Answer
The work done on the trunk by the machine's force is $~~740~J$
Work Step by Step
Since the speed is constant, the machine's force that is directed up the incline must be equal in magnitude to the sum of the frictional force and the component of the gravitational force directed down the incline.
We can find $F_{up}$, the machine's force that is directed up the incline:
$F_{up} = mg~sin~\theta+mg~cos~\theta~\mu_k$
$F_{up} = mg~(sin~\theta+\mu_k~cos~\theta)$
$F_{up} = (40~kg)(9.8~m/s^2)~(sin~40^{\circ}+0.40~cos~40^{\circ})$
$F_{up} = 372~N$
We can find the work done by the machine's force:
$W = F_{up}~d$
$W = (372~N)(2.0~m)$
$W = 740~J$
The work done on the trunk by the machine's force is $~~740~J$
