As shown in the figure, a heavy block rests on a frictionless...

Let $M$ be the mass of the heavy block.

For $t < 0$:

The system is pulled by equal and opposite forces $F$. The net external force is zero, so the acceleration of the system is $a = 0$.

For the left rope of mass $m$, the equation of motion is $F - F_2 = m(0) \Rightarrow F_2 = F$.

For the right rope of mass $m$, the equation of motion is $F - F_1 = m(0) \Rightarrow F_1 = F$.

Thus, for $t < 0$, $F_1 = F_2 = F$.

For $t > 0$:

The left force is removed. The net force on the entire system is $F$ to the right.

The acceleration of the system is $a = \dfrac{F}{M + 2m}$.

For the left rope, the only horizontal force is $F_2$ to the right. Its equation of motion is $F_2 = ma = \dfrac{mF}{M + 2m}$. Since $M > 0 and$m > 0$,$F_2 < F$.<br><br>For the right rope, the forces are$F$to the right and$F_1$to the left. Its equation of motion is$F - F_1 = ma \Rightarrow F_1 = F - ma = F - \dfrac{mF}{M + 2m} = F \left(\dfrac{M + m}{M + 2m}\right). Since $M > 0$ and $m > 0$, $F_1 < F$.

Thus, for $t > 0$, $F_1 < F$ and $F_2 < F$.