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Dynamics/energy problem

  1. Jun 29, 2009 #1
    Hello:
    I was wondering if someone could check if the following is correct: A
    box of mass M is tied to a propeller with an upward force of F. Ignore
    weight of propeller. The box is placed on a vertical spring that is initially
    at equilibrium and then the propeller is ignited. The spring constant is K.
    Find speed after the spring has moved by a distance of X. So would I do:
    K_{initial} = 0; U_{initial} = 0; U_{spring} = 0 initially. Kf = 1/2*mv^2
    ; U_{final} = MgX;U_{spring} = 1/2*kX^2.
    And the total work done by the force F is F.x, or at the distance X, FX.
    So since the total work done is equal to the change in mechanical energy, so
    I write: FX = 1/2*mv^2 +MgX + 1/2*KX^2 and solve for v? Thank you.
     
  2. jcsd
  3. Jun 29, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    I see that the spring is not compressed at the start. So yes, the equation as you have written it looks right.
     
  4. Jun 29, 2009 #3
    Hello:
    Oh right, thanks!
     
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