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#19
Feb513, 08:41 AM

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But do you mean that the question given is wrong? If I was asked to calculate the energy lost, how can I find it? Can you give links to some good resources where I can learn more about these variable mass systems? 


#20
Feb513, 11:41 AM

P: 758

But again, work energy theorem,i.e. W_{net}=ΔK.E. still applies. Did you try applying it ? If work energy theorem fails, then I am sure that this question is somewhat a magic. Work energy theorem applies to every inertial earth system, as per H.C. Verma. 


#21
Feb513, 12:15 PM

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F_{net} = ρu^{2}ρgx Now as x=u^{2}/2g F_{net} = ρu^{2}ρu^{2}/2 F net =ρu^{2}/2 Integrating net F with respect to dx from 0 to x, W=ρu^{2}x/2 So W=ρu^{2}x/2 which is equal to ΔKE. As length x is in air , its incorrect to consider normal reaction. I think this is where you blundered. Edit: As per me the correct answer is D. Then only this question is justified by work energy theorem. 


#22
Feb513, 01:03 PM

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This is not the first time this has happened. Textbooks can be erroneous. Multiple choice: The problem apparently is particularly bad (a) In far too many online automated physics homework problem systems. (b) In far too many Indian physics texts. (c) Both of the above are true. Apparently India has rather weak enforcement of its copyright laws, which leads Indian textbook authors to copy problems from one another. It's much easier to copy some other author's alreadyworked problems rather than to create new ones. That's fine (other than copyright issues) if the problems are well written and the solutions are correct. It's not so fine if the question is poorly written, if the correct answer isn't present in the multiple choice list, or if the solution in the answer guide / worked example is incorrect. We've chased down a number of problems with Indian physics texts at this site, occasionally seeing the exact same bad question replicated in dozens of different texts. In reality, there will be some energy loss due to entropy. (Note: the energy isn't really "lost". It's just converted to unusable energy: Heat.) A cable or chain may heat up a bit due to nonconservative interactions as the cable/chain grows taut, links shift, etc. We can ignore all that messiness by assuming an ideal chain. Even a terribly constructed, nonideal chain will not heat up to the extent implied by the wrong answer to this problem. 


#23
Feb513, 03:17 PM

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I’ll add my 2 cents to the confusion! For any system ##F_{net}^{ext} = M a_{cm}## where M is the total mass of the system.
Let the system be the entire chain alone. The net external force is ##F_{net}^{ext} = F_{lift} – Mg + F_N## where ##F_{lift}## is the lifting force and ##F_N## is the normal force acting on the part of the chain still resting on the platform. When amount ##x## of chain is off the table, the center of mass of the section that is off the table is at ##x/2## above the platform and has mass ##\rho x##. So, relative to the platform, the location of the center of mass of the whole chain is ##x_{cm} = \frac{(\rho x)(x/2)}{M} = \frac{\rho}{2M}x^2##. Taking the second time derivative of this yields ##a_{cm} = \frac{\rho}{M}(\dot{x}^2 + x\ddot{x}) = \frac{\rho}{M}u^2## since ##\ddot{x} = 0## So, we have ##F_{lift} – Mg + F_N = \rho u^2## or ##F_{lift}= \rho u^2 + Mg  F_N## If we take the normal force to equal the weight of the part of the chain resting on the table, then ##F_N = (M\rho x)g## and we get ##F_{lift}= \rho u^2 + \rho g x = \rho (u^2+g x)## 


#24
Feb513, 05:01 PM

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ehild 


#25
Feb513, 05:15 PM

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At any rate, if the energy is right, what is wrong with the F = dp/dt solution? Since that is basically the definition of force, it, too, must be right. Picking the energy to be right because dm/dt is "magic" is not really good enough. 


#26
Feb513, 05:21 PM

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F acts at the lifted end of the rod and its work is ∫Fdx. We do that work on the whole system, and that work increases the energy. Unless something else nonconservative force does also work.
ehild 


#27
Feb513, 05:27 PM

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Regarding energy, I suspect that when each link is jerked off the platform, it's like an inelastic collision causing internal energy to increase. The chain will get warmer!



#28
Feb513, 05:34 PM

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ehild 


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Feb513, 05:38 PM

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Feb513, 05:42 PM

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#31
Feb513, 05:57 PM

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#32
Feb513, 06:02 PM

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#33
Feb513, 06:15 PM

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#34
Feb513, 06:16 PM

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#35
Feb513, 06:16 PM

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I'll be very explicit. Using an F=dp/dt approach and attributing all of that change in momentum to the action by the hoist yields a nonsense answer. This is one of many cases where naively using F=dp/dt is flat out wrong. It's F=ma, not F=dp/dt. The heating implied by that F=dp/dt approach is ludicrous. Even more ludicrous is the cooling that would result from reversing the process were this approach correct. Even yet more ludicrous is the fact that this erroneous approach leads to an over unity device. 


#36
Feb513, 06:28 PM

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