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Maximum extension of a bungee cord

  1. Jun 8, 2015 #1
    1. The problem statement, all variables and given/known data
    (a) An 81 kg student is launched from a bridge by his best friends, some 50 metres above the river surface. Fortunately, he is attached to a 30 m bungee cord with a spring constant of 270 N/m.
    i) What is the equilibrium length of the bungee cord, including the force of gravity? (I managed to solve this, it's 32.9m)
    ii) What velocity would the student have when he reaches the equilibrium position if he fell in free fall with no air resistance? (I can solve this using constant accelaration equations, but can someone tell me how to solve it using energy equations?)
    iii) The student has a velocity of 23 m/s at the equilibrium position. Calculate the maximum extension of the spring from its equilibrium position. Is he safe? (Clueless about this one!)
    2. Relevant equations

    Usp=1/2kx^2
    GPE=mgh
    KE=1/2mv^2

    3. The attempt at a solution
    I've tried everything!
    I have tried taking the equilibrium position of 32.9m as a point of zero GPE (not that I fully understand why you can do this, I just accepted more or less that you can :/)
    Then I used that to say
    KE(at equilibrium)=1/2*k*(delta)x^2+m*g*(delta)x
    This gave me a wrong answer though. The answer is 45.5m. I'm really lost, no one seems to know the answer to this...
     
  2. jcsd
  3. Jun 8, 2015 #2

    Orodruin

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    The delta x you are inserting for both spring potential and gravitational potential is wrong. You are not taking into account that he is falling from the top of the bridge, but working as if he fell from the cord where the cord is unstretched.
     
  4. Jun 8, 2015 #3
    Yeah, I was told you can just say GPE is 0 at equilibrium because it's all relative... But it's not working

    What's the correct way to do it then?:cry:
     
  5. Jun 8, 2015 #4

    ehild

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    The gravitational potential energy decreases, you need minus in front of mgΔx. What do you get for Δx? What is the total length of the cord? Show your work in detail.
     
  6. Jun 8, 2015 #5
    Sorry, this is my first post so I didn't know how much detail to show :oops:
    The total length is 30m (as it says in part (a))
    I don't understand why you need a minus... But even if you do, it still gives me the wrong answer :/
    My friend got the right answer by ignoring GPE, but I don't see how you can simply ignore it...
     
  7. Jun 8, 2015 #6

    ehild

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    Explain your notations. First, what is meant as equilibrium position and equilibrium length of the cord? I think it is 32.9 m , as you got.
    What do you mean on Δx? The deviation from the un-stretched length of the cord, or the deviation from the equilibrium length?
    You can count the gravitational potential energy from that position. The velocity is given. The student falls down, and the gravitational potential energy decreases with decreasing height, therefore it is negative. What is the total energy (KE and elastic energy at that initial position? If you count Δx from the equilibrium position (32.9 m) you will see that the gravitational potential energy cancels, so your friend was right.
    Do not forget to add the result for Δx to the equilibrium length of the cord.
     
  8. Jun 8, 2015 #7

    Orodruin

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    Yes you can put the reference wherever. However, you are interested in the difference in potential and if you take the unstretched cord as a reference point, the potential at the top of the bridge is clearly non-zero. You are working as if you could use two different reference points.
     
  9. Jun 8, 2015 #8
    Δx is deviation from equilibrium.
    How does it cancel?
    My equation is:
    KE(at equilibrium)=1/2*k*(delta)x^2+m*g*(delta)x
    so to account for the gravitational potential decreasing,
    1/2*mv^2=1/2*k*(delta)x^2-m*g*(delta)x
    Given v=23 and m=81kg, 1/2mv^2=21424.5J
    so that gives:
    2142.5=1/2*k*Δx-2mgΔx
    =1/2Δx(k-4mg)
    =1/2Δx(270-3175.2)
    Therefore, 1/2Δx=0.67
    Therefore, Δx=0.33m
    What am I doing wrong? :sorry::sorry::sorry:
     
  10. Jun 8, 2015 #9

    jbriggs444

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    Normal bungee cords have no stored potential energy when the jumper is perched on top of the bridge. Your equation for KE at the equilibrium point seems to assume that the bungee cord is under tension at that point -- that it has stored potential energy and is pulling the jumper downward. Before the jump, the jumper should have 30 meters of un-stretched bungee cord connecting his feet to the bridge.

    When the bungee jumper is at the equilibrium position, by how much will the cord have been stretched from its un-stretched state?
     
  11. Jun 8, 2015 #10

    ehild

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    You were a bit careless. You calculated with 2132.5 kinetic energy instead of 21424.5. And you wrote that (1/2)k(Δx)2-mgΔx= (1/2) k Δx - 2mgΔx. It is strange.
     
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