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In All the Classical Mechanics problems I have come across so far, There's one thing that happens invariably: That the work done by tension is zero. Mostly, It simply happens because the (massless) string is in-extensible(And therefore, no displacement),And the tension in the string is the same.

However, consider:

  1. A pulley with mass, that is free to rotate, has a string(with blocks connected at its free end) wrapped around it.When the system is released, The tension in the Two pieces of string will be different, yet the displacement of the blocks along the string will be equal and opposite. Why is work done by tension zero here? It seems to me that the work done should be $T1*x + (-T2*x)$
  2. A fixed disc free to rotate about its center (a flywheel)has a string wrapped around it, with a block attached to it. As the block falls, The tension in the string makes the disc rotate. Now, When the disc rotates by an angle $\theta$,(assuming the string doesn't slip) a length=$r\theta$ of string unwraps. The length of the piece of string which is attached to the block is changing. Why is work done by tension zero?(on block +disc system)
  3. How can we, (if it is true,that is), generalize that the work done by tension is always zer0? An earlier explanation that I had was that its an internal force.However, I believe internal forces can do work: for example, a spring connected to two blocks does work. An explanation given to me by a friend , was : "A mass-less string cant store energy", which seemed reasonable at first, But i believe the concept of "storing energy" requires the concept of work itself, which makes this argument circular.
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  • $\begingroup$ it is not always true that work done by tension is 0 $\endgroup$ – maverick May 21 '20 at 7:39
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In the first case,the rotating pulley,Tension cannot be 0 as its the differnece of Tension which is providing the torque for the pulley to rotate. .In all the other cases,there are many assumptions like zero gravity, inextensebale string or massless string.As tension is not a conservative force we dont consider the work done by tension.If you have studied wave motion in strings,you will know that we actually measure the work done by tension in pulling the string in order to produce waves,work done by tesnion is not 0 always.

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We need to be careful of context here. If I have two objects joined in some way by a light inextensible string, then the string cannot change the total energy in the system. The tension in the string can do work (add energy) to object A while it does negative work (removes energy) from object B, but it does no work on the system as a whole. Just saying that the tension can (or cannot) do work is not precise enough.

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In the first and second case, you haven't considered the work done by tension on the pulley and flywheel respectively. If you do so, you'll find that the work cancels out perfectly. Of course, if you consider only the block, then tension is definitely doing work on it.

However on an entire system containing a string, tension cannot do work because an ideal string has no properties which allow it to store energy. Which means that any work done on the string can only translate to kinetic energy, and never potential energy.

Edit-

Work done on heavier block = -(T1*x)

Work done on lighter block = (T2*x)

Work done on pulley = τ*θ

= (T1-T2)Rθ (τ=net torque R=radius of pulley)

=(T1-T2)*x

Net work= (-T1*x) + (T2*x) + (T1-T2)x = 0

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  • $\begingroup$ Thats what i thought, But I am unable to calculate work on the pulley by the string $\endgroup$ – satan 29 May 22 '20 at 14:34
  • $\begingroup$ would you consider showing it as part of your answer? $\endgroup$ – satan 29 May 22 '20 at 14:34

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