posted by Jeanne

A box, with a weight of mg = 25 N, is placed at the top of a ramp and released from rest. The ramp measures 4.40 meters horizontally and 3.50 meters vertically. The box accelerates down the incline, attaining a kinetic energy at the bottom of the ramp of 54.0 J. There is a force of kinetic friction acting on the box as it slides down the incline.

a) What is the coefficient of kinetic friction between the box and ramp? & please show work.

  1. drwls

    Equate the work done against friction to the mechanical (potential + kinetic) energy loss. That will allow you to solve for the friction coefficient, Uk.

    The slope of the ramp is arctan 3.5/4.4 , which is 38.5 degrees.

    The hypotenuse of the ramp is 5.622 m

    Show your work for further assistance.

  2. Jeanne

    I know the work done is

    W = ì X mgcosè X d

    and W is Ui+ W= Kf which is 54-(25*3.3)=-33.5

    so W = ì X mgcosè X d

    and so ì= (W/(mgcosè X d))

    and mg=25N so what is cosè? and i think d is the hypotenuse?

  3. Jeanne

    im sorry it should be cos theta for all of them

  4. Jeanne

    and the "i" looking thing is the coefficient of friction

  5. Damon

    It would be easier to do it the way Dr WLS suggested.
    vertical distance down = 3.5 m
    so potential energy lost = mgh=25*3.5 =87.5 Joules
    It has Ke of 54 Joules at the bottom so lost (87.5-54) = 33.5 Joules to friction
    Friction force * length of ramp = 33.5
    length of ramp = sqrt(3.5^2+4.4^2) = 5.62 meters
    so friction force = 33.5/5.62 = 5.96 Newtons
    5.96 = mu*normal force
    5.96 = mu * (25 cos slope)
    cos slope = 4.4/5.62
    mu = 5.96 *5.62/(4.4*25) = .238

  6. Jeanne

    Ohh I see! Thanks a lot!

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