Hooke's law as pertaining to a bungee jumper and the length of tan unstretched cord

  • Thread starter IanL
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  • #1
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I have a problem for an engineering class I am taking. It reads:

Derive an equation for k using the weight of a bungee jumper (170 lbs) the unstretched length of the cord (unknown) and the height of the dam from which the jumper is jumping off of (722 ft)

What I have come up with is:

K=(170lbs*32.2ft/sec^2)/722ft
K=7.58lb ft/sec^2

Is this correct?

What would be the unstretched length of the cord so that the jumper has a speed of 0 when at the bottom of the jump?

If k=7.58, then the unstretched length should be

L=722/7.58
L=95.25

This does not seem possible........
 

Answers and Replies

  • #2
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i don't know how you justified your spring constant from the acceleration of gravity and the mass of the man.

Although

You can right it as a potential U = mgh + .5k(h-b)2

and the kinetic energy T = .5mv2 which i don't think you need

before the guy jumps mgh is at a maximum if the ground if the point of reference, and

the spring potential is 0 because the cord isn't stretched, but when the cord is stretched such that h=0 then the spring potential is at a max. Thus U is a constant

you know the differential of U with respect to h would be force

0 = mg +k(h-b)
 
  • #3
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I understand that. The problem is asking for an equation for K derived from the given information. As I understand it K is the spring constant.

Thus, how would you derive an equation to figure out the unstretched length and the spring constant so that the bungee jumper does not hit the ground?
 

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