# Hooke's Law and bungee

1. Sep 13, 2009

### frisy4

Hi im new to this forum I am currently doing year 11 physics at school nothing to complicated. now i have an assignment which i need to determine the length of the extension for a bungee and im going to drop it from a certain height now i have the spring constant of my bungee and can figure out the equilibrium point using Hooke's law F=kx how do i figure out the rest that includes being able to drop the weight from a point and figure out how far it will stretch?

2. Sep 13, 2009

### arildno

Welcome to PF, frisy4!

Now, it is simplest to split the problem in two parts:

1. Estimation of velocity when the bungee line has become straight, but NOT stretched.
In this phase, we have FREE FALL:

Here, use energy conservation, in that the potential energy associated with the height drop becomes converted into kinetic energy.

(That height drop has a very simple relationship with the length of the bungee line..!)
2. Stretch phase:

Here, the kinetic energy gained at 1. is converted into elastic energy.
Take care to include the effect that you in this time period also lose more potential energy due to falling from a higher to a lower place.

Specifically, use energy conservation principles to solve the problem.

3. Sep 13, 2009

### frisy4

ok so well velocity i will just get a stop watch and do an approximate time of where it passes the point then times by 9.8 after that i sub in KE=1/2mv^2 after that my KE=EPE right? from there were do i go?

4. Sep 13, 2009

### arildno

You do not NEED time explicitly; only think energy conservation!

In the second phase:
a) What is the expression for the elastic energy stored in the line when stretched a distanced x?

b) What is the gravitatitonal loss in potential energy associated with a fall over a distance x?

5. Sep 13, 2009

### frisy4

well the rule for EPE is 1/2kx^2. Im not exactly sure about the gravitational loss in PE over a distance.

6. Sep 13, 2009

### arildno

Oops, stupid formulation on my part, sorry about that.

It should be -mgx.

Thus, if v is the velocity when the bungee line is straight but not stretched, and "x" is used as the distance fallen when the velocity has become 0, energy conservation says:
$$\frac{1}{2}mv^{2}=\frac{1}{2}kx^{2}-mgx$$