#### Answer

(a) W = 15.0 J
(b) W = 0
(c) W = 10 J

#### Work Step by Step

(a) $F = \alpha~x~y~\hat{i}$
$W = \int_{0}^{2.00}F~dx$
$W = \int_{0}^{2.00}\alpha~x~y~dx$
$W = \alpha~y\int_{0}^{2.00}x~dx$
$W = (2.50~N/m^2)(3.00~m)\frac{x^2}{2}\vert_{0}^{2.00}$
$W = (2.50~N/m^2)(3.00~m)(2.00~m^2)$
$W = 15.0~J$
(b) Since the force acts at a $90^{\circ}$ angle to the direction of motion, the work done by the force is zero.
(c) We can find the work done on the object.
$F = \alpha~x~y~\hat{i}$
$W = \int_{0}^{2.00}F~dx$
$W = \int_{0}^{2.00}\alpha~x~y~dx$
$W = \int_{0}^{2.00}\alpha~(x)~(1.5~x)~dx$
$W = (1.5)(2.50~N/m^2)\int_{0}^{2.00}x^2~dx$
$W = (1.5)(2.50~N/m^2)\frac{x^3}{3}\vert_{0}^{2.00}$
$W = (1.5)(2.50~N/m^2)(\frac{8.00}{3}~m^3)$
$W = 10~J$