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Spring Energy Equation

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data

    As a 15000 kg jet plane lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable is attached to a spring with spring constant 60000 N/m. If the spring stretches 30 m to stop the plane, what was the plane's landing speed?

    2. Relevant equations

    m=15000 kg
    k = 60000 N/m
    d = 30 m

    3. The attempt at a solution I assume Hook's Law is involved here somewhere, F = kx = 60000*30 = 1800000.

    Beyond this, I'm lost about where to go. Can anyone help?
    Thanks.
     
  2. jcsd
  3. Oct 13, 2012 #2
    The title of this thread is the clue to the solution.
     
  4. Oct 13, 2012 #3
    Alright. Here's what I've come up with.

    Starting off by calculating the spring potential:

    U = .5kx^2 = .50(60000)(30)^2 = 27000000 J

    Since U = work, than this is the amount of force being applied to the spring by the plane.

    So to convert force to speed, I tried:

    F = ma
    27000000 = 15000a
    a = 1800 m/s/s

    Vfinal^2 = Vinital^2 +2ad
    0 = Vinital^2 + 108000
    Vinital = 328.63 m/s

    This comes back as wrong. Where am I off? Am I just looking at this wrong?
     
  5. Oct 13, 2012 #4

    Astronuc

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    Science Advisor

    How about conservation of energy - neglecting friction.

    The plane has kinetic energy, which is ?

    Where does that energy go?
     
  6. Oct 13, 2012 #5
    KE = 1/2mv^2

    what velocity am I using? Assuming that my above calculations are wrong, then the only velocity left to work with is the final velocity, which is 0. The plane can't have 0 KE at landing.
     
  7. Oct 13, 2012 #6
    Ok. I'm done with this one. Used up all attempts that the homework program allows for. It came back with a correct answer of 60m/s. Clearly there was something that I missed here. Would anyone be willing to clarify for me how this answer works out?

    Thanks. And sorry if I seemed frustrated earlier.
     
  8. Oct 13, 2012 #7
    The plane has certain kinetic energy upon landing. The purpose of the arresting gear is to stop it, meaning bring its energy down to zero. So this energy must be absorbed by the arrestor springs. There is a very simple equation relating the extension of a spring with the energy stored in it.
     
  9. Oct 13, 2012 #8
    You're right, it doesn't have 0J of kinetic energy at landing. It has 0J of kinetic energy after it has transferred all its energy to the spring. Before that, it had some kinetic energy, and the spring was unstretched, with no potential energy.

    Before:
    spring 0J
    plane xJ

    After:
    spring xJ
    plane 0J
     
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