## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can find the work done on the particle as it moves from x = 0 to x = 3.14 m: $W = \int_{0}^{3.14}~F_x~dx$ $W = \int_{0}^{3.14}~(0.250~N)~sin(\frac{x}{2.00~m})~dx$ $W = -(0.50~J)~cos(\frac{x}{2.00~m})~\Big\vert_{0}^{3.14~m}$ $W = 0-(-0.50~J)(1)$ $W = 0.50~J$ The work done on the particle as it move from x = 0 to x = 3.14 m is 0.50 J. Then, we use the work-energy theorem to find the particle's speed at x = 3.14 m; $KE_2 = KE_1+W$ $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W$ $v_2^2 = \frac{mv_1^2+2W}{m}$ $v_2 = \sqrt{\frac{mv_1^2+2W}{m}}$ $v_2 = \sqrt{\frac{(0.15~kg)(2.00~m/s)^2+(2)(0.50~J)}{0.15~kg}}$ $v_2 = 3.27~m/s$ The particle's speed at x = 3.14 m is 3.27 m/s.