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Work done by action-reaction pair of tension force on a system is zero if the string is inextensible

I am trying to understand this statement taking the following arrangement in consideration. Here, a block of mass m initially at rest, is released such that it extends the spring as it moves downwards. The spring is initially at its natural length before the block is released. This is the arrangement before the block is released. Both pulleys and the strings are massless. enter image description here

After the block is released, it extends the spring by say, x metres. The pulley as a result moves to the right by $\frac{x}{2}$ metre, and the block is constrained to move downwards by $\frac{x}{2}$ metre. The tension force $T_1$ equals $kx$ and $T_2$ equals $2kx$.

Now coming back to the statement : Work done by action-reaction pair of tension force on a system is zero if the string is inextensible.

I treat the bllock, both pulleys, the spring, the strings, and the stationary objects the spring and the smaller of the two pulleys are fixed to, as a system.

enter image description here

I want to understand how work done by action-reaction pairs of tension forces acting on this system is zero. I understand that action-reaction pair of $T_2$ does zero work on the system (negative work on the block, equal to $\frac{-1}{2}$$kx^2$. And an equal amount of positive work on the larger of the two pulleys, hence total work done by it on the system is zero)

What I am not able to understand is, how is work done by action-reaction pair of $T_1$ on the system is zero? Should I treat the pink string (a different colour used just for reference) as a string? Or should I treat it as a spring because it is connected to one end of the spring? Hope my question is clear. If it’s not, please let me know and I’ll reframe it.

Edit : If I do the math, work done by the spring on the bigger pulley is twice of $\int{-kx}$$\frac{dx}{2}$ (upper and lower limits are $x$ and $0$ respectively), which equals $\frac{-1}{2}$$kx^2$. And I understand that the bigger pulley does an equal amount of positive work on the spring.

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  • $\begingroup$ Why would you consider work "done on the big pulley"? You could then, consistently, also consider work done on each individual millimeter of string. In other words, if the forces on an object cancel out (action=-reaction) there is no work. $\endgroup$
    – user257090
    Commented May 17, 2020 at 8:33
  • $\begingroup$ @DoctorNuu Why would you consider work done on the big pulley? Because I am treating the arrangement as a system, that includes the big pulley and it is not fixed. I understand that net force acting on the big pulley is zero so work done on it is zero. But as the title of the question says, I am trying to understand how work done by action-reaction pair of tension force on a system add up to zero. Hence I considerd the work done on the moving pulley. $\endgroup$
    – 4d_
    Commented May 17, 2020 at 8:49
  • $\begingroup$ @DoctorNuu and I don't think the forces acting on the moving pulley are action-reaction pairs, are they? Don't action and reaction act on different bodies? I think the tension force acting on the moving pulley and the block form the action-reaction pair $\endgroup$
    – 4d_
    Commented May 17, 2020 at 8:51

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