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
).