Help With Conservation of Energy Problem?

Question

It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 120 m above a raging river and attach a 35-m-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 40 N/m. Assume that your mass is 70 kg. After a long hesitation, you dive off the bridge.

1) How far are you above the water when the cord reaches its maximum elongation?

Answer 1

Since F = ma for an accelerating body...

and F = -kx for the resisting force of the spring...

ma = kx

70(9.81) = 40x

=> x = 17.17m

So the total extension of the rope = 17.17m

Total length of rope after extension = 17.17 + 35 = 52.17m

So u will be 120 - 52.17 = 67.83m above the water...

Answer 2

ans) 67.85 m above the water

force F= weight=70* 9.8 N

restoring force,R of the spring =k * x

where k=spring constant &x=elongation

for the spring,force applied=restoring force

i.e.,F=R

70 *9.8=40 *x

x=17.15 m

my distance from the bridge=35+elongation

=35+17.15 m

= 52.15m

thus heibght above water=120-52.15 m

=67.85m

Answer 3

That first answer looks good... but, I don't see him taking into account that you will overshoot and then bounce back.  I don't know how to calculate this, but this will go past the point of equilibrium.

I think you will have to take into account the acceleration and your current speed.

The force downward is constanst:

Fd = 70*9.81 = 686.7 N

The force upward  is not contstant.  Start from the point where the bungee gets tight:

Fu = kx = 40x.

So the total force start from the point where the bungee gets tight:

Ft = Fu - Fd = 40x - 686.7

So the acceleration from that point is:

Ft/70 = (Fu - Fd)/70 = 40x/70 - 686.7/70 = 0.571x - 9.81

A = 0.571x - 9.81

The maximum elongation will be when the velocity = 0.  So you first need to know the velocity at the point where the bungee gets tight.  We call that Vo:

mgh =  ÃƒÂ‚½mv^2

gh = ½v^2

Vo = √(2gh) = √(2*9.81*35) = 26.2 m/s

I am not sure where to go from here.