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What is the momentum of the box at the bottom?

  1. Oct 14, 2004 #1
    A 98.0 N box slides 25.0 m along a 35.0° incline. If the force of friction along the incline is 32.0 N, and the box starts from rest at teh top, what is the momentum of the box at the bottom?

    This is how i attempted to solve this question. :confused: The answer in the book is 1.1 * 10^2 Kgm/s. Can you please tell me waht error i have made. Thank-you.

    Eptop + Ektop = Epbot + Ekbot
    mgh + 0 = 1/2mv^2 + 0
    (9.81m/s^2)(25.0m*Sin35) = 1/2v^2
    v=16.77 m/s

    m=98.0 N/9.81m/s^2
    m=9.9898 Kg

    p=(9.9898 Kg)(16.77 m/s)
    p=168 Kgm/s
  2. jcsd
  3. Oct 14, 2004 #2


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    You didn't take into account the energy lost to friction.
  4. Oct 14, 2004 #3


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    Lol shaani nu gaal kadh :wink:
  5. Oct 14, 2004 #4
    i think i figured it out:

    Ep=Ek + Ffd
    mgh=1/2mv^s + Ffd

    THen solve for v and use that in p=mv
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