Two Hanging mass's attached to one on the table

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SUMMARY

The discussion centers on calculating the acceleration of a 2.0 kg block on a table, influenced by two hanging masses. The coefficient of kinetic friction is specified as 0.350. The user Charan initially calculated the forces but neglected the left-side tension in their analysis. The correct approach involves considering both tensions and the frictional force to accurately determine the acceleration, which was found to be 6.34 m/s², though this was identified as incorrect due to the oversight.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of kinetic friction and its calculation (F=uK*Force Normal)
  • Ability to analyze forces in a system with multiple masses
  • Familiarity with free-body diagrams for visualizing forces
NEXT STEPS
  • Review the principles of free-body diagrams to visualize forces acting on the block
  • Study the effects of tension in systems with multiple masses
  • Learn about the implications of kinetic friction in dynamic systems
  • Practice similar problems involving multiple masses and friction to reinforce understanding
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to forces, tension, and friction in multi-body systems.

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Homework Statement


The coefficient of kinetic friction between the 2.0 kg block in figure and the table is 0.350.

What is the acceleration of the 2.0 kg block?

Homework Equations


F=uK*Force Normal
Fnet=ma

The Attempt at a Solution


I tried to take the mass's of both the hanging and used them to calculate a net force on the block in the center and then solved for acceleration that way. Here's my work.

3kg*9.8=29.4N Right Side Tension
1kg*9.8=9.8N Left Side Tension
2kg*9.8=19.6N Force Normal

F=.35*19.6N
6.86N

(29.4N-6.86N)-9.8N=2kg*acceleration

a=6.34m/s^2

and it turns out this is wrong please help me out thank you!

-Charan
 

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You forgot the left side tension...
 

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