A box with an initial speed of 8 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 7/4  and an incline of (2 pi )/3 . How far along the ramp will the box go?

Jul 29, 2017

$\text{distance} = 1.87$ $\text{m }$ (if $\theta = \frac{\pi}{3}$)

Explanation:

Since there were differing answers here, I thought I'd settle it:)

I'd like to point out that there can't realistically be an inclined ramp with angle of inclination $\frac{2 \pi}{3}$...it must be between $0$ and $\frac{\pi}{2}$, so I'll choose the closest angle to this, $\frac{\pi}{3}$...

We're asked to find the distance traveled by the box given its initial speed, coefficient of kinetic friction, and angle of inclination. NOTE: the friction force actually acts down the incline as opposed to the image (because the box is traveling up the ramp).

The magnitude of the kinetic friction force ${f}_{k}$ is given by

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

where

• ${\mu}_{k}$ is the coefficient of kinetic friction ($\frac{7}{4}$)

• $n$ is the magnitude of the normal force exerted by the incline plane, equal to $m g \cos \theta$

We must first find the acceleration of the box, using Newton's second law:

$\sum F = m a$

$a = \frac{\sum F}{m}$

The net horizontal force $\sum F$ is

$\sum F = m g \sin \theta + {f}_{k} = m g \sin \theta + {\mu}_{k} m g \cos \theta$

Therefore, we have

$a = \frac{\cancel{m} g \sin \theta + \cancel{m} {u}_{k} m g \cos \theta}{\cancel{m}} = g \sin \theta + {\mu}_{k} g \cos \theta$

Plugging in known values, we have

$a = \left(9.81 \textcolor{w h i t e}{l} {\text{m/s"^2)sin((pi)/3) + 7/4(9.81color(white)(l)"m/s}}^{2}\right) \cos \left(\frac{\pi}{3}\right)$

$= 17.1$ ${\text{m/s}}^{2}$

directed down the incline, so this can also be written as

a = ul(-17.1 ${\text{m/s}}^{2}$

Now, we can use the equation

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

to find the distance it travels up the ramp before it comes to a stop.

Here,

${v}_{x} = 0$ (instantaneously at rest at maximum height)

• ${v}_{0 x} = 8$ $\text{m/s}$

• ${a}_{x} = - 17.1$ ${\text{m/s}}^{2}$

• $\Delta x =$ trying to find

Plugging in known values, we have

$0 = \left(8 \textcolor{w h i t e}{l} {\text{m/s")^2 + 2(-17.1color(white)(l)"m/s}}^{2}\right) \left(\Delta x\right)$

Deltax = color(red)(ul(1.87color(white)(l)"m"

Jul 29, 2017 