Answer
The crate moves a distance of 1.3 meters along the floor.
Work Step by Step
We can find the force of kinetic friction exerted on the crate:
$F_f = F_N~\mu_k$
$F_f = mg~\mu_k$
$F_f = (56.8~kg)(9.80~m/s^2)(0.120)$
$F_f = 66.8~N$
We can find the distance the box moves:
$Work = (124~N)~d-(66.8~N)~d$
$74.4~J = (124~N-66.8~N)~d$
$d = \frac{74.4~J}{57.2~N}$
$d = 1.3~m$
The crate moves a distance of 1.3 meters along the floor.
