# Spring constant problem :/

1. Aug 18, 2009

### the_awesome

Hey guys. I'm trying to figure out the spring constant of a bungee cord.
Weight = 80kg
Gravity = 9.8m/s/s
Length of cord = 10m
Length of jump = 50m

I've been shown two different methods. And I don't know which one is correct because they give two different answers :/

1.
k = F/x = mg/x
Also knows as : k=Fx/x-x0

So i substitute: k= (80kg x 9.8) / (50-10)
k= 19.6 N/M

2.

PE=mgh
KE=0.5mv^2 = 0.5kx^2
F= -kx

mgh=0.5kx^2
80 x 9.8 x 50 = 0.5k x 40^2
therefore k= 49

So which method is correct? One gets 19.6, the other gets 49.

2. Aug 19, 2009

### nvn

Is this a homework question? If the bungee cord stiffness is constant (which is generally not the case), then the second method is correct. The first method is incorrect.

3. Aug 19, 2009

### the_awesome

No it's not a hmwk question. I was at a bungee jumping place and I needed to do some calculations. The fact is that the 1st method is basically the same as the spring constant equation.

k= Fx / x - x0
However I do not know what x or x0 is so I'm confused.

So your saying that the 2nd method would hold true, if I was at a bungee jumping place, calculating the spring constant of the cord? If so then thankyou

4. Aug 19, 2009

### Staff: Mentor

F = k(x-x0) is just Hooke's law for springs. x-x0 is how much the spring stretches from its unstretched length.

When the jumper is at the lowest point, the upward force of the spring is given by that equation, but that force does not equal the weight of the jumper. (The spring force is greater than the jumper's weight at that point.) As nvn says, the first method is incorrect.

5. Aug 19, 2009

### the_awesome

Okay thankyou guys. 49 does seem like a massive spring constant. But thanx