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Work, Energy

  1. Aug 5, 2015 #1
    1. The problem statement, all variables and given/known data

    You hold a 500N barbell 0.5m above a level floor for 100s . During this time, the amount of work you do on the barbell is


    • A

      1000J



    • B

      0J



    • C

      500J



    • D

      50000J
    2. Relevant equations

    W = Fdcos(phi)
    3. The attempt at a solution
    F= 500 N
    d = 0.5 m
    Since the difference in the vector direction is 0, I thought that all you had to do 500*0.5 = 250 J because cos(0) equals 1. But the answer is zero and I don't understand why.
     
  2. jcsd
  3. Aug 5, 2015 #2

    haruspex

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    Equations aren't much use unless you know what all the variables in it represent.
    What is d in this equation? Yes, I know it's a distance, but what distance in relation to F, exactly?
     
  4. Aug 5, 2015 #3
    Its the distance of the barbell above the ground. So is that vector pointing down and the force is pointing up? But still cos(180) is still -1 not 0
     
  5. Aug 5, 2015 #4

    haruspex

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    That's not what I asked.
    Forget this specific problem for the moment. You quoted a standard equation. How is d defined in that equation?
     
  6. Aug 5, 2015 #5
    the displacement?
     
  7. Aug 6, 2015 #6

    haruspex

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    Displacement of what?
     
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