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Bungee Jump problem

  1. Dec 3, 2014 #1
    this problem involves a man jumping off a bridge with a bungee chord holding him. he weighs 102 kg and the stretch constant of the chord is 167n/m. And there is a 12N air resistance. the length of the rope when slack is 20 ft what is the minimum height needed for him to do this?

    my attempt: i had three total attempts which gave me three different answers
    attempt 1:
    i divided the problem into three parts, the initial fall till the rope starts to stretch, then the stretch
    the Ke value that i came up with was mgH, and the sum of the energies is KE-W, which was mg(6.09) -12(6.09). 6.09 m is the length of the chord when it is slack (20ft). the sum of the energies came up to be 6014.484 joules. I then set that equal to 1/2kx^2 12x, which became a quadratic equation: 83.5x^2+12x-6014.484, x being the amount it should stretch, which came up to be about 8.42 meters. The minimum height that i found was 14.51 meters
    attempt 2:
    I set mg(height of the bridge)= 0.5kx=(height-the length of chord/rope)^2 which gave me 22.3 m
    attempt 3:
    i watched a simulation which showed that when the chord is fully stretched, the kinetic energy during the fall converts to elastic potential and gravitational potential. so, i set the KE = mgx+0.5kx^2, x being the distance it is stretched, and i got 4.4 as the answer for the stretch, and the total height came out to be 10.49m.
    I do not know which one to use or whether I'm completely wrong. thank you!
     
  2. jcsd
  3. Dec 3, 2014 #2

    Nathanael

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    Homework Helper

    The first and second methods are almost correct.

    In your first solution you neglected the energy gained from falling (Grav.PE) after the rope begins to stretch.

    In your second method you neglected the air resistance.

    P.S. The height in meters is 6.096 which rounds to 6.10 not 6.09
     
  4. Dec 3, 2014 #3
    thank you very much! but as per what you said, if i account for the energy gained from falling after the rope begins to stretch, i get the same answer i get for my 3rd attempt, which is 4.4 m stretch.
    SUM of Energies during fall = 83.5x^2+12x+mgx (x being the amount stretched)
    if u mean that i should subtract grav.pe , that is:
    SUM of Energies during fall = 83.5x^2+12x-mgx
    I get the TOTAl height to be equivalent to the one i got in the 2nd attempt (the stretch being 16.25, and the length of the rope being 6.09; rounding errors caused my second attempt to be about 22.3m). Which one should I exactly use?
     
  5. Dec 3, 2014 #4
    but as per what you said before, when falling, the KE+grav.pe should be equal to the elastic potential energy, the method should be
    SUM of Energies during fall = 83.5x^2+12x+mgx (x being the amount stretched)
    or
    SUM of Energies during fall -mgx= 83.5x^2+12x(x being the amount stretched) .
    they both are equivalents, but is that approach right?
     
  6. Dec 3, 2014 #5

    Nathanael

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    I would summarize the solution like this:

    [itex](mg-F_{drag})(L+x)=\frac{1}{2}kx^2[/itex]

    Where:
    [itex]L[/itex] is the length of the rope when slack, (6.096 meters)
    [itex]x[/itex] is the amount that the rope gets stretched, (unknown)
    [itex]F_{drag}[/itex] is the force of air resistance, (a constant 12 Newtons)
    [itex]k[/itex] is the "stretch constant" of the rope, (167 Newtons per meter)

    As you can see, if you set [itex]F_{drag}=0[/itex] then this is equivalent to your second approach.

    I think the reason you think the answer is the same as your second method is simply because your second answer is so close to being correct. (This is because air resistance only has a small effect.)
    I calculated that if you use [itex]F_{drag}=0[/itex] instead of [itex]F_{drag}=12[/itex] then the error is only about 0.1 meters (10cm).


    P.S. It should be noted that my equation I wrote for the solution only applies to when the [itex]F_{drag}[/itex] is constant.
     
  7. Dec 3, 2014 #6
    thank you so much!
     
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