Answer
(a) $U_x = (30.0~N/m)~x^2+(6.0~N/m^2)~x^3$
(b) v = 7.85 m/s
Work Step by Step
(a) $U_x = U_0-\int_{0}^{x}F_x~dx$
$U_x = 0-\int_{0}^{x}(-60.0~N/m)~x-(18.0~N/m^2)~x^2~dx$
$U_x = (30.0~N/m)~x^2+(6.0~N/m^2)~x^3$
(b) $K_2+U_2 = K_1+U_1$
$K_2 = 0+U_1-U_2$
$\frac{1}{2}(0.900~kg)v^2 = (30.0~N/m)~(1.00~m)^2+(6.0~N/m^2)(1.00~m)^3 - (30.0~N/m)~(0.50~m)^2-(6.0~N/m^2)(0.50~m)^3$
$(0.450~kg)v^2 = 27.75~J$
$v^2 = \sqrt{\frac{27.75~J}{0.450~kg}}$
$v = 7.85~m/s$
