Work, Energy, Power Question

  • #1
Aurelius120
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Homework Statement:
I came across this question in my textbook
Relevant Equations:
Power=Work/Time=Force×Velocity
I tried to solve it by integrating force from 0 to L
dF=dm.g
where dm=λdx
And then I multiplied it with velocity to get power because velocity is constant


∫(vdF)=v ∫(dF)

But the book used integration to find work done and divided it by time for power

My answer was λlgv(Option B)

Giving (Option C) as answer



Where am I wrong?

Thanks for the help
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  • #2
Dr.D
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Your first statement is
Homework Statement:: I came across this question in my textbook
Relevant Equations:: Power=Work/Time=Force×Velocity

I tried to solve it by integrating force from 0 to L
To solve is a verb that, in most cases, means to isolate one particular quantity on the left side of an equals sign. What are you trying to solve? I don't know, and I rather doubt that you do , so I suggest that you start again and consider where you want to go.
 
  • #3
haruspex
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Your dF expression only gives the force necessary to suspend an element length dx, statically. Multiplying by v then gives the power needed to raise that element steadily at speed v, but you have not considered the work done to get each element up to speed v from rest.
As against that, your integral assumes every element is lifted through height l.

Unfortunately, none of the options provided is strictly correct. The author has wrongly assumed conservation of work. If you don't assume that but instead consider momentum you arrive at ##\lambda v^3+\frac 12\lambda l v g##.
In practice, the answer will be somewhere between the two, its exact value depending on the mechanical properties of the rope. The reason it is not as much as given by the momentum analysis is connected with the chain fountain effect (q.v.).

What is the textbook?
 
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  • #4
Delta2
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I agree with the answer and method of book. It seems very natural to me to use the work energy theorem for this.
On the matter of why your method is wrong I agree with @haruspex.

But I disagree with @haruspex on why we shouldn't use the work energy theorem for this. He seems to imply that some power from the external force is lost as heat and not converted completely to kinetic energy of the rope, but I really can't see why we should view it that way.
 
  • #5
haruspex
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But I disagree with @haruspex on why we shouldn't use the work energy theorem for this. He seems to imply that some power from the external force is lost as heat and not converted completely to kinetic energy of the rope, but I really can't see why we should view it that way.
Try using conservation of momentum, then try to explain the different result.
You could also try modelling what happens as the rope moves from rest. E.g. treat it as a sequence of chain links, each starting vertical. So each link moves up one link length before engaging the next. What happens then? If work is conserved, it will result in a vertical oscillation. That will swiftly decay, but not assist in raising the chain.
 
  • #6
Delta2
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Try using conservation of momentum, then try to explain the different result.
You could also try modelling what happens as the rope moves from rest. E.g. treat it as a sequence of chain links, each starting vertical. So each link moves up one link length before engaging the next. What happens then? If work is conserved, it will result in a vertical oscillation. That will swiftly decay, but not assist in raising the chain.
Oscillation? No clue what you talking about. Anyway I think it is safe to assume that all the energy provided by the external force goes into kinetic energy of the rope, a reasonable assumption from which the book answer follows.

Also no clue how we can apply conservation of momentum here, I think we need additional info on the body or system that applies the external force to the rope.
 
  • #7
haruspex
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Oscillation?
Imagine two chain links, like linked paper clips, standing vertically. The lower end of one sits on the lower end of the other. If you lift the upper its lower end eventually engages the top end of the other. If work is conserved in that collision then it will lead to bouncing. This does not contribute to the overall upward motion of the pair, so is effectively lost energy. In practice, this oscillation will die out quickly.

The situation with a rope may be more like links that start horizontally in a sort of zigzag stack. In this case, as each link is lifted it gains rotational energy. What becomes of that? It is possible it does contribute to lifting the next link. This is what seems to happen in a chain fountain, and in another recent thread concerning a hanging loop of rope when one end is released.
 
  • #8
hutchphd
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Perhaps a reset here makes sense .
One need not integrate anything here. $$P_{ave}=\frac {\Delta (T+U)} {\Delta t}$$where ##\Delta T=\frac {\lambda l v^2} 2##, ##\Delta U= \lambda l g \frac l 2 ##, and ##\Delta t=\frac l v##. I think that does it.
 
  • #9
Delta2
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Perhaps a reset here makes sense .
One need not integrate anything here. $$P_{ave}=\frac {\Delta (T+U)} {\Delta t}$$where ##\Delta T=\frac {\lambda l v^2} 2##, ##\Delta U= \lambda l g \frac l 2 ##, and ##\Delta t=\frac l v##. I think that does it.
I agree with this, the silent assumption you make is that all energy from the external force goes into mechanical energy of the rope, but @haruspex insists that some energy is going into oscillations that die, thus there is loss of energy. Or that's what i can understand from his posts.
 
  • #10
hutchphd
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The issue is what happens if, when the rope is entirely off the table, we stop pulling. Then the Kinetic Energy in the rope will be dissipated by the thrashing of the rope, and only the potential energy remain.
But that is not the question asked here and the correct answer is (C) as discussed.
 
  • #11
haruspex
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The issue is what happens if, when the rope is entirely off the table, we stop pulling. Then the Kinetic Energy in the rope will be dissipated by the thrashing of the rope, and only the potential energy remain.
But that is not the question asked here and the correct answer is (C) as discussed.
That's not my issue. One should never assume conservation of work without justification.

I have seen equivalent problems posted several times over the last ten years on PF. Sometimes the official answer assumes work conservation, sometimes it uses momentum conservation (arriving at half the work being lost).
I used to take the momentum view, but have come to understand that the real behaviour will lie between the two, and where exactly in that range will depend on subtle details of the material involved. What exactly happens at "lift off"?

Depending on the model one uses, you can see how the elements may acquire horizontal or vertical oscillations that do not contribute to the overall upward motion. In the present case, the illustration has the rope starting in a coil laid flat, so at lift off the elements will acquire some horizontal motion. What becomes of that?

At the least, if you reject the momentum result then some explanation is required.
 
  • #12
hutchphd
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I do not fully understand the distinction
Are you saying that the external average power supplied will be more than what I calculated?
 
  • #13
haruspex
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I do not fully understand the distinction
Are you saying that the external average power supplied will be more than what I calculated?
Yes. See post #3. If you consider each element of the rope being 'instantly' accelerated to v then momentum leads to ##\lambda v^3##, not ##\frac 12\lambda v^3##
 
  • #14
hutchphd
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Yes. See post #3.

I missed that (apologies). I'm not yet quite convinced however . Is the implication that there are effectively two quadratic momentum degrees of freedom for the rope because it wiggles both side to side and up and down? I mean the energy still has to go somewhere eventually.
We posit ideal ropes all the time and it seems maybe one could stipulate a no-stretch perfectly flexible rope. But reality is clearly in-between and messy.

You could also try modelling what happens as the rope moves from rest. E.g. treat it as a sequence of chain links, each starting vertical. So each link moves up one link length before engaging the next. What happens then? If work is conserved, it will result in a vertical oscillation. That will swiftly decay, but not assist in raising the chain.
I don't think this argument holds. The key is "this will swiftly decay". Consider a mass on a weightless spring being lifted vertically via the spring by an agent at speed v. The fact that the mass oscillates vertically does not change it being lifted. The energy supplied by the lifting force is still divided equally between T and V. If the energy is not dissipated (to other degrees of freedom internal or geometrical) then (C) holds on average.
Interesting.
 
  • #15
haruspex
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If the energy is not dissipated (to other degrees of freedom internal or geometrical) then (C) holds on average.
That doesn't follow. Not dissipated could just mean the oscillations persist, but the energy therein has not been allowed for in arriving at C.
As I wrote, the challenge in using work conservation is explaining what is invalid in the momentum conservation argument.

Here's a relevant thread: https://physics.stackexchange.com/q...vation-to-the-problem-of-pulling-a-bent-carpe
But in the case there I think I can see how the momentum argument may be wrong.
 
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  • #16
Delta2
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@haruspex suppose that you were given this question in an exam but you weren't allowed to write your own answer and you were forced to choose one of the options, which option would it be?
 
  • #17
haruspex
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@haruspex suppose that you were given this question in an exam but you weren't allowed to write your own answer and you were forced to choose one of the options, which option would it be?
Faced with it prior to a few years back, I would unhesitatingly have used the momentum argument. Since then, I have become aware of the chain fountain effect, and a scenario in which a rope is at first suspended from two adjacent points, then one end released. (I believe there's a link to such a question from the link I put in post #15.). In the second case, experimental evidence is that most of the work is conserved and, in consequence, the free end falls faster than g!
However, as I noted above, I strongly suspect that the extent to which work is conserved depends on the physical attributes of the rope.
In the present case, with the rope in a coil lying on the ground, it is far from clear whether it will emulate the falling rope. In that latter case, it would seem that the rotational inertia as each segment turns around the bend to join the hanging section acts to pull down the next segment. It is not apparent to me how something similar will happen with the lying coil.
 
  • #18
hutchphd
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I am still skeptical.
Yes. See post #3. If you consider each element of the rope being 'instantly' accelerated to v then momentum leads to λv3, not
How does one do this? The rope is a rope because it is continuous. I can't see this as the limit of any realize-able process other than accelerating the entire rope.

The side to side forces supplied by the agent do not strictly "do work" and so there is a semantic issue here also perhaps.
 
  • #19
haruspex
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The rope is a rope because it is continuous. I can't see this as the limit of any realize-able process other than accelerating the entire rope.
You have to model it in some way that makes it amenable to analysis. That is the basis of calculus.
What does a rope look like on the microscopic scale? The fact that a rope can go slack suggests some sort of linked chain. A thread of rubber might be a different story.
My point is that different choices of model may lead to different results, and each might be valid for some material.
The side to side forces supplied by the agent do not strictly "do work"
If side to side motion results, it did work.
 
  • #20
hutchphd
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The nut on a guitar neck is required for there to be side to side motion of the string and yet it does no work. I agree that some initial motion must imparted and energy is needed. In steady state that will be from the tension in the rope. I/m still pondering ( and momentum conservation is a big stretch) .
 
  • #21
haruspex
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In steady state that will be from the tension in the rope
True. I cannot show why that motion should continue to increase in the absence of losses.
momentum conservation is a big stretch
Why? I generally consider it more reliable than work conservation; work can be lost in subtle ways.
Consider this model: two rigid bars, hinged together at one end, lay one on the other. We lift the free end of the upper with a long massless string (so it can be supposed to remain vertical even as the rod tilts up). On approaching vertical, the first rod starts to lift the second rod. The rotational inertia of the first rod carries it past vertical and assists in lifting the second rod. The rotational inertia of the second rod increases the normal force from the ground.
That is one way in which the assumptions of the obvious momentum argument may be violated.
However, I can also dream up models in which that does not happen, e,g, a loose chain with little elasticity.
 
  • #22
Delta2
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@haruspex, I think you are "refining the mosquito" (direct translation from ancient greek phrase, not sure how the english phrase is) and that the energy that goes into oscillations or lost in any other way is really small so what the book does or what is mentioned at post #8 is mostly correct.
 
  • #23
haruspex
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@haruspex, I think you are "refining the mosquito" (direct translation from ancient greek phrase, not sure how the english phrase is) and that the energy that goes into oscillations or lost in any other way is really small so what the book does or what is mentioned at post #8 is mostly correct.
That's not good enough. You have to explain why the momentum argument gives the wrong result.
 
  • #24
Delta2
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That's not good enough. You have to explain why the momentum argument gives the wring result.
I don't know how the momentum solution goes, If it is not too much trouble for you maybe you can pm me the details or make a post in this thread.
 
  • #25
haruspex
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I don't know how the momentum solution goes, If it is not too much trouble for you maybe you can pm me the details or make a post in this thread.
In time dt a further mass ##\rho vdt## is lifted from the ground. That means it goes from rest to speed v, so the impulse is ##Fdt=\rho v^2dt##. Power ##=Fv=\rho v^3##.
As I have remarked, I believe both conservation law approaches have flaws, and the truth will lie somewhere between the two, but where may well depend on details of the configuration and the material properties.
 
  • #26
Delta2
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In time dt a further mass ##\rho vdt## is lifted from the ground. That means it goes from rest to speed v, so the impulse is ##Fdt=\rho v^2dt##. Power ##=Fv=\rho v^3##.
As I have remarked, I believe both conservation law approaches have flaws, and the truth will lie somewhere between the two, but where may well depend on details of the configuration and the material properties.
No I reckon that's not the correct way to apply momentum impulse theorem, cause F is the external force applied at the free edge of the rope, you need to use the tension of the rope at the point where the mass ##\rho dx=\rho v dt## is located, and the problem here is we are not given enough information to calculate the tension of the rope (which I think will vary along the rope, it will not be constant in time or in position).
 
  • #27
haruspex
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No I reckon that's not the correct way to apply momentum impulse theorem, cause F is the external force applied at the free edge of the rope, you need to use the tension of the rope at the point where the mass ##\rho dx=\rho v dt## is located, and the problem here is we are not given enough information to calculate the tension of the rope (which I think will vary along the rope, it will not be constant in time or in position).
The rope between what's immediately being lifted and the top is moving at constant speed, so the applied force exceeds the F in my post by the weight of the suspended length of rope.
 
  • #28
Delta2
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The rope between what's immediately being lifted and the top is moving at constant speed, so the applied force exceeds the F in my post by the weight of the suspended length of rope.
So the correct equation is ##F(t)dt-\rho x(t) gdt=\rho v^2dt##?
 
  • #30
Delta2
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Yes.
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?
 
  • #31
haruspex
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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
hutchphd
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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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