# A box with an initial speed of 2 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 3/2  and an incline of (2 pi )/3 . How far along the ramp will the box go?

Aug 5, 2017

$\text{distance} = 0.126$ $\text{m}$

#### 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).

NOTE: Ideally, the angle of inclination would be between $0$ and $\frac{\pi}{2}$, so I'll choose the corresponding first-quadrant angle of $\frac{\pi}{3}$.

I will also 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

• ${v}_{0 x} = 2$ $\text{m/s}$

• $\theta = \frac{\pi}{3}$

• ${\mu}_{k} = \frac{3}{2}$

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

Plugging these in:

$\textcolor{b l u e}{\Delta x} = \left(\left(2 \textcolor{w h i t e}{l} \text{m/s")^2)/(2(9.81color(white)(l)"m/s"^2)(sin[(pi)/3] + 3/2cos[(pi)/3])) = color(blue)(ulbar(|stackrel(" ")(" "0.126color(white)(l)"m"" }\right) |\right)$

Aug 5, 2017

Here's my own attempt at this problem, using an alternative approach to Nathan's, to see if our answers agree. Like Nathan, I will assume that you mean to have the ramp right-side-up, i.e. an angle of ${60}^{\circ}$ w.r.t. the horizontal...

I also got $\text{0.126 m}$.

My approach here is via conservation of energy:

$\Delta E = \Delta K + \Delta U + {W}_{{\vec{F}}_{k}} = 0$

where:

• My coordinate axes are almost the usual vertical $y$ and horizontal $x$. Rightwards is $+ x$, but downwards is $+ y$.
• $\Delta K = \frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2} = \frac{m}{2} \left({v}_{f}^{2} - {v}_{i}^{2}\right)$
• $\Delta U = m \vec{g} \Delta \vec{y} = m g {y}_{f}$, where ${y}_{i} = 0$, ${y}_{f} < 0$, and $g > 0$. The potential energy becomes more negative due to the sign of ${y}_{f}$.
• ${W}_{{\vec{F}}_{k}} = {\vec{F}}_{k} d$ is the counteracting work due to kinetic friction (negative w.r.t. the box, the system). In this case, I define

$d = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}} - 0 = \sqrt{{x}_{f}^{2} + {y}_{f}^{2}}$ with ${x}_{i} = 0$ and ${x}_{f} > 0$. $d$ is then the final position, $< 0$.

Currently, we know we have a nonzero initial velocity and a zero final velocity when the box comes to a stop:

$\implies \Delta K = - \frac{m}{2} {v}_{i}^{2}$

This so far gives:

$- \frac{m}{2} {v}_{i}^{2} + m g {y}_{f} + {\vec{F}}_{k} d = 0$

or

$\frac{m}{2} {v}_{i}^{2} - m g {y}_{f} - {\vec{F}}_{k} d = 0$

Using the sum of the forces:

${\sum}_{i} {\vec{F}}_{\bot , i} = {\vec{F}}_{N} - m \vec{g} \cos \theta = 0$

So...

$0 = \frac{\cancel{m}}{2} {v}_{i}^{2} - \cancel{m} g {y}_{f} - {\mu}_{k} \cancel{m} \vec{g} \cos \theta d$

$\implies \frac{1}{2} {v}_{i}^{2} - g {y}_{f} - {\mu}_{k} g \cos \theta d = 0$

Now, we can rewrite ${y}_{f}$ in terms of $d$, the distance traveled along the ramp:

${y}_{f} = \mathrm{ds} \int h \eta < 0$

This gives:

$\implies \frac{1}{2} {v}_{i}^{2} - g \mathrm{ds} \int h \eta - {\mu}_{k} g \mathrm{dc} o s \theta = 0$

Solve for $d$ to get:

$d = \frac{\frac{1}{2} {v}_{i}^{2}}{g \sin \theta + {\mu}_{k} g \cos \theta}$

$= \frac{{v}_{i}^{2}}{2 g \left(\sin \theta + {\mu}_{k} \cos \theta\right)}$

(which is also what Nathan found, by the way.)

So, in the end, we get:

$\textcolor{b l u e}{| d |} = \frac{{2}^{2}}{2 \left(9.81\right) \left(\sin \left({60}^{\circ}\right) + \frac{3}{2} \cos \left({60}^{\circ}\right)\right)}$

$=$ $\textcolor{b l u e}{\text{0.126 m}}$