on posting I'll take done my response immediately.

8. Oct 31, 2012

dtseng96

Hi everyone, thanks for all of your help! I'm sorry for not being very clear before, but my main question was the last line in my post: "For spring problems such as these, can anyone clarify for me when to set the force of the spring equal to the force of gravity, or when to use the law of conservation of energy?" My book actually gave the solutions for this problem, so I already know the steps to finish the problem. :)

But anyway, what I realized was that we can't set the force of the spring equal to the force of gravity (kx = mg) because at the maximum compression, the force of the spring is greater than the force of gravity. So I'm supposed to use the law of conservation of energy here.

I'm sorry I wasn't clear enough before. Again, thanks for all of the help :)

9. Oct 31, 2012

frogjg2003

You can set the spring force equal to the gravity force. But you can only do that when the spring is at equilibrium. What happens then is that you get a new equilibrium point. In your case, you're right, you wouldn't set the forces equal. This is really only solvable with conservation of energy.