Comparing work done in five cases with springs and tension

[haruspex]

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.

 

[isukatphysics69]

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

 

[isukatphysics69]

 

[isukatphysics69]

I plugged in some test values and used PE_SPRING = .5T

 

[isukatphysics69]

I also had a lot of sign mistakes when i was first doing the problem 11 that i just realized recently

 

[haruspex]

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 ∫₀^xT(s).ds = ∫₀^xks.ds = ½kx²=½T(x)x.

 

[isukatphysics69]

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

 

[isukatphysics69]

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 ∫₀^xT(s).ds = ∫₀^xks.ds = ½kx²=½T(x)x.

oh yes ok i see

 

[isukatphysics69]

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

 

[isukatphysics69]

i am truly living up to my username good on me

 

[haruspex]

i am truly living up to my username good on me

I see signs of progress