Conservation of Elastic and Gravitational Energy

Click For Summary

Homework Help Overview

The problem involves a bungee jumper who falls from a bridge while attached to a bungee cord. The task is to calculate the spring stiffness constant of the bungee cord using principles of energy conservation, specifically Hooke's law and gravitational potential energy.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial conditions of the bungee cord, questioning whether it is slack or under tension at the start of the jump. There is consideration of how to apply energy conservation principles, particularly regarding the initial and final positions of the jumper and the bungee cord.

Discussion Status

Some participants have provided insights into the initial conditions and the implications for energy calculations. There is an acknowledgment of different interpretations regarding the setup of the problem, particularly concerning the initial stretch of the bungee cord and the corresponding energy states.

Contextual Notes

There is a mention of the jumper falling a total of 37 m and the unstretched length of the bungee cord being 13 m, which raises questions about the energy states before and after the jump. The discussion reflects on the assumptions made about the bungee cord's behavior at the moment of the jump.

PeachBanana
Messages
189
Reaction score
0

Homework Statement



A 70 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m , and falls a total of 37 m .

Calculate the spring stiffness constant of the bungee cord, assuming Hooke's law applies.

Homework Equations



0.5kx^2 (final) + mgy (final) + 0.5mv^2 (final) = mgy (initial) + 0.5kx^2 (initial) + 0.5mv^2 (initial)


The Attempt at a Solution



Her mass won't change throughout the problem but for the sake of keeping everything organized, I'll just share what I plugged in. K is the unknown, but aren't the initial "x" and "y" positions 13 m? That's a little weird to me and that's why I'm thinking I used in the correct equation. I found out I didn't need the 0.5mv^2 at all because kinetic energy is zero at the beginning and end.
 
Physics news on Phys.org
Ask yourself whether the bungee cord is initially slack, tied to bridge at her jumping height, or whether it is tied 13m above her. In one case she'll free-fall for 13 m before the bungee cord starts to act, in the other case it will start immediately to apply a retarding force.
 
PeachBanana said:

Homework Statement



A 70 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m , and falls a total of 37 m .

Calculate the spring stiffness constant of the bungee cord, assuming Hooke's law applies.

Homework Equations



0.5kx^2 (final) + mgy (final) + 0.5mv^2 (final) = mgy (initial) + 0.5kx^2 (initial) + 0.5mv^2 (initial)

The Attempt at a Solution



Her mass won't change throughout the problem but for the sake of keeping everything organized, I'll just share what I plugged in. K is the unknown, but aren't the initial "x" and "y" positions 13 m? That's a little weird to me and that's why I'm thinking I used in the correct equation. I found out I didn't need the 0.5mv^2 at all because kinetic energy is zero at the beginning and end.
x is the amount of stretch (ie. 37-13) and y is the change in vertical position (37). Essentially the decrease in gravitational potential energy results in an equal increase in the elastic potential energy of the bungee cord. When the cord is 13 m. it has no potential energy so treat x0 (unstretched) as 0.

AM
 
I ended up getting the correct answer. This means the bungee cord was initially slack because she hadn't jumped but the moment she DID jump the bungee cord stretched 24 m because it was 13 m in length.
 

Similar threads

Replies
44
Views
7K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
11K
Replies
11
Views
3K
  • · Replies 17 ·
Replies
17
Views
4K