Answer
The work done by the applied force is $~~1980~J$
Work Step by Step
Since the speed is constant, the component of the applied 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 component of the applied force that is directed up the incline:
$F_{up} = mg~sin~\theta+mg~cos~\theta~\mu_k$
$F_{up} = mg~(sin~\theta+\mu_k~cos~\theta)$
$F_{up} = (50~kg)(9.8~m/s^2)~(sin~30^{\circ}+0.20~cos~30^{\circ})$
$F_{up} = 330~N$
We can find the work done by the applied force:
$W = F_{up}~d$
$W = (330~N)(6.0~m)$
$W = 1980~J$
The work done by the applied force is $~~1980~J$
