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

(a) When $x = 1.0$: $U = 20~J$ When $x = 2.0$: $U = 40~J$ When $x = 3.0$: $U = 60~J$ When $x = 4.0$: $U = 70~J$ (b) $x = 2.56~m$
(a) We can write an expression for the potential energy: $U(x) = -\int_{x_0}^{x_f} F_x~dx$ $U(x)$ is the negative of the area under the $F$ versus $x$ graph. We can find $U$ when $x = 1.0$: $U = -(-20~N)(1.0~m) = 20~J$ We can find $U$ when $x = 2.0$: $U = -(-20~N)(2.0~m) = 40~J$ We can find $U$ when $x = 3.0$: $U = -(-20~N)(3.0~m) = 60~J$ We can find $U$ when $x = 4.0$: $U = -[(-20~N)(3.0~m)+(0.5)(-20~N)(1.0~m)] = 70~J$ (b) We can find the initial kinetic energy: $K_0 = \frac{1}{2}mv^2$ $K_0 = \frac{1}{2}(0.100~kg)(25~m/s)^2$ $K_0 = 31.25~J$ We can use conservation of energy to find $x$: $U_f+K_f = U_0+K_0$ $(20~N)(x)+0 = 20~J+31.25~J$ $x = \frac{20~J+31.25~J}{20~N}$ $x = 2.56~m$