Consider the above scenario: In the subsequent motion, we need to find the work done by tension on the (trolley + mass) system.
Solution: Suppose at an instant, the velocity of the trolley (and hence that of point $A$) is $\vec{V_{a}}$, and the velocity of the point mass is $\vec{V_{b}}$. Then the power delivered by tension to the system is $\vec{T}\cdot(\vec{V_{b}}-\vec{V_{a}}$). Now, tension is always directed along the string, and, the velocity of the mass relative to point $A$, (i.e. $\vec{V_{b}}-\vec{V_{a}}$) is always perpendicular to the string. So the dot product is zero, i.e., the power delivered to the string is zero at all times. Hence, the tension does no work on the system.
Consider another scenario: https://physics.stackexchange.com/a/571564/196626. Here, the work done by the normal reaction on the system is zero.
In situation 1, the work was zero, because the relative velocity was constrained to be perpendicular to the string. (string constraint)
In situation 2, the work was zero since the block was constrained to move along the wedge, (and hence perpendicular to the normal force). (contact constraint).
These two situations demonstrate what has often been the case in many classical mechanics questions: The work done by tension and normal forces on the system =0. In these two particular situations, the common thing seems to be that the tension (and normal) are constraining the elements of the system. It seems like the common link is that they are constraint forces. My question attempts at a generalization:
Can we claim (in general) that the work done by constraint forces on a system is always zero?
I somehow feel that the reason somehow lies in the meaning of the term "constraint" itself, but it's just a feeling.