# A box with an initial speed of 7 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 2/5  and an incline of pi /4 . How far along the ramp will the box go?

Mar 27, 2018

The distance is $= 2.53 m$

#### Explanation:

Resolving in 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 of Motion

$F = m \cdot a$

Where $a$ is the of the box

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 ${\mu}_{k} = \frac{2}{5}$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

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

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

$= - 9.7 m {s}^{-} 2$

Apply the equation of motion

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

The initial velocity is $u = 7 m {s}^{-} 1$

The final velocity is $v = 0$

The acceleration is $a = - 9.7 m {s}^{-} 2$

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

$= \frac{0 - 49}{- 2 \cdot 9.7}$

$= 2.53 m$