Question

Potential energy of a particle moving along x-axis is by
`U=(x^(3)/3-4x + 6)`.
here, U is in joule and x in metre. Find position of stable and unstable equilibrium.

Answer 1

Correct Answer - A::B
`U=(x^(3))/3-4x + 6`
`F =-(dU)/(dx)`
`F =0` at `x=+-2m`
.
for `xgt2m, F=-ve` i.e. displacemect is in positive direction and force is negative. Therefore `x=2` is stable equilium position.
for `xlarr2m , F=-ve`
i.e. force and displacement are in negative directions. Therefore, `x = -2m` is unstable equilibrium positions.