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Finding the Kinetic Friction Force using Mass and Acceleration

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data
    While sliding, the only force acting on the block in the horizontal direction is that of friction. From the mass of the block and its acceleration, you can find the frictional force and finally, the coefficient of kinetic friction.

    Okay, so we are sliding a steel block with various added weights on a flat, clean, aluminum surface. Using a motion detector and a logging program we have come to the conclusion that the acceleration(m/s2) was:
    327.6g - 2.853m/s2
    577.6g - 2.8465m/s2
    827.6g - 3.483m/s2
    1077.6g - 2.944 m/s2
    1326.6g - 2.95 m/s2

    We need to find the Kinetic Friction Force using the mass and acceleration and then use the value found to help us find the coefficient of kinetic friction.


    2. Relevant equations
    Normally, the coefficient of kinetic friction can be found by taking µk*Fn

    Convert grams to kg by moving the decimal place 3 places to the left.

    Finally, to find the weight of an object in Newtons you multiply the mass (in kg), by the acceleration due to gravity (9.8m/s2)

    3. The attempt at a solution
    I first converted the 327.6g into kg, .3276kg, and multiplied it by 2.853m/s2. The answer for this was .9346N. I then plugged that value for the Kinetic Friction Force into the equation of
    Fkf = µkFn. I divided .9346N by 3.21(the weight in N of the block) to get .2912.

    I have done this for every value and I get .30 for the coefficient of kinetic friction every time. The problem is, I don't know how to check what the actual coefficient between aluminum and steel is (I'm not sure what kind of Aluminum it is, I saw online that Anodized Aluminum + Steel is .30). This is a percent error lab so we could just be getting different answers due to human error.
     
  2. jcsd
  3. Mar 5, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi Strobot, Welcome to Physics Forums.

    As you say, there are several possibilities for the particular types of metals you've got. That you've found an example where the known value accords with what you're seeing in the lab is encouraging, but not definitive. Being a percent error lab I think you'll be dealing
    with error estimates and error propagation analysis in your calculations, right? Did you take note of error estimates for all your data readings?
     
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