Question

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?

Answer 1

`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.

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...