A box with an initial speed of 8 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 3/4  and an incline of (5 pi )/12 . How far along the ramp will the box go?

Apr 6, 2017

The distance is $= 2.81 m$

Explanation:

Solving in the direction of the plane ↗^+

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

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

$N = m g \cos \theta$

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

The component of the weight is

$= m g \sin \theta$

Applying Newton' second Law

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

Therefore,

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

$= - \frac{3}{4} \cdot g \cdot \cos \left(\frac{5}{12} \pi\right) - g \cdot \sin \left(\frac{5}{12} \pi\right)$

$= - 11.4 m {s}^{-} 2$

The initial velocity is $u = 8 m {s}^{-} 1$

We apply the equation of motion

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

$0 = 64 - 2 \cdot 11.4 \cdot s$

$s = \frac{64}{2 \cdot 11.4}$

$= 2.81 m$