# A box with an initial speed of 3 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 5/2  and an incline of pi /4 . How far along the ramp will the box go?

Jul 9, 2017

The distance is $= 0.19 m$

#### Explanation:

Taking the direction up and parallel to the plane as positive ↗^+

The coefficient of kinetic friction is ${\mu}_{k} = {F}_{r} / N$

Then the net force on the object is

$F = - {F}_{r} - W \sin \theta$

$= - {F}_{r} - m g \sin \theta$

$= - {\mu}_{k} N - m g \sin \theta$

$= m {\mu}_{k} g \cos \theta - m g \sin \theta$

According to Newton's Second Law

$F = m \cdot a$

Where $a$ is the acceleration
So

$m a = - {\mu}_{k} g \cos \theta - m g \sin \theta$

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

The coefficient of kinetic friction is ${\mu}_{k} = \frac{5}{2}$

The incline of the ramp is $\theta = \frac{1}{4} \pi$

$a = - 9.8 \cdot \left(\frac{5}{2} \cos \left(\frac{1}{4} \pi\right) + \sin \left(\frac{1}{4} \pi\right)\right)$

$= - 24.3 m {s}^{-} 2$

We apply the equation of motion

${v}^{2} = {u}^{2} + 2 a s$

$u = 3 m {s}^{-} 1$

$v = 0$

$a = - 24.3 m {s}^{-} 2$

$s = \frac{{v}^{2} - {u}^{2}}{2 a}$

$= \frac{0 - 9}{- 2 \cdot 24.3}$

$= 0.19 m$