# A box with an initial speed of 3 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 5/7  and an incline of ( 5 pi )/12 . How far along the ramp will the box go?

Aug 14, 2017

$\text{distance} = 0.399$ $\text{m}$

#### Explanation:

We're asked to find how far the box will travel up the ramp, given its initial speed, the coefficient of kinetic friction, and the angle of inclination.

I will solve this problem using only Newton's laws and kinematics (i.e. without using work/energy).

I will take the positive $x$-direction as *up the ramp.*

When the box reaches its maximum distance, the instantaneous velocity will be $0$. We are ultimately going to use the constant-acceleration equation

ul((v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)

where

• ${v}_{x}$ is the instantaneous velocity (which is $0$)

• ${v}_{0 x}$ is the initial velocity

• ${a}_{x}$ is the (constant) acceleration

• $\Delta x$ is the distance it travels (what we're trying to find)

Since the velocity ${v}_{x} = 0$, we can also write this equation as

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

We also figure that the acceleration will be negative because it slows down and comes to a brief stop at its maximum height, so we then have

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

And we can move it to the other side:

ul(2a_x(Deltax) = (v_(0x))^2

Rearranging for the distance traveled $\Delta x$:

$\underline{\overline{| \stackrel{\text{ ")(" "Deltax = ((v_(0x))^2)/(2a_x)" }}{|}}}$

We already know the initial velocity, so we need to find the acceleration of the box.

$\text{ }$

Let's use Newton's second law of motion to find the acceleration, which is

ul(sumF_x = ma_x

where

• $\sum {F}_{x}$ is the net force acting on the box

• $m$ is the mass of the box

• ${a}_{x}$ is the acceleration of the box (what we're trying to find)

The only forces acting on the box are

• the gravitational force (acting down the ramp), equal to $m g \sin \theta$

• the friction force (acting down the ramp), equal to ${\mu}_{k} n$

And so we have our net force equation:

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

The normal force $n$ exerted by the incline is equal to

$n = \textcolor{p u r p \le}{m g \cos \theta}$

So we can plug this in to the net force equation above:

ul(sumF_x = mgsintheta + mu_kcolor(purple)(mgcostheta)

Or

ul(sumF_x = mg(sintheta + mu_kcostheta)

Now, we can plug this in for $\sum {F}_{x}$ in the Newton's second law equation:

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

$m {a}_{x} = m g \left(\sin \theta + {\mu}_{k} \cos \theta\right)$

We can cancel the mass $m$ by dividing both sides by $m$, leaving us with

color(green)(ul(a_x = g(sintheta + mu_kcostheta)

$\text{ }$

Now that we have found an expression for the acceleration, let's plug it into the equation

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

And we get

color(red)(ulbar(|stackrel(" ")(" "Deltax = ((v_(0x))^2)/(2g(sintheta + mu_kcostheta))" ")|)

We're given in the problem

• ${v}_{0 x} = 3$ $\text{m/s}$

• $\theta = \frac{5 \pi}{12}$

• ${\mu}_{k} = \frac{5}{7}$

• and the gravitational acceleration $g = 9.81$ $\text{m/s}$

Plugging these in:

$\textcolor{b l u e}{\Delta x} = \left(\left(3 \textcolor{w h i t e}{l} \text{m/s")^2)/(2(9.81color(white)(l)"m/s"^2)(sin[(5pi)/12] + 5/7cos[(pi)/12])) = color(blue)(ulbar(|stackrel(" ")(" "0.399color(white)(l)"m"" }\right) |\right)$