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

  • Thread starter Thread starter isukatphysics69
  • Start date Start date
  • Tags Tags
    rank System Work
Click For Summary
SUMMARY

The discussion revolves around the ranking of work done in various cases involving tension, gravity, and spring forces in a mechanical system. The participants analyze five cases (A to E) using the work-energy principle, concluding that Case B has the highest positive work due to gravitational influence, followed by Case C, Case A, Case E (zero work), and Case D (negative work). Key equations utilized include ΔKE = WTENSION - WSPRING - WGRAVITY and the relationship between potential energy and work done by springs. The absence of friction is confirmed, and the importance of defining variables clearly is emphasized for accurate calculations.

PREREQUISITES
  • Understanding of the work-energy principle in physics.
  • Familiarity with concepts of kinetic energy (KE) and potential energy (PE).
  • Knowledge of forces acting on a system, specifically tension and gravity.
  • Ability to manipulate equations involving work done by various forces.
NEXT STEPS
  • Study the work-energy theorem and its applications in mechanical systems.
  • Learn about the conservation of energy in systems with springs and gravitational forces.
  • Explore the concept of potential energy in elastic materials, specifically Hooke's Law.
  • Investigate how to define and calculate work done by multiple forces in a system.
USEFUL FOR

Students of physics, mechanical engineers, and anyone interested in understanding the dynamics of forces in mechanical systems will benefit from this discussion.

  • #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.
 
  • Like
Likes isukatphysics69
Physics news on Phys.org
  • #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
 

Attachments

  • pd.PNG
    pd.PNG
    10.8 KB · Views: 319
  • #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.
 
  • Like
Likes isukatphysics69
  • #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
 

Attachments

  • FIGUREOUT.PNG
    FIGUREOUT.PNG
    16.1 KB · Views: 286
  • #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:
 
  • Like
Likes isukatphysics69

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
5
Views
8K
  • · Replies 58 ·
2
Replies
58
Views
3K
Replies
8
Views
1K
Replies
29
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K