# A box with an initial speed of 9 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 5/4  and an incline of (3 pi )/4 . How far along the ramp will the box go?

Feb 18, 2017

#### Answer:

The box will go $- 23.36 m$etres along the ramp.

#### Explanation:

If a block is moving up the inclined plane then the distance travelled by the block up the plane is
$s = \left\{\frac{{u}^{2}}{2 g \left(\sin \theta + {\mu}_{k} \cos \theta\right)}\right\}$
Assume box as block and ramp as inclined plane and also consider $g = 9.8 m / {s}^{2}$.
Calculating with given values,
$s = \left\{{\left[9 m / s\right]}^{2} / \left[2 \times \left(9.8 m / {s}^{2}\right) \times \left[\sin \left(\frac{3 \pi}{4}\right) + \left(\frac{5}{4}\right) \cos \left(\frac{3 \pi}{4}\right)\right]\right]\right\}$
$\therefore s = - 23.36 m$.
Here'$-$' sign denotes that the body is travelling up the incline plane.