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

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.
$U(x.y) = (5.80~J/m^2)~x^2-(3.60~J/m^3)~y^3$ $F = -((\frac{dU}{dx})~\hat{i}+(\frac{dU}{dy})~\hat{j})$ $F = (-11.6~J/m^2)(x)~\hat{i}+(10.8~J/m^3)(y^2)~\hat{j}$ 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.