# A box with an initial speed of 6 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 2/5  and an incline of ( 5 pi )/12 . How far along the ramp will the box go?

Jul 12, 2017

Ans: 1.9 m

#### Explanation:

Given:
u = 6m/s.
$\mu$ =2/5 = 0.6
$\theta = \frac{5 \pi}{12} = 75 \mathrm{de} g r e e s$

For an inclined plane,

$a = g \left(\sin \theta - \mu \cos \theta\right)$
$a = 9.8 \left(\sin 75 - 0.4 \cdot \cos 75\right)$
$a = 8.45 \frac{m}{{s}^{2}}$

Again, $v = u - \left(a \sin \theta\right) t$
$\frac{u}{a \sin \theta} = t$
$t = 0.634 s$

Thus,
$s = u t - \frac{1}{2} g \sin \theta \cdot {t}^{2}$

Therefore, $s = 1.9 m$

Thank you!!