A block of mass M is suspended over a smooth and light pulley via a...

Since the string is light and the pulley is smooth and light, the tension is uniform throughout the string. The force $F$ is applied at one end of the string, so the tension $T$ in the string is equal to the applied force $F$.

According to the work-energy theorem, the net work done on the block is equal to the change in its kinetic energy:

$W_{\text{net}} = W_T + W_g = \Delta K = 20 \text{ J}$

Since the block's kinetic energy increases, it has a net acceleration, meaning the net force is non-zero. Therefore, $T \neq Mg$.

The exact values of the work done by tension $W_T$ and the work done by gravity $W_g$ cannot be determined without additional information, but their sum is $20 \text{ J}$. Thus, the only definitively correct statement is that the tension in the string equals $F$.