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Kinetic Friction Coefficient Problem

  1. Jun 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 26.5°, the block starts to slide down the incline, traveling 3.10 m down the incline in 2.00 s. Calculate the kinetic coefficent of friction between the block and the plank.


    2. Relevant equations
    Uk = friction Coefficient
    D = Vot + 1/2at^2
    F= ma
    kinetic Friction = cos(theta)*m*g*Uk

    3. The attempt at a solution
    First i used the kinematic equation to solve for the acceleration,
    3.1 = 0 + 1/2a(2)^2
    a = 1.55 m/s^2

    Then with the acceleration i got here pluged it in the F= ma to get,
    F = m (1.55)
    F = sin(theta)*m*g - cos(theta)*m*g*Uk
    Notice the m cancels out so i get

    Sin(theta)g-Cos(theta)*g*Uk= a
    g = 9.8m/s^2

    I plugged everything in, theta = 26.5, a = 1.55, g = 9.8 and solved for Uk

    Uk turned out to be -.352 which is wrong and makes no sense... but nevertheless, .352 is also wrong.

    Any idea on what i did wrong?
    thanks for the help!
     
  2. jcsd
  3. Jun 25, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Maybe a calculation mistake?

    Using
    Sin 26.5 = .446
    Cosine 26.5 = .895

    I get a different answer.

    The Sine component is larger than the observed acceleration, so it means that the result should be a positive value.

    Check to see that you are not using radians on your calculator ... it's degrees. A common source of error.
     
  4. Jun 25, 2009 #3
    Order of operation mistake... i got .322 which is correct.
    Thanks again for your help!
     
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