Answered

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a small block with mass 0.04 kg is moving in the xy-plane. the net force on the block is describe by the potential energy function U(x,y) = (5.80 j/m2)x2 - (3.60 j/m3)y3. What are the magnitude of the acceleration of the block when it is at the point (x=0.30m , y=0.60m ) ?

Sagot :

deriyan191293

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)

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