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Coefficient of kinetic friction on 2 different parts

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data
    When the mass M is at the position shown, it has a speed v0 = 2.17 m/s and is sliding down the inclined part of a slide. The mass reaches the bottom of the incline and then travels a distance S2 = 2.45 m along the horizontal part of the slide before stopping. The distance S1 = 1.11 m and the angle of the incline is θ = 31.1o. Calculate the coefficient of kinetic friction for the mass on the surface.

    PhysicsGraph.jpg


    2. Relevant equations
    E= KE + U

    E+ W (ext) = 0 -> E = -W (ext) (which is the work done by frictional force)


    3. The attempt at a solution
    So far I've analysed the x and y components of the mass on path S1 and path S2. This is what I've got

    Path S1:
    y-dir: N = mgcosθ
    x-dir: ƩF = -F(friction) + mgsinθ = -μmgcosθ + mgsinθ

    So W1 ( work done on S1) = S1(-μmgcosθ + mgsinθ)

    Path S2:
    W2 = -μmg(S2)

    Since -W (total) = E
    W1+W2 = KE + U (at original position)
    S1(-μmgcosθ + mgsinθ) - μmg(S2) = - [(1/2)mv2 + mgS1sinθ]

    μ(S1gcosθ + gS2) = 2S1gsinθ + v2/2

    μ = (2S1gsinθ + v2/2)/(S1gcosθ + gS2)

    So far I've got μ = .105 but it's wrong!

    I'd really appreciate if anyone helped me out with this cos I totally have no idea how to figure out the right answer!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 27, 2012 #2
    You can begin with part 2 and write an equation that relates the velocity at the bottom of the hill to the distance covered until it stops on the horizontal. To do this, relate the KE (including unknown V at bottom of part 1) to the work done overcoming friction. That gives you an equation with two unknowns, V and mu. There are no potential energy changes when moving horizontally.

    Look at part 1 and write your energy balances. This will give you another equation with two unknowns, again being V and mu. This time you have to consider potential and kinetic energy changes as well as work. Eliminate V and solve for mu.
     
  4. Apr 27, 2012 #3
    I'm not sure if i got u right. Does that mean that on part 1 that i have the potential energy change and the change in the KE too??? (the velocity at the end of S1 is not 0?) and then on part 2 I just have the change in kinetic energy, which also equals to the work done by frictional force on S2?
     
  5. Apr 27, 2012 #4
    I'm not sure if i got u right. Does that mean that on part 1 that i have the potential energy change and the change in the KE too??? (the velocity at the end of S1 is not 0?) and then on part 2 I just have the change in kinetic energy, which also equals to the work done by frictional force on S2?
     
  6. Apr 27, 2012 #5
    Does that mean that on part 1 that i have the potential energy change and the change in the KE too??? (the velocity at the end of S1 is not 0?)

    Correct. The mass has an initial velocity and it will also have a final velocity that is different and unknown but it is the initial velocity for part 2. The initial velocity at the start of part 1 is given.

    On part 2 the kinetic energy change is what it was at the end of part 1 and zero and it equals the work done against friction.
     
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