# A box with an initial speed of 5 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 2/5  and an incline of pi /12 . How far along the ramp will the box go?

Apr 1, 2017

The distance is $= 1.98 m$

#### Explanation:

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N$

Let the mass of the object be $= m k g$

The incline is $= \theta$

Therefore,

${F}_{r} = {\mu}_{k} \cdot N = {\mu}_{k} \cdot m g \cos \theta$

The net force on the box is

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

According to Newtons' second Law

$F = m a$

Where the acceleration is $= a h$

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

$a = - \frac{2}{5} g \cos \left(\frac{\pi}{12}\right) - g \sin \left(\frac{\pi}{12}\right)$

$a = - 6.32 m {s}^{-} 2$

We apply the equation of motion

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

$v = 0$

$u = 5 m {s}^{-} 1$

So,

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

$s = {u}^{2} / \left(2 a\right)$

$= \frac{25}{2 \cdot 6.32} = 1.98 m$