Conservation of Elastic and Gravitational Energy

AI Thread Summary
A 70 kg bungee jumper falls 37 m from a bridge, attached to a bungee cord with an unstretched length of 13 m. To calculate the spring stiffness constant using Hooke's law, it's essential to recognize that the bungee cord is initially slack, allowing for a free fall of 13 m before the cord begins to stretch. The amount of stretch is determined to be 24 m, as the jumper falls beyond the cord's unstretched length. The gravitational potential energy lost during the fall converts into elastic potential energy in the bungee cord. The calculations confirm the relationship between gravitational and elastic potential energy, leading to the correct determination of the spring constant.
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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.
 
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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.
 
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