# Cans of 46.0 N slide down the ramp 24.0 deg above the horizontal at a constant speed of 3.40 m/s and then slide onto a table made of the same material. How far does each slide on the table’s horizontal surface before coming to rest?

Dec 20, 2016

$\frac{1}{2} \cdot {3.4}^{2} / 9.81 \cdot \cot \left({24}^{\circ}\right)$

#### Explanation:

Calling

$\theta = {24}^{\circ}$ ramp declivity.
$v = 3.40$ speed along ramp.
$m = 46.0$ can mass.
$g = 9.81$ gravity acceleration.
$\mu$ kinetic friction coefficient.
$d$ sliding distance.

When sliding with constant speed the dynamic behavior is described by the equation

$- m g \sin \left(\theta\right) + \mu N = 0$

along the ramp surface, where $N = m g \cos \left(\theta\right)$ is the normal force.

After resuming the ramp, the kinetic energy is given by

$K = \frac{1}{2} m {v}^{2}$

This energy is wasted against friction so

$K = \mu m g d$ which is the work produced by the friction force.

Solving for $\mu$ we get

$\mu = \tan \left(\theta\right)$

Solving now for $d$

$d = \frac{1}{2} {v}^{2} / \left(\mu g\right) = \frac{1}{2} {v}^{2} / g \cot \left(\theta\right) = \frac{1}{2} \cdot {\left(3.4\right)}^{2} / 9.81 \cdot \cot \left({24}^{\circ}\right)$