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

The work done on the particle is equal to the area under the force versus position graph. We can find the work done during the interval 0 - 3 m: $W = (15~N)(1~m)+\frac{1}{2}(15~N)(2~m)$ $W = 30~J$ We can use the work-energy theorem to find the particle's velocity at x = 3 m. $KE_2 = KE_1+W$ $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W$ $v_2^2 = \frac{mv_1^2+2W}{m}$ $v_2 = \sqrt{\frac{mv_1^2+2W}{m}}$ $v_2 = \sqrt{\frac{(0.50~kg)(2.0~m/s)^2+(2)(30~J)}{0.50~kg}}$ $v_2 = 11.1~m/s$ The particle's velocity at x = 3 m is 11.1 m/s