Question

A particle moves in a potential region given by \(\sf U=8x^2-4x+400\) J. Its state of equilibrium will be

Answer 1

@Michele_Laino

Answer 2

we have to request that the subsequent condition holds:
\[\Large \frac{{\partial U}}{{\partial x}} = 0\]

Answer 3

Oh, so you mean we have to solve the equation for u=0?

Answer 4

not for U=0, its first derivative with respect to x has to be equal to zero, since, in a field of force coming from a potential, the relationship between force and potential energy is:
\[\Large {\mathbf{F}} = - \nabla U\]

Answer 5

Oh, one min....

Answer 6

Ok, yes. Thanks a bunch c:

Answer 7

thus we get the subsequent condition:
\[\Large {x_0} = \frac{1}{4}\]
as equilibrium position

Answer 8

:)

Answer 9

Yes... c:

Answer 10

:)

Answer 11

now, the sign of the second derivative at this point, will determine whether this equilibrium point is stable or unstable

Answer 12

Oh I see, thanks for the info Felix c: