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

The paint cans glide without acceleration so

`mu m g cos(alpha)=m g sin(alpha)`

where

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

from this relationship we get

`mu = tan(alpha) = 0.445229`

When in the horizontal table the movement has initial velocity `v_0`
so the can carries a kinetic energy of

`1/2mv_0^2` which is lost against the friction losses

`1/2mv_0^2 = mu m g d` where `d` is the glided distance over the table until repose.

so

`d = 1/2 v_0^2/(mu g) = 12.9821/g`

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.
![wikimedia.org]()
If `mu` is the coefficient of kinetic friction and other quantities as shown in the figure
`f=mu N=mu m g costheta`

`=>mu m g costheta=m g sintheta`
We obtain
`mu = tan 24^@ = 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
`=(mu m g)cdot s`
(material of ramp and table being same, has same coefficient of kinetic friction.)
Equating it with the KE we obtain
`(mu m g)cdot s=1/2m u^2`
`=> s=(1/2m u^2)/(mu m g)`
`=> s= u^2/(2mu g)`
Inserting given, calculated quantities and taking `g=9.81ms^-2` we get
`s= (3.40)^2/(2xx0.44523 xx9.81)`
`=>s=1.32m`, rounded to two decimal places.