13. The forces are all constant, so the total work done by them is given by W = F

net

∆x, where F

net

is the

magnitude of the net force and ∆x is the magnitude of the displacement. We add the three vectors,
finding the x and y components of the net force:

F

net x

=

−F

1

− F

2

sin 50

◦

+ F

3

cos 35

◦

=

−3.00 N − (4.00 N) sin 35

◦

+ (10.0 N) cos 35

◦

=

2.127 N

F

net y

=

−F

2

cos 50

◦

+ F

3

sin 35

◦

=

−(4.00 N) cos 50

◦

+ (10.0 N) sin 35

◦

=

3.165 N .

The magnitude of the net force is

F

net

=

F

2

net x

+ F

2

net y

=



2.127

2

+ 3.165

2

= 3.813 N .

The work done by the net force is

W = F

net

d = (3.813 N)(4.00 m) = 15.3 J

where we have used the fact that 

d

 F

net

(which follows from the fact that the canister started from rest

and moved horizontally under the action of horizontal forces – the resultant effect of which is expressed
by 

F

net

).