#### Answer

(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 $6.67~m$.

#### Work Step by Step

(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 $6.67~m$.