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

[isukatphysics69]

Homework Statement

Homework Equations

The Attempt at a Solution

Work = W = ΔKE

Case A: ΔKE = W_TENSION - W_SPRING - W_GRAVITY
Case B: ΔKE = W_GRAVITY - W_TENSION
Case C: ΔKE - ΔPE_SPRING = W_TENSION - W_GRAVITY
Case D: ΔKE + ΔPE_GRAVITY = -W_TENSION
Case E: W_SYSTEM = 0

Velocity and tension for both blocks are equivalent
ΔKE = .5mv^2I am assuming the blocks are moving to the left ( i don't know how i would determine whether the spring force is stronger than the tension force)

C > A (spring force becomes potential energy rather than negative work
E = 0 (No external forces on system means no work)

in the picture i have D>B but i think B > D because positive work from gravity added on is greater than only having negative work from tension ( i tried this and still wrong)
B > D (gravity becomes potential energy leaving only negative tension work

Right now my thinking is that
D Is less than 0 because it has only negative work from the tension
E is 0 because every force is internal, no external forces means no work
C and A are positive and C is greater than A since the spring force changes from negative work to potential energy.
Now i am unsure of B as i am typing this. i will continue trying but if anyone can confirm my reasoning above so far that will be great

 

[isukatphysics69]

I am starting to think that Case B is actually positive. The negative work from the rope is less than the gravitational work from earth.
Big positive work - small negative work is positive

 

[isukatphysics69]

I am unsure how to determing which direction that the blocks are falling, i am making an assumption that the big block is falling downward defeating the spring force and weight of the small block

 

[haruspex]

I think you need to supply more information from the earlier question.
I assume the spring is initially under either greater or lesser tension than corresponds to equilibrium - which?
Is there any friction?
What exactly do the red boxes signify?

 

[isukatphysics69]

So the spring in C becomes negative potential energy,

I think you need to supply more information from the earlier question.
I assume the spring is initially under either greater or lesser tension than corresponds to equilibrium - which?
Is there any friction?
What exactly do the red boxes signify?

The red boxes are the enclosed system
No friction is present

I think you need to supply more information from the earlier question.
I assume the spring is initially under either greater or lesser tension than corresponds to equilibrium - which?
Is there any friction?
What exactly do the red boxes signify?

This is the first part of this homework, there is no friction present. Red boxes are the enclosed system. Example red box in A covers only the small box, it might cover a little of the spring on top but that spring is an external force, the only part of the system internally is the small box

 

[isukatphysics69]

Case A: ΔKE = W_TENSION - W_SPRING - W_GRAVITY > 0
Case B: ΔKE = W_GRAVITY - W_TENSION > 0
Case C: ΔKE - ΔPE_SPRING = W_TENSION - W_GRAVITY > 0
Case D: ΔKE + ΔPE_GRAVITY = -W_TENSION < 0
Case E: W_SYSTEM = 0

Assuming big mass is falling downward

So now i believe i have to use KE = .5mv^2 to compare the positive cases and rank them​

 

[isukatphysics69]

wait a second i think the negative potential energy from the spring in case c is equivelant to the negative work in case a

 

[isukatphysics69]

This is mental

 

[isukatphysics69]

i am completely stumped

 

[haruspex]

Just realized I misread the question - it is a bit fuzzy. I read "relaxed" as "released".

 

[isukatphysics69]

Just realized I misread the question - it is a bit fuzzy. I read "relaxed" as "released".

Wait i am thinking i have to use conservation of momentum now since there is an initial and final state

 

[haruspex]

So now i believe i have to use KE = .5mv^2 to compare the positive cases and rank them

Yes.

i think the negative potential energy from the spring in case c is equivelant to the negative work in case a

Not sure what you mean by that, but what you wrote in post #6 for C is not consistent with what you wrote for A.

 

[haruspex]

Wait i am thinking i have to use conservation of momentum now since there is an initial and final state

You would have to take the momentum of the Earth into account.

 

[isukatphysics69]

You would have to take the momentum of the Earth into account.

That doesn't seem reasonable

 

[haruspex]

That doesn't seem reasonable

Right, don't go there.

 

[isukatphysics69]

Right, don't go there.

One of my main confusions here is ranking something with potential energy against something without. So ranking A against C is going to to be

Case A: ΔKE = W_TENSION - W_SPRING - W_GRAVITY > 0

Case C: ΔKE - ΔPE_SPRING = W_TENSION - W_GRAVITY > 0
using ΔKE = KE_FINAL - KE_INITIAL where kinetic energy initial is 0

it looks like
Case A: .5(1kg)*(1²m/s) = W_TENSION - W_SPRING - W_GRAVITY > 0

Case C: .5(1kg)*(1²m/s) - ΔPE_SPRING = W_TENSION - W_GRAVITY > 0
Is the potential energy equal to the work done by the string in the case A? i am forgetting how that works​

 

[haruspex]

Case A: ΔKE = W_TENSION - W_SPRING - W_GRAVITY > 0

Case C: ΔKE - ΔPE_SPRING = W_TENSION - W_GRAVITY > 0
​

​

You need to define clearly what these variables represent so that you get the signs right. Is W_SPRING the work done by the spring on the mass or the work done on the spring by the mass? Similarly W_GRAVITY.
If you have a list of all the works done on a system by various actors, W₁, W₂, ..., what is the total work done on the system?

 

[isukatphysics69]

You need to define clearly what these variables represent so that you get the signs right. Is W_SPRING the work done by the spring on the mass or the work done on the spring by the mass? Similarly W_GRAVITY.
If you have a list of all the works done on a system by various actors, W₁, W₂, ..., what is the total work done on the system?

The W_SPRING and W_GRAVITY i have as the work done on the mass, same for all cases, The total work will just be the sum of all of the works on the system

 

[isukatphysics69]

Case A: There is positive work from the tension, negative work from the spring, negative work from gravity
Case B: There is positive work from gravity, negative work from tension
Case C: There is positive work from tension, negative work from gravity, potential energy stored in the spring (negative since it wants to snap backwards)
Case D: There is negative work from tension, potential energy from gravity (positive because block is falling towards it)
Case E: there is 0 work because everything is internal to the system

I just want to make sure i am understanding the fundamentals of this

 

[haruspex]

The total work will just be the sum of all of the works on the system

So why the minus signs?

 

[isukatphysics69]

So why the minus signs?

Well isn't there direction for the work? so the tension force is pulling upward, the motion is upward, positive work, the spring force is pulling downward, the motion is upward, negative work

 

[haruspex]

Well isn't there direction for the work?

A work done on a system may have a negative value, but the total work done is the sum of the works.
E.g. if you perform work W_you to lift a mass m up height h the total work done on it is W_you+W_gravity, where W_gravity=-mgh.

 

[isukatphysics69]

A work done on a system may have a negative value, but the total work done is the sum of the works.
E.g. if you perform work W_you to lift a mass m up height h the total work done on it is W_you+W_gravity, where W_gravity=-mgh.

So between just C and A, C has potential energy, A does not, they both have the same work variables, C > A

Actually that isn't even making sense nevermind

 

[haruspex]

So between just C and A, C has potential energy, A does not, they both have the same work variables, C > A

I do not know how that connects with what I wrote.
I am talking about how you assess the work done on a system. It is the sum of the works done by various actors. Sum means addition. It may well be that some of these works turn out to have a negative value, but the total is still the sum.
If one actor does work +5 and another does work -3 then the total is +5+(-3), not +5-(-3).

 

[isukatphysics69]

I do not know how that connects with what I wrote.
I am talking about how you assess the work done on a system. It is the sum of the works done by various actors. Sum means addition. It may well be that some of these works turn out to have a negative value, but the total is still the sum.
If one actor does work +5 and another does work -3 then the total is +5+(-3), not +5-(-3).

Do you agree with this?
​

 

[isukatphysics69]

Ok the answer is B > C > A > E = 0 > D

 

[isukatphysics69]

This took me 4 hours i will drop out of school and become fisherman

 

[haruspex]

Well done.
I assume you had to use that the system is no longer accelerating to determine that B>C.

 

[isukatphysics69]

I'm ashamed to say that i took a guess here.

 

[isukatphysics69]

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..

 

[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