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Elastic Potential Energy Question tough

  1. Nov 13, 2006 #1
    Elastic Potential Energy Question!!! tough

    An 86.0 kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of 1.20 x 10^3 N/m. He accidentally slips and falls freely for 0.750 m before the rope runs out of slack. How much is the rope streched when it breaks his fall and momentarily brings him to rest?

    This is my approach:

    First i found the energies...

    KE = 1/2(86)(0) = 0J
    PE = (86)(9.8)(0) = 0 J
    PE(elastic) = = 1/2ky^2 = 1/2(1.2 x 10^3) (.75)^2 = 338J

    THEN, i put The total energy in this equation to solve for elastic distance..

    E = KE + PE + PE(elastic)

    338J = 1/2(86kg)(0m/s) + (86)(9.8 m/s^2)(.750m) + 1/2(1.2 x 10^3 N/m) (x)

    x = 1.62 m (i know im wrong b/c the book answer is different)
     
    Last edited: Nov 13, 2006
  2. jcsd
  3. Nov 13, 2006 #2
    Be careful about where he is when the rope is at equilibrium, and where you defined the potential energy to be zero. I think that's where the issue is...

    I'd say really look at the potential and kinetic energy at 3 points -- the top of the cliff, where the rope starts to catch him, and where the rope finally streches to its maximum. The energy of each position should equal energy of each other position.
     
  4. Nov 13, 2006 #3
    hmmmm...so like this?

    PE (at the top) = (86)(9.8)(0)
    KE = 0
    PE (catched) = (86)(9.8)(.750) = 632.1 J
    PE (elastic) = 1/2(1.2 x 10^3)(.750)^2 = 338 J<----i think the x-value is
    what im confused about!!

    E(total) = 970J

    so after that step:

    970 = (86)(9.8)(0) + (86)(9.8)(.750) + 1/2(1.2e-3)(x)

    x = same answer

    ^^^i think the confusion part is my prob....so how do i approach that??
     
    Last edited: Nov 13, 2006
  5. Nov 13, 2006 #4
    You ARE corrent in that you are confused about the x.

    I'd choose to define his potential energy to be zero at the place the rope first kicks in (that way you can think of the rope like a normal spring displaced from equilibrium).

    Assuming this -- what is his potential energy at the top? What is his kinetic energy at the top?
     
  6. Nov 13, 2006 #5
    hmmmmm, i know for sure the PE (at the top) is 0 because h = 0. Now for the KE....1/2v(initial)^2 -- thats a toughie! where can i obtain this value with only 2 givens (vfinal = 0 and y=.750m)? is g = 9.8 m/s^2 the third value?
     
  7. Nov 13, 2006 #6
    Is he moving at the top?
     
  8. Nov 13, 2006 #7
    no, not at all...does potential energy have the larger value??
     
  9. Nov 13, 2006 #8
    so what is his KE at the top?
     
  10. Nov 13, 2006 #9
    zero lol.....??

    im really having a hard time figuring the PE elastic though.
     
    Last edited: Nov 13, 2006
  11. Nov 13, 2006 #10
    I think you need to recognize a few important things.

    1) you have THREE points of interest.
    a)top of cliff
    b) point where spring kicks in
    c) point where spring is stretched.

    2) You need to find the KE and PE energy at EACH location.
    KE (top) +PE (top) = KE (middle) +PE (middle) = KE (bottom) +PE (bottom)

    3) YOU get to chose a location where the potential energy is ZERO. You seem to want to do this at the top. I'd personally do it at the middle position to make the rope like a standard "spring" problem -- then finding x is the same as a spring. But ONCE you choose a location, you need to calculate all the potential energies relative to this position, which could even mean that the potential energies are negative (if you chose 0 at the top, and his KE is 0, then you get a positive KE as he falls, balanced out by the negative PE -- conserving energy, make sense?).

    .
     
  12. Nov 13, 2006 #11
    ^I don't even know what the velocity is at the bottom!!?? v =0 as well right?
     
  13. Nov 13, 2006 #12
    :biggrin: :biggrin: YES!!!:biggrin: :biggrin:

    (the rope stopped him!)
     
  14. Nov 13, 2006 #13
    lol ok....this is what i got! but it still doesn't correspond to the book answer

    1/2(86)(3.38 m/s) + (86)(9.8)(0) = 1/2(86)(0) + 1/2(1200)x --- i solved from the middle and the bottom

    145.3 = 600x

    x = .242 m

    (sorry if im not gettin it down...i have a sleeping urge lol)
     
    Last edited: Nov 13, 2006
  15. Nov 13, 2006 #14
    It might be a square on the velocity if things are just like you typed above(I'm not really checking numbers here, just giving you ideas to think about). If you are getting brain-drain, take a break and look at this thread again tomorrow (It's 2 AM here too!).
     
  16. Nov 13, 2006 #15
    haha ok i guess i will...i'll PM you with anymore trouble i have with this prob! what a beastly prob
     
  17. Nov 13, 2006 #16

    OlderDan

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    Homework Helper

    The rope is not stretched at all in the beginning. If you choose the initial position to be the zero of potential energy as you have done (which is as good a choice as any), then the total energy is zero. Since you are calling y the amount by which the rope is stretched, then I suggest you call the position of the climber relative to the starting point something else (I will use y'). As the climber falls, y' becomes increasingly negative, reducing the gravitational potential energy (and increasing kinetic energy). When the rope starts to stretch, GPE continues to get lower, but the elastic energy of the rope increases (decreasing kinetic energy).

    What is the relationship between y (rope stretch) and y' (position of climber relative to starting point) while the rope is stretching? What is the total energy, and the contributing parts of that total when the climber is again at rest at the bottom of the fall?
     
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