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Coefficient of Kinetic Friction

  1. Sep 26, 2008 #1
    SOLVED

    [​IMG]

    1. The problem statement, all variables and given/known data
    Suppose the coefficient of kinetic friction between m1 and the plane in the figure is [tex]\mu[/tex]k= 0.24
    m1 = m2 = 2.5kg

    As m2 moves down, determine the magnitude of the acceleration of m1 and m2 given [tex]\Theta[/tex] = 26.

    2. Relevant equations
    F=ma


    3. The attempt at a solution
    m2g = (2.5kg)(9.8m/s2)= 24.5N
    24.5 - T = 2.5a

    (2.5kg)(9.8m/s2)sin26 = 18.6827N
    (2.5kg)(9.8m/s2)cos26 = 15.8495N
    (.24)(15.8495N) = 3.80388N

    T - 18.6827 - 3.80388 = 2.5a
    24.5 - T = 2.5a
    (-18.6827-3.80388+24.5)= (2.5a+2.5a)

    2.01342 = 5a
    a=.402 m/s2



    Why isn't my answer correct :frown:
    did I do any miscalculations?
     
    Last edited: Sep 26, 2008
  2. jcsd
  3. Sep 26, 2008 #2

    PhanthomJay

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    Science Advisor
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    Gold Member

    You had your calculator set to radians instead of degrees in your sin and cos calcs! Also, round off your answer to 2 decimal places. Otherwise, method looks good!
     
  4. Sep 26, 2008 #3
    Omg. You're totally right.
    I'm taking Calculus also so now I have to remember to check my calculator xD
    THANK YOU VERY MUCH =D
     
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