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Kinetic Friction Lab Question

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    A 4.5 kg block is on an incline with a coefficient of static friction of 0.50. The angle that the ramp makes with the horizontal is increased gradually, until the block begins to slide down the ramp. If you know that the coefficient of kinetic friction between the block and the plane is 0.10, find the time it will take the block to slide down the ramp a distance of 0.65 m starting from rest.


    2. Relevant equations

    Kf = (coef of frict)*m*g*cos(theta)



    3. The attempt at a solution

    tan^-1 (.5) = 26.6
    theta = 26.6 degrees
    F_g = mg
    F_g = (4.5)*(9.81) = 44.1 N
    F_gx = 44.1*sin(26.6) = 19.7
    a_x = 19.7 N / 4.5 kg = 4.39 m/s^2
    4.39 - (coef of frict)*m*g*cos(theta)
    4.39 - (.1)(4.5)cos(26.6)
    4.39 - .402 = 3.99 m/s^2
    d = .65
    d = 1/2at^s
    .65 = 1/2 (3.99) t^2
    .65 = 1.995t^2
    t^2 = .326
    t = .571 seconds.


    I'm really confused, as we haven't covered this stuff in lecture, yet we have labs that deal with it. I'm sorry if I completely fudged up something because I don't know what I'm doing...
     
  2. jcsd
  3. Sep 15, 2013 #2

    PhanthomJay

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    ok so far
    you went astray here. Do not calculate acceleration in x direction until you have first determined all forces acting on the block in the x direction. You should note your units become inconsistent when subtracting force units from acceleration units.
     
  4. Sep 15, 2013 #3
    ok, so I should subtract Kinetic force from my F_gx first before calculating acceleration?
     
  5. Sep 15, 2013 #4

    PhanthomJay

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    Yes.


    I see things haven't changed much in the 40 years since my college days!
     
  6. Sep 15, 2013 #5
    Yep, force due to kinetic friction.

    Then, kinematics.
     
  7. Sep 15, 2013 #6
    Haha I guess not! Thank you all for your help, .61 s is the answer. I will definitely come back to this forum for help (if I need it) again!
     
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