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Bungee Jump - Potential and Kinetic Energy

  1. Sep 30, 2007 #1
    1. The problem statement, all variables and given/known data

    A bungee jumper needs to calculate how much bungee cord to attach to herself so that it will bring her to rest 3m above ground. The spring constant of the bungee cord is 22 N/m, and she has a mass of 55kg. Neglect the bungee cord's mass.
    a. How long a bungee cord is required?
    b. If she uses the length calculated in a but the spring stiffness is 10 percent less than it was advertised to be, how fast will she hit the ground?

    The bungee jumper is standing 70m above the ground.
    2. Relevant equations

    PE(g)=mgh
    PE(sp)=1/2kx^2
    KE=1/2mv^2

    3. The attempt at a solution

    For part a I used this equation:
    PE(g)=PE(sp)+PE(g)
    55(9.81)(70)=1/2(22)*x^2+(55)(9.81)(3)
    x=57.3 m

    For part b I used this equation:
    PE(g)=PE(sp)+KE
    55(9.81)(70)=1/2(19.8)(57.3)^2+1/2(55)v^2
    v=13.835 m/s

    The correct answer is 7.95 m/s for part b.
    So I am not sure if my part a is wrong or if I am not setting up my equation correctly.
     
  2. jcsd
  3. Oct 1, 2007 #2
    Could someone please look over my work? I may have set it up incorrectly.
     
  4. Oct 1, 2007 #3

    mgb_phys

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    For part a, 'x' is the extension in the spring not it's total length.
     
  5. Oct 1, 2007 #4

    Doc Al

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    That's some wacky stuff if its spring constant is independent of its length.
     
  6. Oct 1, 2007 #5

    mgb_phys

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    Yes, but it also has zero mass!
     
  7. Oct 3, 2007 #6
    So, did I go about part a incorrectly or did I substitute the incorrect amount for part b?
     
  8. Oct 3, 2007 #7

    learningphysics

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    Yeah, part a) isn't right.

    She needs to come to rest 3m above ground. So conservation of energy won't work here for part a).

    when she comes to rest, energy has been lost to heat/friction etc...

    What is her acceleration when she comes to rest? hence what is the net force acting on her when she comes to rest? so what is the extension of the bungee cord?
     
  9. Oct 3, 2007 #8
    Since acceleration isn't given, should I use F=ma and find acceleration?
    Then use a constant accelaration equation to find the cord extension?
     
  10. Oct 3, 2007 #9

    learningphysics

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    What is the acceleration of a body at rest?
     
  11. Oct 3, 2007 #10

    Doc Al

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    By "come to rest" I assume they meant momentarily come to rest, not be at equilibrium. (Otherwise she'll probably smack into the ground!) So I see nothing wrong with using conservation of energy.
     
  12. Oct 3, 2007 #11
    Oh...sorry! OK...so the net force acting on her at rest is weight. So would the extension just be 67 m?
     
  13. Oct 3, 2007 #12

    learningphysics

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    Yes, you're right. :smile: For some reason I thought that she does smash into the ground in this problem!

    sorry about that kdm! your initial solution to part a) is fine I believe, except you need to get 70m - 3m - 57.3m = 9.7m I think.

    sorry again.
     
  14. Oct 3, 2007 #13
    ok. so for part a the length of the cord would be 9.7?
    but for part b i understood it as she does hit the ground since the spring stiffness is 10 percent less. however would i use 9.7 as the 'x' in PE(sp) and use 57.3 as the 'h' in PE(g)?
     
  15. Oct 4, 2007 #14

    learningphysics

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    The amount the bungee stretches for part b) is 70m - 9.7m = 60.3m.

    so this would be your equation:

    55(9.81)(70)=1/2(19.8)(60.3)^2+1/2(55)v^2
     
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