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 18 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

mu_k = 0.282

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.orgupload.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" = 7 "m"color(white)(al (distance down ramp)

  • Deltax_"floor" = 18 "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))/(18color(white)(l)"m"+(7color(white)(l)"m")·cos((pi)/3)) = color(blue)(ulbar(|stackrel(" ")(" "0.282" ")|)

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