[Paradox] Spring, work, force problem

[jace1313]

so here is the problem

The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 40 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.7 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. Assume that the stopping point is reached.a) What is the position of the block?

So my original thinking was simplistic, we have two horizontal forces and the block will stop when equilibrium is achieved (when the force applied equals force of the spring):
2.7 = kx
2.7 = 40x
x = 0.0675

Solution teacher gave:
Work done by force = final potential energy of spring
F*x = 1/2kx^2
2.7*x = 1/2(40)x^2
x = 0.135

So I understand this thinking as well, the work done by the force puts energy into the spring (potential energy). But if this method works, shouldn't the force of the spring and the force applied equal each other in the end?

If we calculate the force of the spring (40*0.135) it is 5.4 N and we are given the force applied as 2.7 N. The net force on the block would be 5.4 - 2.7, which is 2.7 N. And if there is a net force on the block, the block is accelerating and not in a final resting position.

Is it just that the block stops temporarily in that position, but will eventually rest in the equilibrium I solved for?

 

[PhysDrew]

I think the force applied in Hooke's law is the restoring force of the spring, not the external force doing the work. In your original thinking, you modeled the restoring force as the external force, and additionally never considered the external force to equal the restoring force (like you stated you waned to do).
I think you had the right idea, just went about it incorrectly. The work involved has to equal the potenial energy in the spring (like the teachers example provided).
Also when you calculated the force on the spring, you're calculating the force imparted by the spring on the block when it is released. Thus it would accelerate towards its' equilibrium position. But this isn't relevant to the question ;-)

 

[Doc Al]

So my original thinking was simplistic, we have two horizontal forces and the block will stop when equilibrium is achieved (when the force applied equals force of the spring)

No, it won't stop there because when it gets to that point it will have kinetic energy due to the work done by the applied force being greater than the increase in spring potential energy. Equilibrium just means there's no net force, not that its necessarily at rest.

So I understand this thinking as well, the work done by the force puts energy into the spring (potential energy). But if this method works, shouldn't the force of the spring and the force applied equal each other in the end?

Ah, but without friction there is no end.

If we calculate the force of the spring (40*0.135) it is 5.4 N and we are given the force applied as 2.7 N. The net force on the block would be 5.4 - 2.7, which is 2.7 N. And if there is a net force on the block, the block is accelerating and not in a final resting position.

Exactly. It accelerates back toward the equilibrium point that you calculated earlier.

Is it just that the block stops temporarily in that position, but will eventually rest in the equilibrium I solved for?

It stops temporarily, swings back through the equilibrium point but doesn't stop there. It oscillates.

But if you gently placed the block at the equilibrium point (with the applied force acting) it would just stay there.