A simple movable pulley not at equilibrium

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In a movable pulley system, pulling harder than the equilibrium force (half the weight of the object) results in the weight accelerating upwards. The tension in the rope remains uniform and equals the weight of the object (mg) when the pulling force equals the object's weight. This creates a total upward force of 2mg, leading to an acceleration of g for the weight. The discussion clarifies that understanding pulleys outside of equilibrium involves applying Newton's second law to analyze forces and accelerations. The conclusion confirms that the described behavior and calculations are accurate.
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Note: This is not a homework question. I have constructed this question to help me understand a concept.

If I have an object hanging off a single (frictionless, negligible mass) movable pulley, the equilibrium system entails me pulling on the free end of the (frictionless, negligible mass) rope with half as much force as the object weighs.

What happens if I pull harder than that?

See the attached GIF. What would be the behavior of the weight? Would it accelerate up? If so, at what acceleration? What would be the tension in the left side of the rope?

Every resource I have access to that mentions pulleys keeps talking about pulleys in this type of setup in equilibrium. I am interested in furthering my understanding in regards to pulleys not in equilibrium.

Thank you for any help.
 

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The tension is uniform throughout the rope. The upward force on the pulley/mass will equal twice the tension. To find the acceleration, just add the vertical forces and apply Newton's 2nd law.
 
So me increasing my pulling force from an equilibrium inducing (1/2)mg to mg will cause the rope to have a uniform tension of mg, thus making the total upward force 2mg (the tension of the rope being a uniform "mg"). This accelerate the weight, which weighs mg, upwards at acceleration = g.

Is that correct?
 
Yes, that's correct.
 

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