Answer
The crate slides across the floor a distance of $~~5.4~m$
Work Step by Step
In part (a), we found that that the speed at the bottom of the ramp is $~~5.46~m/s$
The work done on the crate by the frictional force will be equal in magnitude to the kinetic energy at the bottom of the ramp.
We can find the distance that the crate slides across the floor:
$mg~\mu_k~d = \frac{1}{2}mv^2$
$g~\mu_k~d = \frac{1}{2}v^2$
$d = \frac{v^2}{2g~\mu_k}$
$d = \frac{(5.46~m/s)^2}{(2)(9.8~m/s^2)(0.28)}$
$d = 5.4~m$
The crate slides across the floor a distance of $~~5.4~m$
