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

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Total work for the blocks in a pulley system with friction involves calculating the work done against friction for each block. The initial calculations for work done without friction yielded 5.625 J for the 20.0-N block and 3.375 J for the 12.0-N block. To account for friction, the friction force must be determined by multiplying the normal force by the coefficient of kinetic friction. The discussion emphasizes the need to incorporate friction into the work calculations to find the correct total work done. Accurate calculations are essential for understanding the dynamics of the system.
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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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