Answer
The net work done on the canister by the three forces during the first $4.00~m$ of displacement is $~~8.52~J$
Work Step by Step
We can find the horizontal component of $F_1$:
$F_{1,x} = -3.00~N$
We can find the horizontal component of $F_2$:
$F_{2,x} = -(4.00~N)~sin~50.0^{\circ} = -3.06~N$
We can find the horizontal component of $F_3$:
$F_{3,x} = (10.0~N)~cos~35.0^{\circ} = 8.19~N$
We can find the net horizontal force:
$F = 8.19~N-3.00~N-3.06~N = 2.13~N$
We can find the net work done on the canister by the three forces during the first $4.00~m$ of displacement:
$W = F~d$
$W = (2.13~N)(4.00~m)$
$W = 8.52~J$
The net work done on the canister by the three forces during the first $4.00~m$ of displacement is $~~8.52~J$
