# Conservation of Energy Problem - How far does each can slide?

## In a hardware store, paint cans, which weigh 46.0 N each, are transported from storage to the back of the paint department by placing them on a ramp that is inclined at an angle of 24.0 deg above the horizontal. The cans slide down the ramp at a constant speed of 3.40 m/s and then slide onto a table made of the same material as the ramp. How far does each can slide on the table’s horizontal surface before coming to rest?

Jul 7, 2016

$d = \frac{12.9821}{g}$

#### Explanation:

The paint cans glide without acceleration so

$\mu m g \cos \left(\alpha\right) = m g \sin \left(\alpha\right)$

where

$m$ can mass
$g$ gravity acceleration
$\mu$ kinetic friction coefficient
$\alpha$ ramp angle

from this relationship we get

$\mu = \tan \left(\alpha\right) = 0.445229$

When in the horizontal table the movement has initial velocity ${v}_{0}$
so the can carries a kinetic energy of

$\frac{1}{2} m {v}_{0}^{2}$ which is lost against the friction losses

$\frac{1}{2} m {v}_{0}^{2} = \mu m g d$ where $d$ is the glided distance over the table until repose.

so

$d = \frac{1}{2} {v}_{0}^{2} / \left(\mu g\right) = \frac{12.9821}{g}$

Jul 10, 2016

$1.32 m$, rounded to two decimal places.

#### Explanation:

The paint cans slide down with constant speed. This means there is no acceleration and upwards force due to friction is equal and opposite to the $\sin \theta$ component of gravitational force.

If $\mu$ is the coefficient of kinetic friction and other quantities as shown in the figure
$f = \mu N = \mu m g \cos \theta$

$\implies \mu m g \cos \theta = m g \sin \theta$
We obtain
$\mu = \tan {24}^{\circ} = 0.44523$, rounded to five decimal places

While moving on the horizontal table the can has initial velocity $u$. The KE due to this speed is lost in doing work against the force of friction.

If $s$ is the distance moved on the table before coming to rest, work done against the force of friction
$= \left(\mu m g\right) \cdot s$
(material of ramp and table being same, has same coefficient of kinetic friction.)
Equating it with the KE we obtain
$\left(\mu m g\right) \cdot s = \frac{1}{2} m {u}^{2}$
$\implies s = \frac{\frac{1}{2} m {u}^{2}}{\mu m g}$
$\implies s = {u}^{2} / \left(2 \mu g\right)$
Inserting given, calculated quantities and taking $g = 9.81 m {s}^{-} 2$ we get
$s = {\left(3.40\right)}^{2} / \left(2 \times 0.44523 \times 9.81\right)$
$\implies s = 1.32 m$, rounded to two decimal places.