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Kinetic friction coefficient

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4.94 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 12.1 N at an angle θ = 15.0° above the horizontal, as shown in the Figure.

    The coefficient of kinetic friction between the block and the floor is 0.100. What is the speed of the block 3.70 s after it starts moving?


    2. Relevant equations
    F=ma


    3. The attempt at a solution

    I understand to find the acceleration I need F=ma, so cos(15)12.1 = 4.94(a).
    And I know that when I find the (a), I can just * the 3.7 seconds. I just don't know where to fit in the kinetic friction of 0.1. I'm guessing it has to be minus from the total F but something isn't going right, any ideas, thanks
     
  2. jcsd
  3. Sep 21, 2009 #2
    Erm not exactly...draw the free body diagram and resolve the x and y component of the force applied and you can find the static friction by using uN where u is static coefficient and N is the normal force on the block.
     
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