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Work-Energy Question

  1. Oct 30, 2004 #1
    Hi,

    I am having diffculty with a physics problem. If anyone could help me out it would be greatly appreciated.

    Question:
    One end of a long elastic cord is tied to the railing of a high bridge. The other end is tied to a sandbag with a mass of 23.2 kg. The unstretched length of the elastic cord is 5.44 m and its force constant is 382.0 N/m. The sandbag is dropped from the railing. How far does it drop before coming to rest momentarily before being pulled back up by the cord.

    My answer, so far:
    Known: k=382.0 N/m; Xinitial = 5.44 m; m of bag = 23.2 kg.
    Unknown: Xfinal = ?; W =?

    I know that W = F(Xfinal) and that W = Efinal - Einitial
    Therefore,
    F(Xfinal) = Efinal - Einitial
    F(Xfinal) = 1/2k(Xfinal)^2 - 1/2k(Xinitial)^2
    228 N (Xf) = 1/2(382.0 N/m)(Xf^2) - 1/2(382.0 N/m)(5.44 m)^2
    By rearranging and solving for Xf
    Xf = -4.87 m; +6.09 m

    Rejecting -4.87 m, my answer is 6.09 m, however it is not right.
    Does anyone see what I have done wrong??

    If anyone could help me out, it would be greatly appreciated.
    Thanks
     
  2. jcsd
  3. Oct 30, 2004 #2

    Doc Al

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    Staff: Mentor

    One thing to point out before discussing the physics: By "How far does it drop" I assume they mean how far below the railing. (Don't forget the unstretched length.)

    I find it a bit hard to follow what you've done. (Not enough coffee, perhaps?) In any case, here's how to analyze this. At its lowest point call the gravitational PE = 0, so all you have is spring PE. At the initial point (the railing) all you have is gravitational PE (measured from the lowest point--don't forget the unstretched length). Since energy is conserved, set them equal.
     
  4. Jan 8, 2005 #3

    Eus

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    Hi Ho! ^_^

    I found a question like this in Fundamental Physics 6th Edition by Halliday/Resnick/Walker.
    Page 148-149, Sample Problem 8-4.
    Therefore, I can answer your question.
    1. Let's take an isolated system consists of: earth-bridge-sandbag-cord.
    2. Because it's an isolated system without any external force acts on it, there's no work done to this isolated system. (Work is act of transferring energy)
    Thus, Work = (delta)Energy of this system = 0
    (delta)K + (delta)U gravitation + (delta)U elastic = 0
    3. For the first condition, that's before the sandbag is thrown,
    K = 0 (it is at rest)
    Ug = 0 (taking the bridge as the referential point, y = 0)
    Ue = 0 (the rope is at rest)
    4. For the last condition,
    K = 0 (it stops momentarily before being pulled back up by the cord)
    Ug = mg(h+z) where h is the last position before the cord is scretched by the sandbag and z is the last scretched position of the rope (when it momentarily stops).
    Ue = 0.5(k)(z^2)
    5. (delta)K + (delta)U gravitation + (delta)U elastic = 0
    0 + mg(h+z)+0.5(k)(z^2) = 0
    23.2(9.8)(-(5.44+z))+0.5(382.0)(z^2) = 0
    z = 3.208578894
    I use -(5.44+z) because I use the bridge as y = 0. So all position below the bridge will be negative in sign.
    6. Therefore, the answer is 8.648578894 m. That's 8.6 m.
     
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