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Newton's Third Law-two masses on a wedge

  1. Oct 7, 2009 #1
    3. The attempt at a solution

    8 kg block:

    T-m2g = -m2a

    T-m1g sin(theta)- u m1g cos (theta) = m1a
    T -m2g=-m2a

    subtract equations:

    -m1gsin(theta)-umigcos(theta) + m2g =
    (m1+m2)a

    or a = g(m2-m1sin(theta)-m2cos(theta)/(m1+m2)

    substituting values:

    a=g(8-10sin(30)-0.2(10)cos30)/(18)

    a=0.69m/s/s

    b)

    vf^2=v0^2+2ad to find final velocity, knowing v0=0, a=0.69m/s/s and d=0.4m
     
    Last edited: Oct 8, 2009
  2. jcsd
  3. Oct 7, 2009 #2
    Your 1st part looks good so therefore i think the second part should be alright since its just subsituting values inside. So whats your question haha:biggrin:
     
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