# Answer on Mechanics | Relativity Question 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)

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment