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Acceleration down an inclined plane

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A 25-kg box slides, from rest, down a 9.0-m-long incline that makes an angle of
    15° with the horizontal. The speed of the box when it reaches the bottom of the
    incline is 2.4 m/s. (a) What is the coefficient of kinetic friction between the box
    and the surface of the incline? (b) How much work is done on the box by the
    force of friction? (c) What is the change in the potential energy of the box?

    2. Relevant equations

    Fx = mg * sin15
    Fn = mg * cos15
    Ffr = uk * Fn = uk * mg * cos15
    Vf2-Vo2 = 2 *a*d

    3. The attempt at a solution

    Having an issue with kinetic friction calculation here

    Taking the force components and applying Newton's 2nd law -> ƩF = ma
    Fx - Ffr = max
    mg * sin 15 - uk (mg * cos 15) = m * ax

    Since I want uk and the final velocity is given, I calculate accel -> Vf2 - Vo2 = 2*a*d so a = 2.4m/s / (2*9.0m) = 0.32 m/s^2
    Can I put it into mg * sin 15 - uk (mg * cos 15) = m * ax and end up with uk = 0.2345 ?

    I can't get over that ax isn't constant acceleration and I don't think I can use it in this case....
     
    Last edited: Nov 28, 2011
  2. jcsd
  3. Nov 28, 2011 #2
    I would assume constant acceleration, since there is no external forces acting on the box and it's sliding under its own mass. Therefore, I believe your method is correct.
     
  4. Nov 28, 2011 #3

    gneill

    User Avatar

    Staff: Mentor

    What makes you think that the acceleration is not constant? Can you identify a force that changes during the slide?

    If you're leery about the acceleration you could always check the result by using an energy conservation approach. You know the potential energy (due to gravity) at the top of the slope, and the kinetic energy at the bottom (i.e. velocity). The difference of the two will be the energy lost to friction.
     
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