# A box with an initial speed of 4 m/s is moving up a ramp. The ramp has a kinetic friction coefficient of 1/3  and an incline of (2 pi )/3 . How far along the ramp will the box go?

Jun 14, 2016

$\Delta x = \frac{4 \sqrt{3}}{g}$[m]

#### Explanation:

The initial speed is ${v}_{0}$ so the initial kinetic energy is given by

${E}_{k} = \frac{1}{2} m {v}_{0}^{2}$

The box friction work loses along the ramp are given by

$\mu \times m g \times \sin \left(\theta\right) \Delta x$

with $\mu$ the kinetic friction coefficient, $\theta$ the ramp incline, and $\Delta x$ the covered distance along the ramp.

So we can equate

$\frac{1}{2} m {v}_{0}^{2} = \mu \times m g \times \sin \left(\theta\right) \Delta x + m g \sin \left(\theta\right) \Delta x$

Solving for $\Delta x$

$\Delta x = \frac{\frac{1}{2} {v}_{0}^{2}}{\left(\mu + 1\right) \sin \left(\theta\right) g} = \frac{4 \sqrt{3}}{g}$[m]