# Answer to Question #4243 in Mechanics | Relativity for neha

Question #4243

The potential energy U(x)of a particle moving along x-axix is given by U(x)=ax-bx2. Find the equilibrium position of particle.

Expert's answer

The equilibrum position is the point in which U'(x)=0.

U' = a - 2bx =

0

Hence

x = a/(2b)

U' = a - 2bx =

0

Hence

x = a/(2b)

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