Work done by motor against gravity

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SUMMARY

The discussion centers on the work done by a motor against gravity in a pulley system, specifically analyzing the equation derived from the conservation of energy: ∑T1 + ∑U1-2 = T∑2. The user attempts to solve for the work done by the motor and notes a discrepancy where the book's solution includes a factor of 2 for the motor's work term. This discrepancy is attributed to the influence of the pulley on the forces involved, necessitating a free body diagram to clarify the motor's effect on the weight being lifted.

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  • Understanding of conservation of energy principles in physics
  • Familiarity with work-energy equations
  • Basic knowledge of pulley systems and their mechanics
  • Ability to draw and analyze free body diagrams
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Students in physics, particularly those studying mechanics, engineers working with mechanical systems, and educators teaching concepts related to work and energy in pulley systems.

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Homework Statement


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Homework Equations


Conservation of Energy: ∑T1+∑U1-2=T∑2

The Attempt at a Solution



∫[/B](600+2s2)ds - 100*9.8*15 = (1/2)mv2.

However, the term of work done by the force is less than the term of work done by gravity! The book has a solution, which is exactly this except the term of work done by the motor is multiplied by 2. Could it have something to do with the pulley? Otherwise, I don't get it.

Thanks.
 
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Well, the motor is shown connected to the weight by means of the pulley, so you can't assume that the pulley is not there.

You should draw a free body of the sheave and see how its presence affects the force provided by the motor and on the weight being lifted.
 

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