Item four but use English

Let $T$ be the tension in the string. We define the downward direction as positive for acceleration.

For mass $m_1$:

$$m_1 g - T = m_1 a_1 \quad (1)$$

For mass $m_3$:

$$m_3 g - T = m_3 a_3 \quad (2)$$

For mass $m_2$ (the movable pulley): It is pulled downwards by gravity and upwards by two segments of the string, each with tension $T$.

$$m_2 g - 2T = m_2 a_2 \quad (3)$$

The total length of the string is constant. If $y_1, y_2, y_3$ are the vertical positions of $m_1, m_2, m_3$ from a fixed reference (e.g., the ceiling), the length constraint is:

$$y_1 + 2y_2 + y_3 = \text{constant}$$

Differentiating this equation twice with respect to time gives the relationship between accelerations:

$$a_1 + 2a_2 + a_3 = 0 \quad (4)$$

From (1), $T = m_1(g - a_1)$.
From (2), $T = m_3(g - a_3)$.
Equating these two expressions for $T$:

$$m_1(g - a_1) = m_3(g - a_3)$$

Substitute $m_1=2$ and $m_3=3$:

$$2(g - a_1) = 3(g - a_3)$$ $$2g - 2a_1 = 3g - 3a_3$$ $$3a_3 - 2a_1 = g \quad (A)$$

From (4), we can express $a_3 = -a_1 - 2a_2$. Substitute this into (A):

$$3(-a_1 - 2a_2) - 2a_1 = g$$ $$-3a_1 - 6a_2 - 2a_1 = g$$ $$-5a_1 - 6a_2 = g \quad (B)$$

Now, substitute $T = m_1(g - a_1)$ into (3):

$$m_2 g - 2m_1(g - a_1) = m_2 a_2$$

Substitute $m_1=2$ and $m_2=12$:

$$12g - 2(2)(g - a_1) = 12a_2$$ $$12g - 4g + 4a_1 = 12a_2$$ $$8g + 4a_1 = 12a_2$$

Divide by 4:

$$2g + a_1 = 3a_2$$

From this, express $a_1$:

$$a_1 = 3a_2 - 2g \quad (C)$$

Substitute (C) into (B):

$$-5(3a_2 - 2g) - 6a_2 = g$$ $$-15a_2 + 10g - 6a_2 = g$$ $$-21a_2 = -9g$$ $$a_2 = \frac{9g}{21} = \frac{3g}{7}$$

Now find $a_1$ using (C):

$$a_1 = 3\left(\frac{3g}{7}\right) - 2g = \frac{9g}{7} - \frac{14g}{7} = -\frac{5g}{7}$$

And find $a_3$ using (A):

$$3a_3 - 2\left(-\frac{5g}{7}\right) = g$$ $$3a_3 + \frac{10g}{7} = g$$ $$3a_3 = g - \frac{10g}{7} = -\frac{3g}{7}$$ $$a_3 = -\frac{g}{7}$$

Using $g = 9.8 \text{ m/s}^2$:

$$a_1 = -\frac{5 \times 9.8}{7} = -5 \times 1.4 = -7 \text{ m/s}^2$$ $$a_2 = \frac{3 \times 9.8}{7} = 3 \times 1.4 = 4.2 \text{ m/s}^2$$ $$a_3 = -\frac{9.8}{7} = -1.4 \text{ m/s}^2$$

The negative signs indicate acceleration in the upward direction, as we defined downward as positive.
The acceleration of each mass is:

• $a_1 = \boxed{7 \text{ m/s}^2 \text{ (upwards)}}$
• $a_2 = \boxed{4.2 \text{ m/s}^2 \text{ (downwards)}}$
• $a_3 = \boxed{1.4 \text{ m/s}^2 \text{ (upwards)}}$

We can use equation (1):

$$T = m_1 g - m_1 a_1$$

Substitute the values for $m_1$ and $a_1$:

$$T = 2 \text{ kg} \times 9.8 \text{ m/s}^2 - 2 \text{ kg} \times (-7 \text{ m/s}^2)$$ $$T = 19.6 \text{ N} + 14 \text{ N}$$ $$T = 33.6 \text{ N}$$

The tension in the connecting string is $\boxed{33.6 \text{ N}}$.