What is the acceleration of a falling block attached to a pulley system?

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SUMMARY

The acceleration of the falling block (mass m3 = 4 kg) in the pulley system is calculated to be 3.9 m/s². The system includes a block on a table (mass m1 = 2 kg) with a coefficient of friction (µ = 0.75) and a uniform disk pulley (mass m2 = 0.5 kg, radius 15 cm). The tension in the string must account for both the gravitational force acting on m3 and the frictional force acting on m1. The net force on the 4 kg mass is determined to be 15.6 N, which is influenced by the tension in the string and the forces acting on the system.

PREREQUISITES
  • Understanding of Newton's second law (F = MA)
  • Knowledge of frictional force calculations (Frictional Force = µ x normal)
  • Concept of rotational dynamics as applied to pulleys
  • Basic principles of tension in a string system
NEXT STEPS
  • Study the dynamics of pulley systems in detail
  • Learn how to calculate tension in a string with multiple masses
  • Explore the effects of friction on motion in mechanical systems
  • Investigate the principles of rotational inertia and its impact on acceleration
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to explain pulley systems and forces in motion.

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Homework Statement


A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.75. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.5 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.
With what acceleration does m3 fall?

Homework Equations


Frictional Force = mew x normal
F = MA
Newtons' second law

The Attempt at a Solution


If I could treat the pulley as a point and didn't have to pay attention to it this would be a cinch. To be honest, I don't know what to do with it at all, and I'm looking for some conceptual help here.
 
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Much time later, I finally got the acceleration to be 3.9. The next question in this set is: what is the tension in the string above the table, and below the pulley. (They will be separate.)
Now, since F=MA, I thought this would be easy. 3.9 x 4 = 15.6. It itsn't that easy, it turns out. Any ideas?
 
Well, 15.6 N is the net force on the 4 kg mass.

This force is due to the string tension and what else?
 

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