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Kinetic friction on smooth then rough surface

  1. Jun 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Given a 2.0 kg mass at rest on a horizontal surface at point zero. For 30.0 m, a constant horizontal force of 6 N is applied to the mass.

    For the first 15 m, the surface is frictionless. For the second 15 m, there is friction between the surface and the mass.

    The 6 N force continues but the mass slows to rest at the end of the 30 m. The coefficient of friction between the surface and the mass is _____.

    2. Relevant equations
    F = ma
    Work-kinetic energy theorem
    F_friction = mu * m * g

    3. The attempt at a solution


    To solve this, I found the velocity at 15 m by first using F=ma to find that the acceleration for the first 15 m is 3 m/s^2.

    Then I used a kinematic equation to find that the velocity at 15 m is sqrt(90).



    So then, for the second 15 m, I drew a force diagram and saw that for the mass to decelerate to 0 in the exact same distance as it took to accelerate, then the net force must be the same magnitude but in the reverse direction to slow it down.

    So I thought that since it was previously just 6 N to the right, I thought the force of friction would have to be 12 N to the left so that the net force is 6 N to the left.



    Force of friction = coeff_fric * m * g

    So that means that 12 = coeff_fric * m * g

    Solving for coeff_fric, I got 0.6.

    But that is apparently wrong since it's supposed to be 0.3.



    But the only way to get 0.3 is if Force of friction = 6 N to the left. But I don't see why it should be 6 N instead of 12 N to the left.
     
  2. jcsd
  3. Jun 3, 2015 #2
    total force on the object in rough surface is, [(coffee-of friction)mg-F] coffee of friction = u
    by work energy theorem, kinetic energy is lost in doing work against friction,
    so F.x=1/2mv^2
    F is = [(coffee-of friction)mg-F]
    putting values we get
    (20u-6)*15=1/2*90*2 ------>15 meters is length of rough surface

    20u=12
    u=.6
    [[[[[[[[correct answer is .6]]]]]]]
    if u=.3 then both forces will balance each other so there will be no deceleration
    and body will continue
    .3 is WRONG answer
     
  4. Jun 4, 2015 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    @Sagar Singh coffee of friction = coefficient of friction? :wink:

    I agree with 0.6
     
  5. Jun 4, 2015 #4
    haha typing mistake, coefficient of friction
    i want to write, (coeff) but it auto corrected to (coffee) to coffee
     
  6. Jun 4, 2015 #5
    Thank you everyone! Just wanted to confirm.
     
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