Answer
(a) $a=3.68m/s^2$
(b) $a=11.8m/s^2$
(c) The difference in the results comes from the difference in mass in 2 cases.
Work Step by Step
(a) The heavier block will fall down while the lighter block will be pulled up. As these blocks act as a system, their movement has similar acceleration $a$.
The weight of the heavier block pulls it downward, while the tension $T$ in the rope holds it upward. According to Newton's 2nd Law, as the block falls with acceleration $a$, $$W-T=Ma$$ $$908-T=\frac{908}{9.8}a=92.65a (1)$$
Similarly, the weight of the lighter block pulls it downward, while the tension $T$ in the rope holds it upward. According to Newton's 2nd Law, as the block is pulled up with acceleration $a$, $$T-W=ma$$ $$T-412=\frac{412}{9.8}a=42.04a(2)$$
We solve equations (1) and (2) and get $a=3.68m/s^2$
(b) The removal of the heavier block means we have to revisit equation (1) $W-T=Ma$
We assume the pulling hand has zero mass, which means $M=0$ and $W=0$. If we call the pulling force $P$, we get, as a result $$P-T=0$$ $$P=T=908N$$
The tension in the rope is $908N$. Apply this back to equation $T$: $$42.04a=908-412=496N$$ $$a=11.8m/s^2$$
(c) The difference in the results comes from the difference in mass in 2 cases as we have seen. This leads to different tension values in the rope, which lead to different acceleration in the end.
