A box with an initial speed of 6 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 1/2  and an incline of (3 pi )/8 . How far along the ramp will the box go?

Jun 10, 2016

I found $1.6 m$ (but double check my maths!)

Explanation:

Consider the diagram: We can use Newton's Second Law: $\Sigma \vec{F} = m \vec{a}$
Where:
along the horizontal (along the ramp):
$- m g \sin \left(3 \frac{\pi}{8}\right) - f = m a$
along the vertical:
$N - m g \cos \left(5 \frac{\pi}{8}\right) = 0$
$f = \mu N$ is friction
$m g \sin \left(5 \frac{\pi}{8}\right)$ and $m g \cos \left(5 \frac{\pi}{8}\right)$ are the horizontal and vertical components of weight.

We also need to change $a$ into an expression involving the distance $s$;
we can use kinematics and:
${v}_{f}^{2} = {v}_{i}^{2} + 2 a s$
$a = \frac{{v}_{f}^{2} - {v}_{i}^{2}}{2 s}$
with
${v}_{i} = 6 \frac{m}{s}$
${v}_{f} = 0$

So we get:
$- m g \sin \left(3 \frac{\pi}{8}\right) - f = m a$
$- m g \sin \left(3 \frac{\pi}{8}\right) - \mu N = m \frac{{v}_{f}^{2} - {v}_{i}^{2}}{2 s}$
$- \cancel{m} g \sin \left(3 \frac{\pi}{8}\right) - \cancel{m} \mu g \cos \left(5 \frac{\pi}{8}\right) = \cancel{m} \frac{{v}_{f}^{2} - {v}_{i}^{2}}{2 s}$
$- 9.8 \sin \left(3 \frac{\pi}{8}\right) - \left(\frac{1}{2}\right) 9.8 \cos \left(5 \frac{\pi}{8}\right) = \frac{{0}^{2} - {6}^{2}}{2 s}$
$- 9.1 - 1.8 = - \frac{36}{2 s}$
$s = \frac{1}{2} \left(\frac{36}{10.9}\right) = 1.6 m$