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Bungee Jumping Energy (with drag)

  1. Dec 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A 102kg man bungee jumps with a 20ft (~6m) cord (k=167 N). Air resistance is 12N.
    1. What should be the minimum height of the platform the man jumps off?
    2. What is the maximum velocity he will reach?
    3. How high will he bounce up the first time?

    2. Relevant equations
    Uelastic_initial+ KEinitial + Ugravity = Uelastic_final + KEfinal + Ugravity
    Elastic Potential Energy= 1/2kx^2
    Kinetic Energy= 1/2mv^2
    Gravitational Potential Energy= mgh
    mgh = (1/2)k(h-L)^2, where L is unstretched distance

    3. The attempt at a solution
    I've seen a lot of threads on similar bungee jumping problems, but I'm still not clear how the 12N air resistance would factor in. I know W=Fd so the air resistance would be multiplied by displacement, but beyond that I'm not sure where to add/subtract when applying conservation of energy.
    Am I right in thinking that the equation mgh = (1/2)k(h-6)^2 could be used if air resistance is negligible? If so, how could I adjust for the 12N air resistance?
    (Also, any additional help on the three questions would be much appreciated... I feel like the correct thinking to get the answers is way beyond me)
     
  2. jcsd
  3. Dec 2, 2014 #2
    Update: Okay, so here's what I attempted to do. Could anyone let me know if this thinking/my answers make sense???


    to find height: mgh = (1/2)k(h-6.1)^2+Wdrag
    h= 22.37m.

    to find height bounced up to:
    mgh=.5k(22.37-6.1)^2-12h
    h=21.388 m

    to find velocity:
    1/2mv^2+Wdrag=mgh
    .5*102*v^2+12*6.1=9.8*6.1*102... since height here would just be unstretched length of cord, right?
    v=10.8685 m/s

    Any feedback would be much appreciated!
     
  4. Dec 2, 2014 #3

    haruspex

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    I get a slightly larger number (less than 22).
    Will the jumper stop getting faster as soon as the cord is taut?
    Looks like you made a decimal place error before square rooting.
     
  5. Dec 2, 2014 #4
    Oops, I actually have 21.838 written down. Was it just a typo or should I completely redo the calculation?

    I thought maximum velocity would be at this point because kinetic energy starts being transferred to elastic energy as the cord then stretches.

    I'm doing sqrt (118.1247059) to get the 10.8685 answer...can't quite locate my mistake.
     
  6. Dec 2, 2014 #5

    haruspex

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    That's what I get.
    Yes, but gravity is still doing work. How do you normally find the maximum of a continuous variable?
    No, my mistake - sorry. But the answer is wrong as noted just above.
     
  7. Dec 2, 2014 #6
    I have no idea... I've never heard of a continuous variable before and a quick search doesn't really return how you would find the maximum.
    Oh wait, is this integrals? If so, it's way beyond the math my class level should capable of.
    Would it make sense to set KE+Wdrag+GPE equal to the initial GPE?
    51v2+73.2+(102*9.8*(22.37-6.1))=22361.052 ? Edit: never mind, this just gives the same answer.
     
    Last edited: Dec 2, 2014
  8. Dec 2, 2014 #7

    haruspex

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    I was thinking in terms of calculus, yes, but differentiation. Anyway, we don't really need calculus here.
    If an object accelerates up to some speed, then stops getting faster, what can you say about the acceleration now?
     
  9. Dec 2, 2014 #8
    Well if its velocity is now constant, acceleration would be 0. But if the object is slowing down then acceleration is negative.
     
  10. Dec 2, 2014 #9

    haruspex

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    Right, so what is the acceleration at the moment of maximum speed?
     
  11. Dec 2, 2014 #10
    Acceleration would be 0. I'm not exactly sure where you're going with this though.
     
  12. Dec 2, 2014 #11

    haruspex

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    So think about forces when the acceleration is zero.
     
  13. Dec 2, 2014 #12
    Well when acceleration is zero there's no net force. However there's still gravity acting on it along with elastic potential energy...? So these would all be equal?
     
  14. Dec 2, 2014 #13

    haruspex

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    "elastic potential energy" is not a force. Other than that, you're on the right lines.
     
  15. Dec 2, 2014 #14
    So force of the "spring" instead? F=-kx 167*x =mg
    x=999.6/167= 5.99? I"m confused as to what I'm solving for at this point? Is it displacement/height?
     
  16. Dec 2, 2014 #15

    haruspex

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    You left out drag.
    Well, what does x represent in F=-kx?
     
  17. Dec 2, 2014 #16
    Fspring-Wdrag=mg
    I'm always confused as to whether I should add or subtract drag.
    167x-12x=mg
    x=6.45 m

    Well x represents displacement from equilibrium, so it's the extra distance that the cord has stretched beyond the 6.1m?

    Using this I get velocity to be about 15.59 m/s, unless I"m still missing something.
    Edit: also getting 11.17 if I use 6.45m as height.
     
    Last edited: Dec 2, 2014
  18. Dec 3, 2014 #17

    haruspex

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    At max velocity, which way is the jumper moving, so which way is drag? Is that the same way as mg or opposite?

    That's not right. What would the dimension of 12x be?
    Yes.
    Much closer.
     
  19. Dec 3, 2014 #18
    Drag would be in opposite direction of mg, so it would be addition then? I'm not very good at envisioning these things.

    Should it just be N then, and not N m?

    Is 15.25 m/s any closer? 167x+12=mg, x=5.9
     
  20. Dec 3, 2014 #19

    haruspex

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    Right equation and number for x, but not what I get for v. How did you calculate v? Are you forgetting drag again?
     
  21. Dec 3, 2014 #20
    KE+Wdrag=GPE
    .5mv^2+12(12.01)=102*9.8*12.01
    v^2=232.5701176
    am I using the wrong value of h or rounding prematurely?
     
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