Answer:
Explanation:
Use that:
[tex]F=-\vec{\nabla}U(x,y)=-\left(\hat{i} \frac{\partial U}{\partial x}+\hat{j}\frac{\partial U}{\partial y}\right)=-11.6x\hat{i}+10.8y^{2}\hat{j}[/tex]
Then use the 2nd Newton's Law of Motion:
[tex]\vec{a}=\frac{\vec{F}}{m}=\frac{-11.6x\hat{i}+10.8y^{2}\hat{j}}{0.04}=-290x\hat{i}+270y^{2}\hat{j}[/tex]
At x = 0.3 and y = 0.6, we can find the acceleration as:
[tex]\vec{a}=-87\hat{i}+97.2\hat{j}[/tex] (in SI unit)
Then the magnitude of the acceleration on that point is:
[tex]a=\sqrt{(-87)^{2}+(97.2)^{2}}\approx 130.44[/tex] (SI Unit)
