## University Physics with Modern Physics (14th Edition)

(a) v = 16.9 m/s (b) The magnitude of the acceleration is $17.8~m/s^2$ in the direction of the +x-axis. (c) The maximum value of $x$ reached by the object is $6.67~m$.
(a) When $x$ is very small, the total energy in the system is approximately equal to zero. $K+U = 0$ $K = -U$ $\frac{1}{2}mv^2 = -(-\alpha x^2+\beta x^3)$ $v^2 = \frac{2\alpha x^2-2\beta x^3}{m}$ $v = \sqrt{\frac{2\alpha x^2-2\beta x^3}{m}}$ $v = \sqrt{\frac{(2)(2.00~J/m^2)(4.00~m)^2-(2)(0.300~J/m^3)(4.00~m)^3}{0.0900~kg}}$ $v = 16.9~m/s$ (b) We can find the force $F$ at $x = 4.00~m$. $F = -\frac{dU}{dx}$ $F = 2\alpha x - 3\beta x^2$ $F = (2)(2.00~J/m^2)(4.00~m) - (3)(0.300~J/m^3)(4.00~m)^2$ $F = 1.6~N$ We can find the acceleration. $ma = F$ $a = \frac{F}{m} = \frac{1.6~N}{0.0900~kg}$ $a = 17.8~m/s^2$ The magnitude of the acceleration is $17.8~m/s^2$ in the direction of the +x-axis. (c) When $x$ is very small, the total energy in the system is approximately equal to zero. We can find the maximum value of $x$ where the total energy in the system is zero. $U(x) = 0$ $-\alpha x^2+\beta x^3 = 0$ $-(2.00~J/m^2) x^2+(0.300~J/m^3) x^3 = 0$ $(0.300~J/m^3) x^3 = (2.00~J/m^2) x^2$ $x = \frac{2.00~J/m^2}{0.300~J/m^3}$ $x = 6.67~m$ The maximum value of $x$ reached by the object is $6.67~m$.