Solving Atwood/Pulley Problems: Masses, Acceleration & Distance

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AI Thread Summary
The discussion focuses on solving Atwood pulley problems involving a 140 g and a 110 g mass connected by a string over a frictional pulley. Key questions include determining the system's acceleration after release, the distance the 140 g mass falls in 1.5 seconds, and the minimum mass difference required for the system to accelerate when the total mass is 250 g. The approach involves using the equation T1 + T2 = (m1 + m2)g + (m2 - m1)a to find acceleration and tensions. It is noted that friction reduces the resultant force acting on the system, affecting the calculations. Understanding these principles is essential for accurately solving the problems presented.
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Homework Statement



140 gram mass & a 110 g mass are connected by a masless string passin over a massles pully which has a frictional resist. f=0.147 N when M (total weight of 2 masses) = 250 g.

-What's the acceleration of the system after release?
-How far does the 140 g mass fall during a 1.5 s interval following release from rest
-what's the minimum mass differnce for which the pulley syst. will have a non-0 acceler. when released from rest when M = 250 g

Homework Equations



Somethin like this?

T1 + T2 = (m1 + m2)g + (m2-m1)a

The Attempt at a Solution



The third question's my problem. For the first one, I'm guessin it'd be right just to use the equation to solve for a and T1/T2 would be the weight of each mass multiplied by gravity's constant? Would you use kinetic equations once you find acceleration to find how far the mass would fall during 1.5 s?
 
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If there were no friction the resultant force acting on the system would have been the difference between the weight of the larger mass and the smaller mass (the tensions would have been the same and canceled each other out since they are opposing each other). With friction the resultant force is reduced according to the magnitude of the friction (that is the one tension is less than the other by the size of the friction).
 
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