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Block moving up an incline with friction

  1. Aug 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A block of mass m is moving along a horizontal frictionless surface with a speed of 5.7 m/s. The block encounters a slope which it travels up. If the slope is 11 degrees and the coefficient of kinetic friction between the block and the incline is 0.26, how far does the block travel up the incline?


    2. Relevant equations
    fk = uk*Fn

    K1 + U1+ Wother = K2 + U2


    3. The attempt at a solution

    Using a free body diagram I found the normal force, Fn, to be

    Fn - mgcos(11) =0
    Fn = 2.50119

    Then using K1 + U1 + Wother = K2 + U2

    1/2*m*v^2 - 2.50119m = 9.8*m*h

    I solved for h to find the height that the block travels up the ramp
    h = 1.40243 m

    Then using SOHCAHTOA I solved for the length of the ramp that the block traveled:
    1.40243/sin(11) = 7.34992m

    I know the answer should be 3.7 meters but I don't know where I'm going wrong.
     
  2. jcsd
  3. Aug 9, 2012 #2

    PeterO

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    Homework Helper

    I think your "value" of Fn should have an m in it [you had it in later so probably just a typo] HOWEVER, 11 degrees is quite a slight slope, so I think the normal force should be much closer to that from a horizontal surface. not the full 9.8m, but probably not below 8m. Was your calculator set to degrees when you found cos(11)?

    I would probably calculate how much energy is lost to friction and converted to Potential energy [mgh] for each metre travelled up the slope, then by comparison to the initial Kinetic energy. compute how many metres along the slope it would go.
     
  4. Aug 9, 2012 #3
    I've made a mistake with my labeling. Fn = 2.50119m is actually Fn*uk = fk. So I actually mean that fk = 2.50119.

    I Fn = 9.61995m

    If I labelled the distance that the block goes up the slope D, do I need to multiple my value of fk by D?
     
  5. Aug 9, 2012 #4

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah. The work done by friction is equal to the frictional force multiplied by the distance over which it acts. In this case, that's the distance along the incline that the object travels.
     
  6. Aug 9, 2012 #5
    I've got it now:

    1/2*m*(5.7)^2 - 0.26*m*9.8*cos(11)*d= 9.8*m*dsin(11)
    16.245 = 4.37112d

    d=16.245/4.37112
    d= 3.7 m

    where d is the distance the block travels up the ramp.

    Thanks for your help.
     
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