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Possible velocity question?

  1. Nov 26, 2009 #1
    hey everyone! I think this question can probably be solved using a constant acceleration formula, but since we are doing the momentum and work units in class i figured i was suppose to use those formulas to solve my question. anyways here it is :

    Paul lives on the sixth floor of an apartment complex. His window is 20.2 m above the ground. Paul notices a 7.25 kg object falling past his window at 8.50 m/s. If Ken's window is 5.00m above the ground level, how fast will that same object be falling as it passes by Ken's window?

    KE = 1/2(7.25)(8.50)^2
    KE = 261.91
    KE = W, W = FD, FD = 1/2mv^2final - 1/2mv^2initial

    261.91 = 1/2(7.25)(v^2) - 1/2(7.25)(8.50)^2
    261.91 = 3.625v^2 - 261.91
    523.82/3.625 = v^2
    vfinal = 12 m/s.
    i know that is wrong, because when i put the numbers into the formula vf^2 = vi^2 + 2ad i get 19.2 m/s which seems more reasonble. Is there anyway to figure this question out properly by using momentum/work ?
    KE = W, W = FD, FD = 1/2mv^2final - 1/2mv^2initial ( this doesnt seem to make much sense either, but i thought i saw it on my formula sheet )
    thanks for any help!
     
  2. jcsd
  3. Nov 27, 2009 #2

    Redbelly98

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    Welcome to PF :smile:

    Work W is not equal to the initial KE, so it's wrong to simply plug in the initial KE of 262 J for W.

    Instead, calculate what W=FD is by figuring out the values of F and D.
     
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