## College Physics (4th Edition)

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.