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Homework Help: Bungee Cord Physics

  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A daredevil plans to bungee jump from a balloon 59.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 13.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.65 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.

    2. Relevant equations
    Hookes Law = 1/2 kx^2
    Potential Energy = mgx

    3. The attempt at a solution
    Okay so for starters I said when he falls out of the balloon he falls a distance x at which he is now not falling.
    x = 59-13 = 46
    Then I set out to find k for Hookes law. I said mg = ky where y is the 1.65 m. So
    k = (mg)/y
    I then said r is the actual length of the chord, and L is the length that the chord stretches. So
    r + L = x
    I then said the total potential energy of the devil is mgx, and the amount of work in opposition to him by the chord is 1/2KL^2 = 1/2 * (mgL^2)/y
    I set this equal to mgx: 1/2 * (mgL^2)/y = mgx
    = (L^2)/(2y) = x
    = L = sqrt(2yx)
    solving for L and subtracting it from x to get r (the actual length of the chord) I get 37.3 m. Which apparently is wrong. What am I missing? Thanks in advance
  2. jcsd
  3. Oct 2, 2014 #2


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    k=mg/y Is the spring constant of an 5 m long chord. What is K for an L m long chord of the same kind?

    Also, the potential energy of a stretched spring depends on the change of length (r) instead of the unstretched length (L) as you wrote.

  4. Oct 2, 2014 #3
    Hi and thank you for responding
    I did set L as the change of length and r as the unstretched length.

    Also I just assumed K = mg/y. Does K change because of length?
  5. Oct 2, 2014 #4


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    Imagine a 0.5 m length of chord. Would it stretch to 2.15 m if the fellow hung from it ?
  6. Oct 2, 2014 #5
    Lol no I suppose not. Ok so K =mg/(stretched length) for a cord. I will try it with this.
  7. Oct 2, 2014 #6
    Ok so I put K = mg/L instead of mg/y for the chord, and I get L=2x which is obviously incorrect. I'm lost on what to do now...
  8. Oct 2, 2014 #7


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    No, it is still the case that for a given piece of elastic K = mg/(length of stretch). Your problem is how to convert the K for one length of material to the K for a greater length. It's the innate property of the material that's constant. If a given load stretches a rope length L by x, by how much would it stretch a rope of twice the length?
  9. Oct 2, 2014 #8


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    So k=mg/1.65 for the 5 m long chord. Assuming it is r m long, what is the spring constant?

    Think: 5 m chord stretches 1.65 m. You bind two 5 m chords together. How much does that 10 m chord stretch?
    If the chord is r m long, how much does it stretch by the weight of the man?

  10. Oct 2, 2014 #9
    YES! I got it. Okay so from what I understood, if a 5 m chord stretches 1.65 m, a 10 m chord should stretch 3.3 m (=1.65*2). So to do the conversion I need to find the factor to multiply by and that factor is given by f = (r/5) where 5 is the length of the small cord. So mg/fy = K. Putting that back into the original stuff, I get 2.5L^2 = yx(x-L), and therefore L = 25.156 m. So to find the length of the cord r, x- l = r which is about 20.8 which the website accepts as correct. Thank you!
  11. Oct 2, 2014 #10


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    Well done! :)
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