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Homework Help: Varying coefficient of friction

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at .100 at P and increases linearly with a distance past P, reaching a value of .600 at 12.5m past point P. a) Use the work energy theorem to find how far this
    box slides before stopping.

    2. Relevant equations
    Work = .5mV1-.5mV2
    K= .5mv^2

    3. The attempt at a solution
    I am not really sure where to begin, if someone could steer me in the right direction, I would be grateful. I know that the friction area will cause a deceleration until the box stops. I know that the net force of Fx will be the F force in the positive x direction minus the Friction force in the negative x direction. I feel like I need the mass of the box and since it is not given, it must not be necessary and that means I am missing something.
    I believe the s vector will be 12.5 since that is the distance of the friction area.


    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 14, 2007 #2


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    Staff Emeritus
    Science Advisor

    Since the friction force is related to the coefficient of friction [itex]\mu[/itex], one can use the fact that the kinetic energy is dissipated by the friction over the distance from P to the point that it stops, where the kinetic energy is of course zero. The friction force is simply [itex]\mu[/itex](x)*mg.

    So the initial KE = 1/2 mv2 = energy dissipated (or work done by friction). Note the mg is constant, but the friction coefficient is a function of x or the distance traveled.
  4. Nov 14, 2010 #3
    Alright so I did relate that Work is equal to the change in kinetic energy which is also equal to the magnitude of Force times the magnitude of the change in distance.

    By doing so, I get that the initial Kinetic Energy equals mg(0.04x + 0.1)x

    The equation for u(x) = 0.04x + 0.1. The distance travelled during the process is x.

    When I solve for x, I get the answer 3.984 m, but this seems to be the wrong answer. Does anyone see where I went wrong?
  5. Nov 15, 2010 #4
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