How Does Releasing an Object Affect Tension in a Pulley System?

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SUMMARY

The discussion centers on the effects of releasing an object (m1) in a pulley system where it is connected to another mass (m2) hanging off a frictionless ramp. When m1 is held in place and then released, the tension in the rope decreases due to the acceleration of m2. The tension is calculated using the equations T = M2*g when stationary and T = M2*(g - a) when m2 is accelerating, confirming that tension is less when m2 is in motion. This is further illustrated by the scenario where cutting the rope results in T=0, demonstrating the relationship between tension and acceleration.

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Redfire66
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So I've been thinking about something for some time now, suppose you have an object (m1) on a frictionless ramp. There is an ideal pulley where the object is attached to one another and hangs down off the ramp (m2). And suppose that someone is holding the first block in place (m1) . If the person stops holding onto it, the first block will be pulled by the second block (hence m2 > m1).
I heard that the tension should be less than before

From what I assume, the net force on the first object should be greater than the net force on the second object (to keep it in place, however they should not equal each other right? Since that could mean there is a velocity)
Anyhow, if the tension of the rope is affected by the second object pulling on it (m2) then shouldn't the tension be the same since releasing the object doesn't change the tension applied from the block pulling on it?
I have a diagram here, I assume that the tension on block 1 should be the same on block 2.
 

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I heard that the tension should be less than before.

That's correct. Consider the forces on mass M2 on it's own.

When stationary:
M2*g - T = 0
so
T = M2*g

When mass M2 is accelerating:
M2*g - T = M2*a
so
T = M2*(g - a)
which is less.

It might seem more obvious if you consider what happens if the rope is cut so that M2 is in freefall (a=g). Clearly T=0 which is less than M2*g.
 

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