An object, previously at rest, slides #3 m# down a ramp, with an incline of #(3pi)/8 #, and then slides horizontally on the floor for another #2 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

1 Answer
Aug 3, 2017

Answer:

#mu_k = 0.880#

Explanation:

WARNING: LONG-ish ANSWER!

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

upload.wikimedia.org

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" = 3# #"m"color(white)(al# (distance down ramp)

  • #Deltax_"floor" = 2# #"m"color(white)(aa# (distance along floor)

  • #theta = (3pi)/8color(white)(aaaaaa.# (angle of inclination)

Plugging these in:

#color(blue)(mu_k) = ((3color(white)(l)"m")*sin((3pi)/8))/(2color(white)(l)"m"+(3color(white)(l)"m")·cos((3pi)/8)) = color(blue)(ulbar(|stackrel(" ")(" "0.880" ")|)#

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