Work/Energy of Two Bungee Jumpers

[kcolman]

Homework Statement

Exercise 1. Two bungee jumpers jump from a bridge. Jumper A is fastened to a bungee cord and holding on to jumper B. The cord has a stiffness k = 80 lb/ft. Find the unstretched length of the cord if the two students just reach the surface of the water at the maximum displacement. If student A let's go of student B at the surface of the water. Is there a problem for student A on the return? The distance between the bridge and water is 120 ft.

Homework Equations

T1 + ƩU_1-2 = T2
U_1-2 = -WΔy Work of a weight
U_1-2 = -ksΔs² Work of a Spring Force

The Attempt at a Solution

I solved the first part by using the principle of work and energy, setting T1 and T2 to zero and solving for the s-initial. However, I'm having trouble with the second part. I know that the potential energy from the first part gets transferred to the bungee jumpers kinetic energy in his ascension, and that his final height will be higher than if he didn't let go of the other bungee jumper. Also, I know his final kinetic energy will be zero at his maximum height, and I think the question wants to know if this height will be greater than the height of the bridge over water. Any thoughts?

 

[tiny-tim]

welcome to pf!

hi kcolman! welcome to pf!

… I know that the potential energy from the first part gets transferred to the bungee jumpers kinetic energy in his ascension, and that his final height will be higher than if he didn't let go of the other bungee jumper. Also, I know his final kinetic energy will be zero at his maximum height, and I think the question wants to know if this height will be greater than the height of the bridge over water.

that analysis looks correct …

so will it be greater? (if so, by roughly how much?) ​

 

[kcolman]

Okay, I think I figured it out. I realized that the energy from the spring is equal to the total original gravitational energy, and set that equal to the work done by gravity with the weight cut in half. So my work looks like:

(1/2)kΔs = (W1+W2)y1
(1/2)kΔs = W1*y2
y2 = ((W1+W2)y1)/W1
y2 = (300 lb * 120 ft) / (150 lb)
y2 = 240 ft

So he doubles his initial height on his ascension... Probably hits his head pretty brutally. Does this look right?

 

[tiny-tim]

hi kcolman!

yes, it will certainly rise well above the bridge

but it's not clear to me whether you've taken into account the downward force that the bungee rope will exert once he gets above the bridge

(and now I'm off to sleep :zzz: …)​