A particle located in a 1 dimensional potential field has potential energy function U (x)= (a/x^4) - (b/x^2), where a and b are positive constants. The position of equilibrium x- corresponds to?

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piyush gera · Answer

simply we know that -du/dx=F. simply differentiate this expreesion by putting f=0(in equlibrium no force acte net force is 0) and get answer.(u= potential energy)

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A particle located in a 1 dimensional potential field has potential energy function U (x)= (a/x^4) - (b/x^2), where a and b are positive constants. The position of equilibrium x- corresponds to?

A particle located in a 1 dimensional potential field has potential energy function U (x)= (a/x^4) - (b/x^2), where a and b are positive constants. The position of equilibrium x- corresponds to?

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A particle located in a 1 dimensional potential field has potential energy function U (x)= (a/x^4) - (b/x^2), where a and b are positive constants. The position of equilibrium x- corresponds to?

A particle located in a 1 dimensional potential field has potential energy function U (x)= (a/x^4) - (b/x^2), where a and b are positive constants. The position of equilibrium x- corresponds to?

1 Answers

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Other Related Questions on Mechanics

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