## Essential University Physics: Volume 1 (3rd Edition)

a) 0 b) $-F_0a$
a) We found in problem one that figure a is a conservative vector field. The net work on a closed path in a conservative vector field is 0, so we see that the work done is 0. b) Since one side of the box is on the y-axis, where there is no force, the work along this side is 0. Two other sides are perpendicular to the vector field, so the work done along these sides is 0. Thus, the net work is the work done when the path goes along the final side: $W=F\times d = \frac{-F_0a}{a}\times a = -F_0a$