Calculating Work, Energy, and Power: Integrating Force and Velocity

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SUMMARY

The discussion centers on calculating work, energy, and power by integrating force and velocity in a physics problem involving a rope. The user attempted to derive power using the equation P = Work/Time and integrating force from 0 to L, but their approach neglected the work done to accelerate the rope to velocity v. The correct method involves using the work-energy theorem, leading to the conclusion that the average power supplied is given by Pave = (ΔT + ΔU) / Δt, where ΔT and ΔU represent changes in kinetic and potential energy, respectively. The final answer aligns with the textbook's solution, which is option C.

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  • Understanding of the work-energy theorem
  • Familiarity with integration techniques in physics
  • Knowledge of kinetic and potential energy concepts
  • Basic grasp of momentum conservation principles
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  • #31
Delta2 said:
Ok I see and then we arrive at the answer you post at #2, which differs by a factor 1/2 on the first term from the book answer. Well, it beats me, according to your opinion why it differs? OK I know that conservation of energy does not hold always but why using the momentum approach we still don't get the right answer cause as you say the right answer will be something in between?
Maybe the flaw in the momentum calculation is P=Fv. The element being accelerated from rest only averages v/2 in that process, so it should be P=Fv/2.
 
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  • #32
I think I like it. And if there are discrete lengths (like a chain link) there will be some fluctuations but the average work will be as you say. And if we allow other degrees of freedom to bleed off the speed during this process we end up with drag forces which I am certain will be difficult to characterize and more difficult to quantify.
 

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