Annabel Zamora

2023-04-01

An object moving in the xy-plane is acted on by a conservative force described by the potential energy function $U\left(x,y\right)=\alpha \left(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}\right)$ where $\alpha$ is a positive constant. Derivative an expression for the force expressed terms of the unit vectors $\stackrel{\to }{i}$ and $\stackrel{\to }{j}$.

Jeffery Romero

To find the force $\stackrel{\to }{F}$ associated with the given potential-energy function $U\left(x,y\right)=a\left(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}\right)$, we can use the relationship between force and potential energy. The force $\stackrel{\to }{F}$ can be expressed as the negative gradient of the potential energy function:
$\stackrel{\to }{F}=-\nabla U\left(x,y\right)$
Taking the partial derivatives of $U\left(x,y\right)$ with respect to $x$ and $y$, we obtain:
$\frac{\partial U}{\partial x}=\frac{\partial }{\partial x}\left(a\left(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}\right)\right)=-\frac{2a}{{x}^{3}}$
$\frac{\partial U}{\partial y}=\frac{\partial }{\partial y}\left(a\left(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}\right)\right)=-\frac{2a}{{y}^{3}}$
Now, we can express the force $\stackrel{\to }{F}$ in terms of the unit vectors $\stackrel{^}{i}$ and $\stackrel{^}{j}$:
$\stackrel{\to }{F}=\frac{2a}{{x}^{3}}\stackrel{^}{i}+\frac{2a}{{y}^{3}}\stackrel{^}{j}$
Therefore, the expression for the force $\stackrel{\to }{F}$ in terms of the unit vectors $\stackrel{^}{i}$ and $\stackrel{^}{j}$ is:
$\stackrel{\to }{F}=\frac{2a}{{x}^{3}}\stackrel{^}{i}+\frac{2a}{{y}^{3}}\stackrel{^}{j}$

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