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

We can find the work done by the force between x = 0 and x = 14.0 meters. $W = \int_{0}^{14.0}F_x~dx$ $W = \int_{0}^{14.0}(18.0~N)-(0.530~N/m)~x~dx$ $W = (18.0~N)~x-(0.265~N/m)~x^2 ~\vert_0^{14.0}$ $W = (18.0~N)(14.0~m)-(0.265~N/m)(14.0~m)^2$ $W = 200~J$ We can use the work done by the force to find the speed of the box. $\frac{1}{2}mv^2= 200~J$ $v^2 = \frac{400~J}{m}$ $v = \sqrt{\frac{400~J}{6.00~kg}}$ $v = 8.16~m/s$ The speed of the box after it has traveled 14.0 meters is 8.16 m/s.