Why is the net work done in a pulley-string system zero? In any pulley system, where the pulleys and strings are massless and frictionless, why is the net work done by Tension zero?
 A: In the systems you describe, each string connects always two masses. The tension force exerted on these two masses by the string is equal in magnitude and opposite in direction with respect to the displacement. Hence, the work done by each string on the two masses attached have opposite sign. 
As a consequence, if you sum up all contributions from each string and each mass, the net work done by tension is zero.
A: In a two mass dependent motion systems with pulleys, tension displacement ratios are inverse of each other. That is, given blocks A and B, where A is attached to a certain number $n$ of cables by pulleys, and B is attached to $m$ number of cables. The ratio of forces would be $\frac{n}{m}$ and the ratio of displacements would be $\frac{m}{n}$. From here you might be able to guess that they could be equal, but to be sure, here's some work you can look at. HA, PUNS!
Dependent motion in pulley systems tells us that 
$${\Delta}s_B = -{\Delta}s_A  \frac{n}{m}$$ 
Work on A is simply 
$${\int_0^{x_A}}\vec{F_A}\cdot d\vec{x_A} = W_A = T n{\Delta}s_A$$ 
compared to the work on B, 
$${\int_0^{x_B}}\vec{F_B}\cdot d\vec{x_B} = T m{\Delta}s_B =$$
    $$T m(-{\Delta}s_A  \frac{n}{m}) = $$
    $$W_B = -T n {\Delta}s_A = -W_A$$
and $${\Sigma}{\int}\vec{F}\cdot d\vec{x_i} = 0$$
You can then generalize this to a multiple pulley system by saying the work of a pulley sub-system is equivalent to $W_A$ and the other sub-system is equivalent to $W_B$.
A: It is a consequence of total length and tension being constant throughout the taut rope. Here is how it extends to work done:
Let two ends of a segment of a taut rope be at $x_1$ and $x_2$. Then, $d|x_1-x_2|=d\sqrt{(x_1-x_2)\cdot(x_1-x_2)}=\frac{1}{2}\frac{2\cdot(dx_1-dx_2)\cdot(x_1-x_2)}{|x_1-x_2|}=dx_1\cdot n + dx_2 \cdot (-n)$
where $n$ is unit vector along the rope segment. The sum represents dot product of displacement of ends of rope to the opposite of unit vector of tension at each point. Multiplying by constant value of tension and summing up for all segments of the rope, the LHS will become tension multiplied by total change in length of rope and the RHS will become the negative work done. LHS being zero, the work done by rope is also zero.
A: You will have to go back to the equation F = m a.
If the mass of the string is 0, then F = 0.
So yea, the net force of the tension by the string is 0.
To add on, in order for the net tension of the string to be zero, the tension everywhere has to be equal to cancel out.
