Answer
(a) F = 73.5 N
(b) W = 331 J
(c) W = -331 J
(d) These forces do zero work.
(e) The total work done on the crate is zero.
Work Step by Step
(a) The worker's force $F$ must be equal in magnitude to the force of kinetic friction.
$F = F_f$
$F = mg~\mu_k = (30.0~kg)(9.80~m/s^2)(0.25)$
$F = 73.5~N$
(b) $W = F\cdot d = (73.5~N)(4.5~m)$
$W = 331~J$
(c) $W = F_f\cdot d = (73.5~N)(4.5~m)~cos(180^{\circ})$
$W = -331~J$
(d) The normal force and gravity act at a $90^{\circ}$ angle to the direction of motion. Therefore the work these forces do is zero.
(e) The total work done on the crate is zero. Note that the kinetic energy of the crate does not change.
