Two bars connected by a spring on a floor with friction

[Null_Void]

You need to think about what happens after the equilibrium point.

Is this the equation we should be concerned with?
$Fx - μm_2gx - 1/2kx^2 = -1/2m_2v^2$

 

[PeroK]

Is this the equation we should be concerned with?
$Fx - μm_2gx - 1/2kx^2 = -1/2m_2v^2$

No. That's no use, as you yourself have repeatedly complained about the $v^2$ term being a nuisance.

 

[Null_Void]

No. That's no use, as you yourself have repeatedly complained about the $v^2$ term being a nuisance.

Will it perform simple harmonic motion about the equilibrium point?

 

[PeroK]

Will it perform simple harmonic motion about the equilibrium point?

No.

 

[haruspex]

But an extra velocity term $v$ is still present? What value should $v$ take in this case

Suppose there is still a velocity term at that point. What would happen if you were to make F just a tiny bit less?

 

[Null_Void]

No.

$Fx - 1/2kx^2 - μm_2gx= 0$
Is this the right equation?

 

[PeroK]

$Fx - 1/2kx^2 - μm_2gx= 0$
Is this the right equation?

It is. It's useful because that's the point at which $m_2$ stops. And that's the point at which the force in the spring is at its maximum.

 

[Null_Void]

It is. It's useful because that's the point at which $m_2$ stops. And that's the point at which the force in the spring is at its maximum.

Ah I get it now. Combining this with the equation @haruspex pointed out, I'll get
$F = μ(m_1g/2 + m_2g)$ which is the correct answer, Thanks for your help everyone!