# Hooke's law

1. May 7, 2010

### wangking

Hey, i need some help here
I'm trying to figure out the spring constant for bungee jump

Say, I'm 735N and the rope will apparently double in length when i jump off it.
I don't know the length of the rope thought.
but i do know it does have a spring constant.
How to find out is the spring constant using the Hooke's Law F=-KX

2. May 7, 2010

### wangking

oh, we are trying to find the N/m ratio
how could i do that

3. May 7, 2010

### zachzach

F= -k/x --> |k| = F/x you cannot find it without some information on its stretching. O I just read that it doubles in length.

4. May 7, 2010

### wangking

The rope doubles in length when 735N is added, so does that mean that
k would be equal to 735N/the length of the rope.
k = 735 / L

5. May 7, 2010

### zachzach

yes. When it has doubled its length it has displaced L from it equilibrium position.

6. May 7, 2010

### wangking

Thanks homie, so is the
(the change in F) proportional to (the change in x) = k

7. May 7, 2010

### wangking

If i were using the f = k x equation, and then use
EPE formula = half k x squared
Which is also equal to GPE so could i say
mgh = half k x squared,
and is the x still L

8. May 7, 2010

### zachzach

Yes, this is the point where all the kinetic energy was converted into elastic potential energy.

9. May 8, 2010

### Studiot

If this is more that just a thought exercise you should be aware that at the bottom of the jump impulsive forces equal to approximately twice your weight act on the bungee cord.

There have been fatalities due to cord breakages from this cause.

Anyone who has had dealings with rope rigging will be familiar with snatch forces.