# An object, previously at rest, slides 9 m down a ramp, with an incline of (pi)/3 , and then slides horizontally on the floor for another 6 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Aug 5, 2017

${\mu}_{k} = 0.742$

#### 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)

• $\Delta {x}_{\text{floor}} = 6$ "m"color(white)(aa (distance along floor)

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

Plugging these in:

$\textcolor{b l u e}{{\mu}_{k}} = \left(\left(9 \textcolor{w h i t e}{l} \text{m")*sin((pi)/3))/(6color(white)(l)"m"+(9color(white)(l)"m")·cos((pi)/3)) = color(blue)(ulbar(|stackrel(" ")(" "0.742" }\right) |\right)$

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