## University Physics with Modern Physics (14th Edition)

(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. $F = \alpha~x~y~\hat{i}$ $W = \int_{0}^{2.00}F~dx$ $W = \int_{0}^{2.00}\alpha~x~y~dx$ $W = \int_{0}^{2.00}\alpha~(x)~(1.5~x)~dx$ $W = (1.5)(2.50~N/m^2)\int_{0}^{2.00}x^2~dx$ $W = (1.5)(2.50~N/m^2)\frac{x^3}{3}\vert_{0}^{2.00}$ $W = (1.5)(2.50~N/m^2)(\frac{8.00}{3}~m^3)$ $W = 10~J$