An object, previously at rest, slides #7 m# down a ramp, with an incline of #(pi)/3 #, and then slides horizontally on the floor for another #14 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Answer 1

#mu_k = 0.346#

WARNING: LONG-ish ANSWER!

We're asked to find the coefficient of kinetic friction, #mu_k#, between the object and the ramp.

We'll split this problem into two parts: the first part is where the object is sliding down the incline, and the second part is where it is sliding across the floor.

#-----------bb("INCLINE")-----------#

The only two forces acting on the object as it slides down the ramp are

  1. The gravitational force (its weight; acting down the ramp)

  2. The kinetic friction force (acting up the ramp because it opposes motion)

    The expression for the coefficient of kinetic friction #mu_k# is

    #ul(f_k = mu_kn#

    where

    • #f_k# is the magnitude of the retarding kinetic friction force acting as it slides down (denoted #f# in the above image)

    • #n# is the magnitude of the normal force exerted by the incline, equal to #mgcostheta# (denoted #N# in the above image)

      The expression for the net horizontal force, which I'll call #sumF_(1x)#, is

      #ul(sumF_(1x) = overbrace(mgsintheta)^"gravitational force" - overbrace(color(red)(f_k))^"kinetic friction force"#

      Since #color(red)(f_k = mu_kn)#, we have

      #sumF_(1x) = mgsintheta - color(red)(mu_kn)#

      And since the normal force #n = mgcostheta#, we can also write

      #ul(sumF_(1x) = mgsintheta - color(red)(mu_kmgcostheta)#

      Or

      #ul(sumF_(1x) = mg(sintheta - mu_kcostheta))#

      #" "#

      Using Newton's second law, we can find the expression for the acceleration #a_(1x)# of the object as it slides down the incline:

      #ul(sumF_(1x) = ma_(1x)#

      #color(red)(a_(1x)) = (sumF_(1x))/m = (mg(sintheta - mu_kcostheta))/m = color(red)(ul(g(sintheta - mu_kcostheta)#

      #" "#

      What we can now do is apply a constant-acceleration equation to find the final velocity as it exits the ramp, which we'll call #v_(1x)#:

      #ul((v_(1x))^2 = (v_(0x))^2 + 2(a_(1x))(Deltax_"ramp")#

      where

      • #v_(0x)# is the initial velocity (which is #0# since it was "previously at rest")

      • #a_(1x)# is the acceleration, which we found to be #color(red)(g(sintheta - mu_kcostheta)#

      • #Deltax_"ramp"# is the distance it travels down the ramp

        Plugging in these values:

        #(v_(1x))^2 = (0)^2 + 2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp")#

        #" "#
        #ul(v_(1x) = sqrt(2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp"))#

        This velocity is also the initial velocity of the motion along the floor.

        #-----------bb("FLOOR")-----------#

        As the object slides across the floor, the plane is perfectly horizontal, so the normal force #n# now equals

        #n = mg#

        The only horizontal force acting on the object is the retarding kinetic friction force

        #f_k = mu_kn = mu_kmg#

        (which is different than the first one).

        The net horizontal force on the object on the floor, which we'll call #sumF_(2x)#, is thus

        #ul(sumF_(2x) = -f_k = -mu_kmg#

        (the friction force is negative because it opposes the object's motion)

        Using Newton's second law again, we can find the floor acceleration #a_(2x)#:

        #color(green)(a_(2x)) = (sumF_(2x))/m = (-mu_kmg)/m = color(green)(ul(-mu_kg#

        #" "#

        We can now use the same constant-acceleration equation as before, with only a slight difference in the variables:

        #ul((v_(2x))^2 = (v_(1x))^2 + 2(a_(2x))(Deltax_"floor")#

        where this time

        • #v_(2x)# is the final velocity, which since it comes to rest will be #0#

        • #v_(1x)# is the initial velocity, which we found to be #sqrt(2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp")#

        • #a_(2x)# is the acceleration, which we found to be #color(green)(-mu_kg#

        • #Deltax_"floor"# is the distance it travels along the floor

          Plugging in these values:

          #(0)^2 = [sqrt(2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp"))color(white)(l)]^2 + 2(color(green)(-mu_kg))(Deltax_"floor")#

          Rearranging gives

          #2(color(green)(mu_kg))(Deltax_"floor") = 2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp")#

          #" "#

          At this point, we're just solving for #mu_k#, which the following indented portion covers:

          Divide both sides by #2g#:

          #mu_k(Deltax_"floor") = (sintheta - mu_kcostheta)(Deltax_"ramp")#

          Distribute:

          #mu_k(Deltax_"floor") = (Deltax_"ramp")sintheta - (Deltax_"ramp")mu_kcostheta#

          Now, we can divide all terms by #mu_k#:

          #Deltax_"floor" = ((Deltax_"ramp")sintheta)/(mu_k) - (Deltax_"ramp")costheta#

          Rearrange:

          #((Deltax_"ramp")sintheta)/(mu_k) = Deltax_"floor" + (Deltax_"ramp")costheta#

          Finally, swap #mu_k# and #Deltax_"floor" + (Deltax_"ramp")costheta#:

          #color(red)(ulbar(|stackrel(" ")(" "mu_k = ((Deltax_"ramp")sintheta)/(Deltax_"floor" + (Deltax_"ramp")costheta)" ")|)#

          #" "#

          The question gives us

          • #Deltax_"ramp" = 7# #"m"color(white)(al# (distance down ramp)

          • #Deltax_"floor" = 14# #"m"color(white)(aa# (distance along floor)

          • #theta = pi/3color(white)(aaaaaa.# (angle of inclination)

            Plugging these in:

            #color(blue)(mu_k) = ((7color(white)(l)"m")*sin((pi)/3))/(14color(white)(l)"m"+(7color(white)(l)"m")·cos((pi)/3)) = color(blue)(ulbar(|stackrel(" ")(" "0.346" ")|)#

            Notice how the coefficient doesn't depend on the mass #m# or gravitational acceleration #g# if the two surfaces are the same...

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Answer 2

To find the coefficient of kinetic friction, you can use the equation:

[ \mu_k = \frac{{m \cdot g \cdot d}}{{m \cdot g \cdot h + m \cdot g \cdot d}} ]

Where:

  • ( \mu_k ) is the coefficient of kinetic friction.
  • ( m ) is the mass of the object.
  • ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )).
  • ( d ) is the horizontal distance the object slides on the floor (14 m).
  • ( h ) is the vertical height of the ramp (given by ( h = d \cdot \sin(\pi/3) )).

Given that the object slides down a ramp with an incline of ( \pi/3 ) and then slides horizontally on the floor, the vertical height of the ramp is ( h = d \cdot \sin(\pi/3) ).

Now, you can substitute the values into the equation and solve for ( \mu_k ).

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