Interpretation of the setup

The three blocks of masses $M_1=4\text{ kg},\quad M_2=6\text{ kg},\quad M_3=10\text{ kg}$ are hanging vertically one below another using ropes 1, 2, and 3.

Since the system is moving upward with acceleration $a=2\text{ m s}^{-2},$ each block has upward acceleration $2\text{ m s}^{-2}$.

We need the tension in rope 1, i.e. $T_1$.

Key idea

Rope 1 is the topmost rope, so it supports all three masses together.

Hence, consider the three blocks as a single system of total mass: $M = M_1+M_2+M_3 = 4+6+10 = 20\text{ kg}.$

Apply Newton's second law to the whole system

Forces on the combined system:

• Upward force = $T_1$
• Downward force = total weight = $Mg = 20\times 10 = 200\text{ N}$

Since acceleration is upward, $T_1 - Mg = Ma$

Substituting values: $T_1 - 200 = 20\times 2 = 40$

Therefore, $T_1 = 200 + 40 = 240\text{ N}.$

Final answer

$\boxed{240}$

Comparison with stored answer

Stored correct answer = $240$.

This matches our derived answer.