An object, previously at rest, slides #1 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
Jul 23, 2017

Answer:

#mu_k = 0.388#

Explanation:

WARNING: Long-ish answer

We're asked to find the coefficient of kinetic friction (#mu_k#) of an object sliding down a 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.

Incline:

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

#f_k = mu_kn#

where

  • #f_k# is the magnitude of the retarding 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 #sumF_(1x)# is

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

Since #f_k = mu_kn#, we have

#sumF_(1x) = mgsintheta - mu_kn#

And since the normal force #n = mgcostheta#,

#sumF_(1x) = mgsintheta - mu_kmgcostheta#

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

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

#a_(1x) = (sumF_(1x))/m = (mgsintheta - mu_kmgcostheta)/m = gsintheta - mu_kgcostheta#

What we can now do is apply a kinematics equation for constant acceleration to find the final velocity #v_(1x)#:

#(v_(1x))^2 = (v_(0x))^2 + 2a_(1x)Deltax#

Here, the initial velocity #v_(0x)# is #0#, as we take it to start from rest (no value is given otherwise), and the distance #Deltax# it travels is the length of the incline (#1# #"m"#), so we have

#(v_(1x))^2 = (0)^2 + 2(gsintheta - mu_kgcostheta)(1color(white)(l)"m")#

#v_(1x) = sqrt(2(gsintheta - mu_kgcostheta)(1color(white)(l)"m"))#

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

Floor:

As the object slides across the floor, the plane is perfectly horizontal, so the normal force #n# now equals #mg#. The only horizontal force acting on the object is the kinetic friction force #f_k = mu_kn = mu_kmg# (which is different than the first one).

The net horizontal force #sumF_(2x)# is thus

#sumF_(2x) = overbrace(-f_k)^"negative because it opposes motion" = -mu_kmg#

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

#a_(2x) = (sumF_(2x))/m = (-mu_kmg)/m = -gmu_k#

We can now use the same constant-acceleration equation as before, but this time the initial velocity

#v_(1x) = sqrt(2(gsintheta - mu_kgcostheta)(1color(white)(l)"m"))color(white)(aaa)# (#v_(1x)# from above)

and the final velocity #v_(2x) = 0# (it comes to rest).

Plugging in known values, we have

#(0)^2 = (sqrt(2(gsintheta - mu_kgcostheta)(1color(white)(l)"m")))^2 + 2(-gmu_k)(2color(white)(l)"m")#

Rearranging gives

#2(gmu_k)(2color(white)(l)"m") = 2(gsintheta - mu_kgcostheta)(1color(white)(l)"m")#

At this point, we're just solving for #mu_k#. Dividing both sides by #2g#:

#mu_k(2color(white)(l)"m") = (sintheta - mu_kcostheta)(1color(white)(l)"m")#

Now, we can divide all terms by #mu_k#, as well as eliminate the unit #"m"#:

#2 = (sintheta)/(mu_k) - costheta#

#(sintheta)/(mu_k) = 2+costheta#

Therefore,

#color(red)(mu_k = (sintheta)/(2+costheta)#

Plugging in the angle #theta = (3pi)/8#, we have

#mu_k = (sin((3pi)/8))/(2+cos((3pi)/8)) = color(blue)(0.388#

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