Question

An object, previously at rest, slides `6 m` down a ramp, with an incline of `(3pi)/8`, and then slides horizontally on the floor for another `16 m`. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Answer 1

`mu_k = 0.303`

Explanation:

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" = 6` `"m"color(white)(al` (distance down ramp)

• `Deltax_"floor" = 16` `"m"color(white)(aa` (distance along floor)

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

Plugging these in:

`color(blue)(mu_k) = ((6color(white)(l)"m")*sin((3pi)/8))/(16color(white)(l)"m"+(6color(white)(l)"m")·cos((3pi)/8)) = color(blue)(ulbar(|stackrel(" ")(" "0.303" ")|)`

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