An object moving in the xy-plane is acted on by a conservative force described by the potential-energy function U(x,y)=α(1x2+1y2) where α is a positive constant. Derivative an expression for the force expressed terms of the unit vectors i→ and j→.

Rewrite the potential energyU(x,y)=α(1x2+1y2)&nbsp;F→=−(dUdXi^+dU&nbsp;dy&nbsp;j^)&nbsp;F→=−αd&nbsp;dx&nbsp;(1x2+1y2)i^−αd&nbsp;dy&nbsp;(1y2)j^&nbsp;=−αd&nbsp;dx&nbsp;(1x2)i^−αd&nbsp;dy&nbsp;(1y2)j^&nbsp;=2α1x3i^+2α1y3j^&nbsp;F→=2αx3i^+2αy3j^

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Globokim8

Answered question

2021-09-26

An object moving in the xy-plane is acted on by a conservative force described by the potential-energy function

U (x, y) = α (\frac{1}{x^{2}} + \frac{1}{y^{2}})

where

α

is a positive constant. Derivative an expression for the force expressed terms of the unit vectors

\vec{i}

and

\vec{j}

Answer & Explanation

l1koV

Skilled2021-09-27Added 100 answers

Rewrite the potential energy $U (x, y) = α (\frac{1}{x^{2}} + \frac{1}{y^{2}})$
$\vec{F} = - (\frac{d U}{d X} \hat{i} + \frac{d U}{d y} \hat{j})$
$\vec{F} = - α \frac{d}{d x} (\frac{1}{x^{2}} + \frac{1}{y^{2}}) \hat{i} - α \frac{d}{d y} (\frac{1}{y^{2}}) \hat{j}$ $= - α \frac{d}{d x} (\frac{1}{x^{2}}) \hat{i} - α \frac{d}{d y} (\frac{1}{y^{2}}) \hat{j}$ $= 2 α \frac{1}{x^{3}} \hat{i} + 2 α \frac{1}{y^{3}} \hat{j}$
$\vec{F} = \frac{2 α}{x^{3}} \hat{i} + \frac{2 α}{y^{3}} \hat{j}$

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