Lifting a rope

Classical Mechanics Level 5

A rope of length $l$ and mass density $\lambda$ lies in a heap on a floor. You grab one end of the rope and pull upwards with a force such that the rope moves at a constant speed $v$ . What is total work you do, by the time the rope is completely off the floor? Assume that the rope is greased.

Details and Assumptions

$\lambda =\si{2\ \kilo\gram/\meter}$
$L=\si{1\ \meter}$
$v=\si{1\ \meter/\second}$
$g=\si{10\ \meter/\second^2}$

The answer is 12.

1 solution

Deepanshu Gupta
Jan 13, 2015

Let The at t=t , 'x' length of rope as our system which is uplift by us in presence of gravity , and at the Lower end of rope there is heap of rope so no Tension act's at the end of heap.

$\displaystyle{{ F-\lambda gx=\cfrac { d(\lambda xv) }{ dt } \\ F=\lambda gx+\lambda { v }^{ 2 }\quad (\because \quad v=const.)\\ w=\int _{ o }^{ l }{ (\lambda gx+\lambda { v }^{ 2 })dx } \\ \boxed { w=\lambda ({ v }^{ 2 }l+\cfrac { g }{ 2 } { l }^{ 2 }) } }}$ .

I have a small doubt. @Deepanshu Gupta @Satvik Pandey

At the end of the lifting, the rope will be up in the air. Work done by gravity = $\text{Change in position of centre of mass} \times mg$

So work done by gravity = $-\frac{l}2 (\lambda l) g$

Its KE will be $\frac12 (\lambda l)v^2$

Since the rope is greased, no frictional forces act on the rope, so:

By work energy theorem

$W_{external force}+W_{gravity} = \text{Change in KE}$

So we get $W_{external force}$ as $11J.$

Help please! Please, can someone tell me the mistake in my approach?

Kp Govind - 6 years, 4 months ago

The crux of the matter is that all the collisions happening at the bottom of the rope(i.e. in the heap) are inelastic in nature.

Thus, there is a loss of energy, and conservation of energy (which is equivalent to the work energy theorem) gives an incorrect answer.

Shashwat Shukla - 6 years, 4 months ago

Ah. Okay. But how come the correct answer (12 J) arises without ever considering that?

The answer would be the same whether you took that the rope was lying in a straight line (where no such collisions are possible) or a heap.

Kp Govind - 6 years, 4 months ago

@Kp Govind – The difference is that, if it were lying in a straight line, then all parts of the chain would have the same acceleration and would be moving with the same velocity(this is the only way in which the chain can always be 'straight')...

This is clearly very different from what's happening in a heap, wherein, only a fraction of the rope is in motion at any given instant.

I am saying this because I think that your argument is based on the premise that because the final state of motion is the same(whether a heap or a straight line), and because the initial states are equivalent in kinetic and potential energy, the difference in their energy must also be the same.

The problem is that the aforementioned inelastic collisions are non conservative in nature.
An analogy can be as follows:

Consider a block that is pulled down from the same height, along two inclined wedges: one with, and one without friction till it reaches the ground. We can then consider what happens when it has the same velocity at the bottom of the incline. Clearly, the initial and final states are the same in K.E, P.E but the work done is different. Internal configuration does matter.

Shashwat Shukla - 6 years, 4 months ago

@Shashwat Shukla – Ah. Very interesting. Yeah, I did indeed think that if the final state was the same and initial energy was the same, you could apply CoE. Now, I realize that it was fallacious. Thank you for clearing that up. :)

Kp Govind - 6 years, 4 months ago

@Kp Govind – You're welcome :)

Shashwat Shukla - 6 years, 4 months ago

@Shashwat Shukla – A nice explanation indeed!

Miraj Shah - 5 years, 1 month ago

Nice solution Deepanshu.

satvik pandey - 6 years, 5 months ago

Thanks buddy !, Hey If we do slight variation in this problem , such that we apply constant force , instead of constant velocity . Then ask Time Taken when whole chain is pulled up ?

I'am getting answer as $\displaystyle{T=\cfrac { 3 }{ g } (\sqrt { \cfrac { { F }_{ o } }{ \lambda } } -\sqrt { \cfrac { { F }_{ o } }{ \lambda } -\cfrac { 2gl }{ 3 } } )}$ .

In opinion You should have also post this question , and I gave you Privileged To make this question . Choice is urs ! $\ddot\smile$

Deepanshu Gupta - 6 years, 5 months ago

Thanks bro. I will do that tomorrow. Happy MakarSankranti. :D

satvik pandey - 6 years, 5 months ago

@Satvik Pandey – No problem , and same to u ! $\ddot\smile$

Deepanshu Gupta - 6 years, 5 months ago

@Deepanshu Gupta – Please take a look. I have posted it just now. Is it fine?

satvik pandey - 6 years, 5 months ago

Lifting a rope

The answer is 12.

1 solution

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