Ranking the Work on Each System: Which Case Has the Most Positive Work?

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Homework Help Overview

The discussion revolves around ranking the work done on various systems in different cases, specifically focusing on the positive work associated with each case. The subject area includes concepts of work, kinetic energy, potential energy, and forces acting on a system.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations of work done in different cases, questioning the signs and values associated with tension, gravity, and spring forces. Some participants express uncertainty about the direction of forces and the implications of potential energy in their rankings.

Discussion Status

There is ongoing exploration of the relationships between the various cases, with some participants attempting to clarify their reasoning and assumptions. Questions about the definitions of variables and the effects of potential energy are being raised, indicating a productive dialogue without a clear consensus yet.

Contextual Notes

Participants note the absence of friction in the scenarios and the need for more information regarding the initial conditions of the spring and the system setup. There is also mention of potential confusion stemming from the wording of the problem statement.

  • #31
isukatphysics69 said:
I just had no idea how to do it my mind was completely blank like an unfillable void there is a gap in my knowledge of energy that prevented me from finishing off the problem and i took what i had and made a guess based off of instinct idk how to explain i just figured the tension forces are the same in all cases and ranked them accordingly using the fact that the tension forces are equal in magnitude and the gravity's will vary idk..
I found it helpful to substitute the usual sorts of expressions for the Ws. If the system moved a distance y then gravitational works are terms like mgy, Mgy; since the system is no longer accelerating the change in spring PE is ½Ty and T=Mg; etc.
 
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  • #32
haruspex said:
I found it helpful to substitute the usual sorts of expressions for the Ws. If the system moved a distance y then gravitational works are terms like mgy, Mgy; since the system is no longer accelerating the change in spring PE is ½Ty and T=Mg; etc.
So it is a fact that change in potential energy of a spring is half of the tension? That just helped me solve the next question
 
  • #33
pd.PNG
 

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  • #34
I plugged in some test values and used PESPRING = .5T
 
  • #35
I also had a lot of sign mistakes when i was first doing the problem 11 that i just realized recently
 
  • #36
isukatphysics69 said:
So it is a fact that change in potential energy of a spring is half of the tension? That just helped me solve the next question
Energy cannot equal tension.
When a spring of modulus k is extended by x the final tension is T(x)=kx. The work done is ∫0xT(s).ds = ∫0xks.ds = ½kx2=½T(x)x.
 
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  • #37
FIGUREOUT.PNG

I was able to figure this one out because i realize that work is change in kinetic energy so the kinetic energys will be 0, there are two cases that have stored potential energy the D case being negative stored potential and the C case being positive stored potential
 

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  • #38
haruspex said:
Energy cannot equal tension.
When a spring of modulus k is extended by x the final tension is T(x)=kx. The work done is ∫0xT(s).ds = ∫0xks.ds = ½kx2=½T(x)x.
oh yes ok i see
 
  • #39
Wait a second i see what you mean now on the last problem the question stated "maximum speed" meaning acceleration would be 0 that went over my head at the time
 
  • #40
i am truly living up to my username good on me
 
  • #41
isukatphysics69 said:
i am truly living up to my username good on me
I see signs of progress :smile:
 
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