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

Aug 1, 2017

$\text{distance} = 0.959$ $\text{m}$

Explanation:

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 ($\frac{7}{5}$)

• $n$ 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 = (cancel(m)gsintheta + cancel(m)u_kmgcostheta)/(cancel(m)) = ul(gsintheta + mu_kgcostheta

Plugging in known values, we have

$a = \left(9.81 \textcolor{w h i t e}{l} {\text{m/s"^2)sin((5pi)/12) + 7/5(9.81color(white)(l)"m/s}}^{2}\right) \cos \left(\frac{5 \pi}{12}\right)$

$= 13.0$ ${\text{m/s}}^{2}$

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

a = ul(-13.0color(white)(l)"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)

• ${v}_{0 x} = 5$ $\text{m/s}$ (given initial velocity)

• ${a}_{x} = - 13.0$ ${\text{m/s}}^{2}$

• $\Delta x =$ trying to find

Plugging in known values, we have

$0 = \left(5 \textcolor{w h i t e}{l} {\text{m/s")^2 + 2(-13.0color(white)(l)"m/s}}^{2}\right) \left(\Delta x\right)$

Deltax = color(red)(ul(0.959color(white)(l)"m"