## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The work done by the force is equal to the area under the force versus position graph. We can find the work done by the force from x = 0 to x = 1 m. $W = \frac{1}{2}(5~N)(1~m)$ $W = 2.5~J$ We can use the work-energy theorem to find the velocity at x = 1 m; $KE_f = KE_0+W$ $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_0^2+W$ $v_f^2 = \frac{mv_0^2+2W}{m}$ $v_f = \sqrt{\frac{mv_0^2+2W}{m}}$ $v_f = \sqrt{\frac{(0.50~kg)(2.0~m/s)^2+(2)(2.5~J)}{0.50~kg}}$ $v_f = 3.7~m/s$ At x = 1 m, the velocity is 3.7 m/s. We can find the work done by the force from x = 0 to x = 2 m. $W = \frac{1}{2}(10~N)(2~m)$ $W = 10~J$ We can use the work-energy theorem to find the velocity at x = 2 m; $KE_f = KE_0+W$ $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_0^2+W$ $v_f^2 = \frac{mv_0^2+2W}{m}$ $v_f = \sqrt{\frac{mv_0^2+2W}{m}}$ $v_f = \sqrt{\frac{(0.50~kg)(2.0~m/s)^2+(2)(10~J)}{0.50~kg}}$ $v_f = 6.6~m/s$ At x = 2 m, the velocity is 6.6 m/s. We can find the work done by the force from x = 0 to x = 3 m. $W = \frac{1}{2}(15~N)(3~m)$ $W = 22.5~J$ We can use the work-energy theorem to find the velocity at x = 3 m; $KE_f = KE_0+W$ $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_0^2+W$ $v_f^2 = \frac{mv_0^2+2W}{m}$ $v_f = \sqrt{\frac{mv_0^2+2W}{m}}$ $v_f = \sqrt{\frac{(0.50~kg)(2.0~m/s)^2+(2)(22.5~J)}{0.50~kg}}$ $v_f = 9.7~m/s$ At x = 3 m, the velocity is 9.7 m/s.