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Change in Kinetic Energy for a Sliding Box

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A 5.0 kg box slides up a 10 m long frictionless incline at an angle of 20 degrees with the horizontal, pushed by a 40 N force parallel to the incline. What is the change in kinetic energy?

    2. Relevant equations
    Ek = 1/2 mv2
    Ep = mgh


    3. The attempt at a solution
    I tried to assume that vi was zero. Was that incorrect?

    1/2 mvf2 + mgh = 0
    2.5vf2 = -(5)(9.8)(10/(sin(20)))
    vf2= -25.89130...
    ∴ Ekf = 1/2mvf2 = 64.7 N and ΔEk = 64.7 N

    The correct answer is supposed to be 232.4 J. Help?
     
  2. jcsd
  3. Mar 2, 2014 #2
    Well, a couple things. You forgot to involve the 40 N force that is also in the problem -- this will affect the kinetic energy. That's what's messing up your conservation equation and giving you a negative velocity -- it's saying that both kinetic energy and potential energy increase, but that can't be.

    To make it easier to solve for the change in kinetic energy, try writing the conservation of energy equation like this: $$\Delta K + \Delta U_{g} = W_{app}$$
    This comes from using both energy conservation and the work-kinetic energy theorem.
     
  4. Mar 2, 2014 #3
    Sorry, what does ΔUg represent?
     
  5. Mar 2, 2014 #4
    Oh, that's potential energy from gravity -- mgh. It's just another notation for it. Wapp also means the applied work from the 40 N force, just to clarify that too.
     
  6. Mar 2, 2014 #5
    Okay, so then:
    If ΔEk + Epf = W
    Then:
    ΔEk = (40N)(10m) - (5kg)(9.8m/s/s)(10(sin(20)))
    ΔEk = 400 - 167.58
    ΔEk = 232.410... = 232.4 J

    That's the correct answer! Thank you very much. I hadn't thought of just finding the total change in kinetic energy instead of finding them separately.
     
  7. Mar 2, 2014 #6
    No problem -- glad you could work it out. And don't worry, finding the separate energies is perfectly fine too. Using the change just saves you a few extra steps :)
     
  8. Mar 2, 2014 #7
    Wouldn't there be a net force up the plane though due to the presence of the 40N force and the horizontal component of the weight in the opposite direction, ((mg)sinθ)???

    How come you guys used the work done as Fx instead of ([itex]F{net}[/itex])x???
     
  9. Mar 2, 2014 #8
    Easy with the question marks. Yes, there are forces involved here, and a net force up the plane, but all the problem asks for is the change in energy, so you can summarize the actions of the forces in energy conservation.
     
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