The potential energy of an object of mass m m | vedclass

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Question

(A) The potential energy is given by $U = ax + by$. The force $\overrightarrow{F}$ acting on the object is the negative gradient of the potential ener

Vedclass · Accepted Answer

$\frac{\sqrt{a^2 + b^2}}{m}$

Vedclass · Answer

(A) The potential energy is given by $U = ax + by$. The force $\overrightarrow{F}$ acting on the object is the negative gradient of the potential energy: $\overrightarrow{F} = -
abla U = -\left( \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} ight)$. Calculating the partial derivatives: $\frac{\partial U}{\partial x} = a$ and $\frac{\partial U}{\partial y} = b$. Thus,$\overrightarrow{F} = -a \hat{i} - b \hat{j}$. The magnitude of the force is $|\overrightarrow{F}| = \sqrt{(-a)^2 + (-b)^2} = \sqrt{a^2 + b^2}$. Using Newton's second law,$F = ma$,the magnitude of acceleration is $a_{acc} = \frac{|\overrightarrow{F}|}{m} = \frac{\sqrt{a^2 + b^2}}{m}$. Therefore,the correct option is $A$.

The potential energy of an object of mass $m$ moving in the $xy$ plane in a conservative field is given by $U = ax + by$,where $x$ and $y$ are position coordinates of the object. Find the magnitude of its acceleration.

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