# An object, previously at rest, slides 9 m down a ramp, with an incline of (pi)/12 , and then slides horizontally on the floor for another 32 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Aug 3, 2017

${\mu}_{k} = 0.0572$

#### Explanation:

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.

$- - - - - - - - - - - \boldsymbol{\text{INCLINE}} - - - - - - - - - - -$

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 $m g \cos \theta$ (denoted $N$ in the above image)

The expression for the net horizontal force, which I'll call $\sum {F}_{1 x}$, is

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

Since $\textcolor{red}{{f}_{k} = {\mu}_{k} n}$, we have

$\sum {F}_{1 x} = m g \sin \theta - \textcolor{red}{{\mu}_{k} n}$

And since the normal force $n = m g \cos \theta$, we can also write

ul(sumF_(1x) = mgsintheta - color(red)(mu_kmgcostheta)

Or

$\underline{\sum {F}_{1 x} = m g \left(\sin \theta - {\mu}_{k} \cos \theta\right)}$

$\text{ }$

Using Newton's second law, we can find the expression for the acceleration ${a}_{1 x}$ 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)

$\text{ }$

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 ${v}_{1 x}$:

ul((v_(1x))^2 = (v_(0x))^2 + 2(a_(1x))(Deltax_"ramp")

where

• ${v}_{0 x}$ is the initial velocity (which is $0$ since it was "previously at rest")

• ${a}_{1 x}$ is the acceleration, which we found to be color(red)(g(sintheta - mu_kcostheta)

• $\Delta {x}_{\text{ramp}}$ is the distance it travels down the ramp

Plugging in these values:

${\left({v}_{1 x}\right)}^{2} = {\left(0\right)}^{2} + 2 \textcolor{red}{g \left(\sin \theta - {\mu}_{k} \cos \theta\right)} \left(\Delta {x}_{\text{ramp}}\right)$

$\text{ }$
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.

$- - - - - - - - - - - \boldsymbol{\text{FLOOR}} - - - - - - - - - - -$

As the object slides across the floor, the plane is perfectly horizontal, so the normal force $n$ now equals

$n = m g$

The only horizontal force acting on the object is the retarding kinetic friction force

${f}_{k} = {\mu}_{k} n = {\mu}_{k} m g$

(which is different than the first one).

The net horizontal force on the object on the floor, which we'll call $\sum {F}_{2 x}$, 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 ${a}_{2 x}$:

color(green)(a_(2x)) = (sumF_(2x))/m = (-mu_kmg)/m = color(green)(ul(-mu_kg

$\text{ }$

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

• ${v}_{2 x}$ is the final velocity, which since it comes to rest will be $0$

• ${v}_{1 x}$ is the initial velocity, which we found to be sqrt(2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp")

• ${a}_{2 x}$ is the acceleration, which we found to be color(green)(-mu_kg

• $\Delta {x}_{\text{floor}}$ 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

$2 \left(\textcolor{g r e e n}{{\mu}_{k} g}\right) \left(\Delta {x}_{\text{floor") = 2color(red)(g(sintheta - mu_kcostheta))(Deltax_"ramp}}\right)$

$\text{ }$

At this point, we're just solving for ${\mu}_{k}$, which the following indented portion covers:

Divide both sides by $2 g$:

${\mu}_{k} \left(\Delta {x}_{\text{floor") = (sintheta - mu_kcostheta)(Deltax_"ramp}}\right)$

Distribute:

${\mu}_{k} \left(\Delta {x}_{\text{floor") = (Deltax_"ramp")sintheta - (Deltax_"ramp}}\right) {\mu}_{k} \cos \theta$

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)" ")|)

$\text{ }$

The question gives us

• $\Delta {x}_{\text{ramp}} = 9$ "m"color(white)(al (distance down ramp)

• $\Delta {x}_{\text{floor}} = 32$ "m"color(white)(aa (distance along floor)

• theta = pi/12color(white)(aaaaaa. (angle of inclination)

Plugging these in:

$\textcolor{b l u e}{{\mu}_{k}} = \left(\left(9 \textcolor{w h i t e}{l} \text{m")*sin((pi)/12))/(32color(white)(l)"m"+(9color(white)(l)"m")·cos((pi)/12)) = color(blue)(ulbar(|stackrel(" ")(" "0.0572" }\right) |\right)$

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