Answer
The answer to part (a): $~~U = 11.3~J~~$ at $~~x = 2.0~m$
The answer to part (b): The speed is $~~6.4~m/s~~$ as it passes through the origin.
Work Step by Step
We can find the potential energy at $~~x = 2.0~m$:
$\Delta U = -\int_{0}^{2.0}~F(x)~dx$
$\Delta U = -\int_{0}^{2.0}~(-3.0~x-5.0~x^2)~dx$
$\Delta U = \int_{0}^{2.0}~(3.0~x+5.0~x^2)~dx$
$\Delta U = (1.5~x^2+\frac{5.0~x^3}{3})\Big \vert_{0}^{2.0}$
$\Delta U = [1.5~(2.0)^2+\frac{(5.0)~(2.0)^3}{3}]-0$
$\Delta U = 19.3~J$
Since $~~U=-8.0~J~~$ at $~~x = 0,~~$ then $~~U = 11.3~J~~$ at $~~x = 2.0~m$
The answer to part (a) is $~~U = 11.3~J~~$ at $~~x = 2.0~m$
We can find the potential energy at $~~x = 5.0~m$:
$\Delta U = -\int_{0}^{5.0}~F(x)~dx$
$\Delta U = -\int_{0}^{5.0}~(-3.0~x-5.0~x^2)~dx$
$\Delta U = \int_{0}^{5.0}~(3.0~x+5.0~x^2)~dx$
$\Delta U = (1.5~x^2+\frac{5.0~x^3}{3})\Big \vert_{0}^{5.0}$
$\Delta U = [1.5~(5.0)^2+\frac{(5.0)~(5.0)^3}{3}]-0$
$\Delta U = 245.8~J$
Since $~~U=-8.0~J~~$ at $~~x = 0,~~$ then $~~U = 237.8~J~~$ at $~~x = 5.0~m$
We can use conservation of energy to find the speed when it passes through the origin:
$K_f+U_f = K_0+U_0$
$K_f = K_0+U_0-U_f$
$K_f = K_0+237.8~J-(-8.0~J)$
$\frac{1}{2}mv_f^2 = \frac{1}{2}mv_0^2+245.8~J$
$v_f^2 = v_0^2+\frac{(2)(245.8~J)}{m}$
$v_f^2 = v_0^2+\frac{(2)(245.8~J)}{20~kg}$
$v_f = \sqrt{v_0^2+\frac{(2)(245.8~J)}{20~kg}}$
$v_f = \sqrt{(-4.0~m/s)^2+\frac{(2)(245.8~J)}{20~kg}}$
$v_f = 6.4~m/s$
The answer to part (b) is $~~6.4~m/s$
The speed when it passes through the origin is $~~6.4~m/s$
