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Block placed on spring

  1. Apr 10, 2005 #1
    A vertical spring with k = 490 N/m is standing on the ground. You are holding a 5.0 kg block just above the spring, not quite touching it.

    A)How far does the spring compress if you let go of the block suddenly?

    B)How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?


    A) This part was simple enough,
    U(potential) = U(spring)
    where the h in potential is the height from where the spring would compress to...
    mg(deltaS) = 1/2 k*(deltaS)^2
    plug #'s and solve for deltaS. 0.2m

    B)I don't know where to start this any differently. 0.2m isn't the correct answer. I just don't know how this differs...
    help?

    Thanks
     
  2. jcsd
  3. Apr 10, 2005 #2

    Doc Al

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    Staff: Mentor

    Hint: If you can move your hand without disturbing it, what must be the net force on the block?
     
  4. Apr 10, 2005 #3
    the net force must be 0.

    So the force upward by the spring = mg

    But how to i relate that to figure out what the displacement is?
     
  5. Apr 10, 2005 #4

    Doc Al

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    Staff: Mentor

    How does the spring force depend on its displacement from equilibrium? (What is Hooke's law?)
     
  6. Apr 10, 2005 #5
    oh...

    mg=-kdeltaS
    5(-9.81)=-490deltaS
    :biggrin: :biggrin:


    so they're different because it occilates when dropped and not when lowered, right?
     
  7. Apr 10, 2005 #6

    Doc Al

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    Staff: Mentor

    Right! (Here's a tip: Don't get hung up on the signs. The minus sign in Hooke's law just means that the spring always exerts a force opposite to its displacement from equilbrium. And g is always a positive number, by the way.)


    Exactly right!
     
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