How Is Total Work Calculated for Blocks on a Pulley System with Friction?

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SUMMARY

The total work done on the 20.0-N block in a pulley system with friction is calculated using the coefficients of static and kinetic friction, specifically u_s=0.500 and u_k=0.325. The work done without friction was determined to be 5.625J for the 20.0-N block and 3.375J for the 12.0-N block. To find the work done with friction, the friction force must be calculated by multiplying the normal force by the coefficient of kinetic friction. This approach is essential for accurately determining the total work in a system where friction is present.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of work-energy principles
  • Familiarity with friction coefficients (static and kinetic)
  • Basic algebra for calculating forces and work
NEXT STEPS
  • Calculate work done with friction using the formula: Work = Friction Force x Distance
  • Explore the implications of different coefficients of friction on work calculations
  • Study the effects of pulley systems on force distribution
  • Investigate energy conservation principles in systems with friction
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Students studying physics, particularly those focusing on mechanics and friction in pulley systems, as well as educators looking for practical examples of work calculations in real-world applications.

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Homework Statement


Two blocks are connected by a very light string passing over a massless and frictionless pulley. The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward.

1. Find the total work done on 20.0-N block if u_s=0.500 and u_k=0.325 between the table and the 20.0-N block.
2. Find the total work done on 12.0-N block if u_s=0.500 and u_k=0.325 between the table and the 20.0-N block.

Homework Equations



The Attempt at a Solution


Well the first thing we were supposed to do was find the work done if there was no friction and for the first I got the answer to correctly be 5.625J and the second to be 3.375J. So since the system was already moving I tried to just take the normal force multiplied by the coefficient of kinetic energy but that answer was incorrect. PLEASE HELP.
 
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lil_50 said:
So since the system was already moving I tried to just take the normal force multiplied by the coefficient of kinetic energy but that answer was incorrect.
Not sure what you did. Multiplying the normal force times the coefficient of kinetic friction will give you the friction force. What then?

How did you find the answer to part 1?
 

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