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?

Jul 23, 2017

${\mu}_{k} = 0.388$

Explanation:

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

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

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

The expression for the net horizontal force $\sum {F}_{1 x}$ is

$\sum {F}_{1 x} = {\overbrace{m g \sin \theta}}^{\text{gravitational force" - overbrace(f_k)^"kinetic friction force}}$

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

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

And since the normal force $n = m g \cos \theta$,

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

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:

$\sum {F}_{1 x} = m {a}_{1 x}$

${a}_{1 x} = \frac{\sum {F}_{1 x}}{m} = \frac{m g \sin \theta - {\mu}_{k} m g \cos \theta}{m} = g \sin \theta - {\mu}_{k} g \cos \theta$

What we can now do is apply a kinematics equation for constant acceleration to find the final velocity ${v}_{1 x}$:

${\left({v}_{1 x}\right)}^{2} = {\left({v}_{0 x}\right)}^{2} + 2 {a}_{1 x} \Delta x$

Here, the initial velocity ${v}_{0 x}$ is $0$, as we take it to start from rest (no value is given otherwise), and the distance $\Delta x$ it travels is the length of the incline ($1$ $\text{m}$), so we have

${\left({v}_{1 x}\right)}^{2} = {\left(0\right)}^{2} + 2 \left(g \sin \theta - {\mu}_{k} g \cos \theta\right) \left(1 \textcolor{w h i t e}{l} \text{m}\right)$

${v}_{1 x} = \sqrt{2 \left(g \sin \theta - {\mu}_{k} g \cos \theta\right) \left(1 \textcolor{w h i t e}{l} \text{m}\right)}$

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 $m g$. The only horizontal force acting on the object is the kinetic friction force ${f}_{k} = {\mu}_{k} n = {\mu}_{k} m g$ (which is different than the first one).

The net horizontal force $\sum {F}_{2 x}$ is thus

$\sum {F}_{2 x} = {\overbrace{- {f}_{k}}}^{\text{negative because it opposes motion}} = - {\mu}_{k} m g$

Using Newton's second law again, we can find the floor acceleration ${a}_{2 x}$:

${a}_{2 x} = \frac{\sum {F}_{2 x}}{m} = \frac{- {\mu}_{k} m g}{m} = - g {\mu}_{k}$

We can now use the same constant-acceleration equation as before, but this time the initial velocity

${v}_{1 x} = \sqrt{2 \left(g \sin \theta - {\mu}_{k} g \cos \theta\right) \left(1 \textcolor{w h i t e}{l} \text{m}\right)} \textcolor{w h i t e}{a a a}$ (${v}_{1 x}$ from above)

and the final velocity ${v}_{2 x} = 0$ (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

$2 \left(g {\mu}_{k}\right) \left(2 \textcolor{w h i t e}{l} \text{m") = 2(gsintheta - mu_kgcostheta)(1color(white)(l)"m}\right)$

At this point, we're just solving for ${\mu}_{k}$. Dividing both sides by $2 g$:

${\mu}_{k} \left(2 \textcolor{w h i t e}{l} \text{m") = (sintheta - mu_kcostheta)(1color(white)(l)"m}\right)$

Now, we can divide all terms by ${\mu}_{k}$, as well as eliminate the unit $\text{m}$:

$2 = \frac{\sin \theta}{{\mu}_{k}} - \cos \theta$

$\frac{\sin \theta}{{\mu}_{k}} = 2 + \cos \theta$

Therefore,

color(red)(mu_k = (sintheta)/(2+costheta)

Plugging in the angle $\theta = \frac{3 \pi}{8}$, 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...