Bungee Jumping: Min Spring Constant to Support Jorge (75kg)

[renegade05]

Homework Statement

Jorge is going to bungee jumpy from a bridge that is 55.0m over the river below. The bungee cord has an unstretched length of 27.0m. To be safe, the bungee cord should stop Jorge's fall when he is at least 2.00 m above the river. If Jorge has a mass of 75.0kg what is the minimum spring constant of the bungee cord?

Homework Equations

I used ki + Ui + Wf + WF = Ki + Ui

Us = 1/2kx^2

and F = -kx

The Attempt at a Solution

My question is this:

I got the potential inital energy by mgh = 75x9.8x53 ( 53 because jorge is stopping 2 m for the bottom).

Then the Potential spring engery by Us = 1/2k(26)^2 (26 because that is how far the cord is stretching 55-2-27)

I made them equal to each other and solved for K (115 Newtons/meter)

But i also tried it a different way using F = -kx since at the bottom the netforces are zero then -mg = -kx = 75 x 9.8 / 26 = (28.3 Newtons /meter)

So i got two completely different answers. Which one is right (if any) and where did i screw up so i didnt get the same answer?

Thanks!

 

[renegade05]

bump...

 

[Dick]

The PE solution looks right to me. The second solution isn't. At the bottom v=0. The acceleration isn't zero. It's not true the forces balance just because an object is momentarily stationary.