Modeling potential energy equations of a balloon bungee jumper

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Homework Help Overview

The discussion revolves around modeling the potential energy equations for a bungee jumper descending from a balloon. The problem involves gravitational potential energy and elastic potential energy, with specific parameters such as mass, height, and spring constant provided.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of potential energy equations, questioning the correct representation of elastic potential energy and the total potential energy of the system. There is uncertainty about how to determine the minimum height of the jumper during the plunge.

Discussion Status

The conversation is ongoing, with participants offering insights into the relationship between potential energy and kinetic energy. Some guidance has been provided regarding the total mechanical energy of the system, but there is still confusion about the implications of the equations and the conditions for minimum height.

Contextual Notes

Participants are navigating assumptions about energy conservation and the role of conservative forces, while also addressing the implications of their calculations regarding the minimum height of the jumper.

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Homework Statement


Starting from rest, a 64.0 kg person bungee jumps from a tethered balloon 65.0 m above the ground. The bungee cord has negligible mass and unstretched length 25.8 m. One end is tied to the basket of the balloon and the other end to the person's body. The cord is modeled as a spring that obeys Hooke's law with a spring constant of 81.0 N/m, and the person is considered to be a particle. The balloon does not move. Find an equation for the total potential energy of the system as a function of height y above the ground and determine the minimum height the person will be above the ground during the plunge.

Homework Equations


Gravitational potential energy is modeled as U = mgy and elastic potential energy is modeled as U = .5ky^2

The Attempt at a Solution


So I thought the gravitational potential equation was straightforward; U = (64)(9.8)y. For the elastic potential equation I am not so sure; I wrote down: U = .5(81)(39.2 - y)^2 since the tether hangs unstretched 39.2 m above the ground. And again, I am not totally sure but the total potential energy equation should simply be the sum of those two equations; U = 627.2y + 40.5(39.2 - y)^2. Even if this equation is correct, I do not know how to find the minimum height of the jumper; is it when the potential energy is 0 J? Any help would be appreciated.
 
Last edited:
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The U expression looks good to me.
I think the minimum height will occur when the kinetic energy is zero, and you'll have to model the total energy in order to find it.
 
Okay, and since there is no air resistance, the person is only subjected to conservative forces, and therefore the total mechanical energy of the system is equal to the sum of kinetic and potential energy. So if kinetic energy has to equal 0 J, the minimum height is when the total mechanical energy is equal to the potential energy. But now I'm stuck on how to find the total energy.
 
Initially the only energy is mgy, so that is the total.
 
Ah, okay, that makes sense. Thank you. :)

Edit: But now looking at it, if the total energy is mgy (which is 627.2y), then the equation becomes 0 = 40.5(39.2 - y)^2, making the y= 39.2, which is means the minimum height is when the rope is unstretched. That does not make sense. I was actually thinking, would the derivative of the potential energy equation have to equal zero since the total energy is always constant (the mechanical is never converted into internal energy)?
 
Last edited:
Not 627.2y, but m*g*65 = 40810 J.
0 = 40.5(39.2 - y)^2
doesn't make sense. The spring will have a lot of the energy after it stops the falling jumper. I think you want to say:
total E = mgy + spring energy + KE
and consider the point where KE = 0.
 
Ah okay, I think I understand now. Thanks again.
 
Most welcome! An interesting problem.
 

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