Conservation of Energy - Bungee Jumping

[jtw2e]

Conservation of Energy -- Bungee Jumping

Homework Statement

Unstretched length = 139 ft
k = 39 lb/ft
Weight of jumper = 122 lbs

Homework Equations

Assuming he hasn't hit the ground, how much has the bungee chord stretched when the professor is at the bottom of the jump? (ft)

The Attempt at a Solution

I attempted to setup a CoE equation narrowing it down to:
KE₁ = U_elas2

I came up with ~29.5 ft of stretch in the rope. This was wrong.

Does the jumper have some potential energy as he is falling? I cannot see why I'm getting this one wrong. Thanks for any help you could offer.

 

[SammyS]

Where is the fixed end of the cord relative to the professor's location at his jumping off point?

 

[jtw2e]

Where is the fixed end of the cord relative to the professor's location at his jumping off point?

Well, I guess the fixed end would be attached to the bridge at the point where he jumped off. But I chose x (for Kx1) to be at the position where the rope is neither stretched nor unstretched; i.e. 139 ft.

I found his velocity at the time the rope begins to stretch to be 94.6 ft/s which was correct, but I still am having trouble finding how much the rope stretched at it's maximum point before it started to pull him upwards.

Should it be KE₁ + U_elas1 = U_elas2 ?

 

[SammyS]

Since you already calculated KE, you don't need mgh for the 139ft.

How ever you need to include -mgx as a loss of potential energy as the cord is stretched.

 

[rwisz]

I would like to clarify some concepts related to the conservation of energy and its application in bungee jumping. The conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the case of bungee jumping, we can apply this principle to understand the behavior of the bungee cord and the jumper's energy.

Firstly, it is important to note that the potential energy of the jumper is not relevant in this situation as it is negligible compared to the potential energy stored in the bungee cord. Therefore, we can focus on the energy stored in the bungee cord, which is in the form of elastic potential energy.

The equation you have used, KE1 = Uelas2, is not applicable in this situation as it only considers the kinetic energy and elastic potential energy at a specific point in time. In the case of bungee jumping, the energy is constantly changing as the jumper falls and the bungee cord stretches.

To accurately calculate the stretch of the bungee cord, we need to consider the total energy of the system at the start and end of the jump. At the start of the jump, the jumper has gravitational potential energy, which is converted into kinetic energy as they fall. As the bungee cord stretches, the kinetic energy is converted into elastic potential energy in the bungee cord. At the bottom of the jump, the elastic potential energy is at its maximum, and the kinetic energy is at its minimum.

Therefore, to calculate the stretch of the bungee cord, we need to consider the difference in elastic potential energy between the start and end of the jump. This can be calculated using the equation Uelas = 1/2kx^2, where k is the spring constant and x is the stretch of the bungee cord. We can then equate this with the gravitational potential energy at the start of the jump, mgh, where m is the mass of the jumper, g is the acceleration due to gravity, and h is the height of the jump.

In summary, to accurately calculate the stretch of the bungee cord, we need to consider the total energy of the system and its conversion between kinetic, potential, and elastic potential energy. I hope this helps in finding the correct solution.