# Homework Help: Spring constant problem

1. Apr 7, 2010

### TyErd

1. The problem statement, all variables and given/known data
The natural length of the bungee cord is 10m
Sam stops falling and first comes to rest momentarily when the length of the bungee cord is 18metres. Height of the tower was calculated to be 19.6metres

What is the spring constant of the bungee cord?

2. Relevant equations
mgh
0.5kx^2

3. The attempt at a solution
after free falling 10metres the gravitational potential would be 70*10*9.6=6720. then equate that to 0.5kx^2, so 6720=0.5k*8^2, thus k=210N/m.
I was absolutely confident my solution was correct but turns out it isnt.

File size:
34.2 KB
Views:
128
2. Apr 7, 2010

### PinkCrayon

Hey there, I'm not sure if I'm qualified to help you, but I'll try my best.

It seems you are trying to equate the potential elastic energy of the bungee cord when it is fully stretched with Sam's potential gravitational energy from when he is at the top of the tower with respect to when the cord starts stretching. I believe your answer is wrong because Sam is still converting gravitational potential to elastic potential as the cord is stretching. If you add that little bit you should get the right answer. Just remember:

When equating...
Eg = Ee

You are actually equating...
Etotal1 = Etotal2
Eg1 + Ee1 = Eg2 + Ee2

But at the very top of the tower, Ee1 = 0, and at Sam's lowest point Eg2 = 0 so...
Eg1 = Ee2

Bear in mind that this only holds true if you set your "zero" (Point of reference for gravitational potential) at Sam's lowest point on his jump. Since you set your point of reference at the point where the cord begins stretching, you need to account for the "negative" gravitational potential that Sam owes to your point of reference when the cord is fully stretched because he has gone past that point.

Hope this helps!