91Ó°ÊÓ

The potential energy of the particle is $U(x,y)=(5.80 J/m^2)x^2-(3.60J/m^3)y^3$

The position of the particle is $(x,y)=(0.300 m,0.600 m)$

The mass of the block is m = 0.400 kg

The variation in the speed of a particle with time is called acceleration. It may be uniform or non-uniform.

The magnitude of the force is given as:

$\stackrel{→}{F}=-\left(\frac{dU}{dx}\hat{i}+\frac{dU}{dy}\hat{j}\right)$

$\stackrel{→}{F}$is the force on the block

Substitute all the values in the above equation.

$$\begin{gathered} \stackrel{→}{\mathrm{F}}=-\left(\frac{\mathrm{d}\left(\left(5.80\mathrm{J}/{\mathrm{m}}^2\right){\mathrm{x}}^2-\left(3.60\mathrm{J}/{\mathrm{m}}^3\right){\mathrm{y}}^3\right)}{\mathrm{dx}}\hat{\mathrm{i}}+\frac{\mathrm{d}\left(\left(5.80\mathrm{J}/{\mathrm{m}}^2\right){\mathrm{x}}^2-\left(3.60\mathrm{J}/{\mathrm{m}}^3\right){\mathrm{y}}^3\right)}{\mathrm{dy}}\hat{\mathrm{j}}\right) \\ \stackrel{→}{\mathrm{F}}=-\left[\left(11.60\mathrm{J}/{\mathrm{m}}^2\right)\mathrm{x}\hat{\mathrm{i}}+\left(-10.8\mathrm{J}/{\mathrm{m}}^3\right){\mathrm{y}}^2\hat{\mathrm{j}}\right] \\ \stackrel{→}{\mathrm{F}}=-\left(11.60\mathrm{J}/{\mathrm{m}}^2\right)\left(0.300\mathrm{m}\right)\hat{\mathrm{i}}+\left(10.80\mathrm{J}/{\mathrm{m}}^3\right){\left(0.600\mathrm{m}\right)}^2\hat{\mathrm{j}} \\ \mathrm{F}=5.22\mathrm{J}/\mathrm{m} \end{gathered}$$

The acceleration of the block is given as:

$a=\frac{F}{m}$

Substitute all the values in the above equation and we get,

$$\begin{gathered} a=\frac{5.22 J/m}{0.0400kg} \\ a=130 m/s^2 \end{gathered}$$

The direction of the acceleration of the block is the counter-clockwise direction.

Therefore, the acceleration of the block is $130 m/s^2$in the counter-clockwise direction.