Answer
The magnitude of the acceleration is $131~m/s^2$ and it is directed at an angle of $48.2^{\circ}$ above the (-x)-axis.
Work Step by Step
$U(x.y) = (5.80~J/m^2)~x^2-(3.60~J/m^3)~y^3$
$F = -((\frac{dU}{dx})~\hat{i}+(\frac{dU}{dy})~\hat{j})$
$F = (-11.6~J/m^2)(x)~\hat{i}+(10.8~J/m^3)(y^2)~\hat{j}$
We can find the force acting on the block at the point (0.300 m, 0.600 m)
$F = (-11.6~J/m^2)(0.300~m)~\hat{i}+(10.8~J/m^3)(0.600~m)^2~\hat{j}$
$F = (-3.48~N)~\hat{i}+ (3.89~N)~\hat{j}$
We can find the magnitude of the force F.
$F = \sqrt{(-3.48~N)^2+(3.89~N)^2}$
$F = 5.22~N$
We can find the acceleration of the block.
$a = \frac{F}{m} = \frac{5.22~N}{0.0400~kg}$
$a = 131~m/s^2$
We can find the angle above the (-x)-axis.
$tan(\theta) = \frac{3.89}{3.48}$
$\theta = arctan(\frac{3.89}{3.48})$
$\theta = 48.2^{\circ}$
The magnitude of the acceleration is $131~m/s^2$ and it is directed at an angle of $48.2^{\circ}$ above the (-x)-axis.
