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

Answer:

at stable potential energy is minimum

use calculas to find x for minima and maxima

Answer 2

The position of stable equilibrium is x = 2 and that of unstable equilibrium is x = -2 .

Given the potential energy of a particle moving along x-axis , U = (x³/3) - 4x +6

F = -dU/dx = force is equal to the negative of the differential of potential energy w.r.t x.

F = -x² + 4

For equilibrium , F = 0

=> -x² + 4 = 0

=> x = 2,-2

For X>2m,F = -ve

displacement is in positive direction and force is negative. Therefore X=2 is stable equilibrium position.

For X<-2 , F = -ve

force and displacement are in negative directions. Therefore X=−2 is unstable equilibrium position.