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Two Hanging mass's attached to one on the table

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data
    The coefficient of kinetic friction between the 2.0 kg block in figure and the table is 0.350.

    What is the acceleration of the 2.0 kg block?

    2. Relevant equations
    F=uK*Force Normal
    Fnet=ma
    3. The attempt at a solution
    I tried to take the mass's of both the hanging and used them to calculate a net force on the block in the center and then solved for acceleration that way. Here's my work.

    3kg*9.8=29.4N Right Side Tension
    1kg*9.8=9.8N Left Side Tension
    2kg*9.8=19.6N Force Normal

    F=.35*19.6N
    6.86N

    (29.4N-6.86N)-9.8N=2kg*acceleration

    a=6.34m/s^2

    and it turns out this is wrong please help me out thank you!

    -Charan
     

    Attached Files:

  2. jcsd
  3. Oct 7, 2008 #2
    You forgot the left side tension....
     
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