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Energy and Power Question

  1. Apr 2, 2006 #1
    Question is.. how much power does a 45 kg person develop climbing up a vertical ladder 7.0 m high if the average acceleration is 0.25 m/s^2?

    This is where I'm at... (so far anyways)

    calculate t from d = vt + 1/2at^2 and get value of 7.5 s to climb ladder... but now, how do I calculate Work in order to sub into P=W/t equation...

    I have tried this... W = (Fg + Fa)*d = [(45)(9.81) + (45)(0.25)]*7 = 3168.9

    and then P= W/t = 3168.9/7.5 = 423 W

    Answer in book is 423 W but I am confused as to why I am right.. don't Fg and Fa act in opposite directions on the FBD??

    Any help is appreciated
     
  2. jcsd
  3. Apr 2, 2006 #2

    nrqed

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    The correct way to think about this is [itex] F_{person} - m g = m a_y [/itex] (where I am working in the y direction). So the force applied by the person is [itex] F_{person} = mg + m a_y [/itex] (makes sense, the person must apply a larger force than mg in order to accelerate upward). Now you multiply this by the distance to get the work done by the person.

    Patrick
     
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