Tension Force and Pulley Block System Questions in English

(A) This is an Atwood machine system. The acceleration $a$ of the system is given by:
$a = \left(\frac{M-m}{M+m}\right) g$
Substituting the values $M = 3 \ kg$,$m = 2 \ kg$,and $g = 10 \ m/s^2$:
$a = \left(\frac{3-2}{3+2}\right) \times 10 = \frac{1}{5} \times 10 = 2 \ m/s^2$
The distance $s$ covered by the $3 \ kg$ block in $t = 2 \ s$ starting from rest $(u = 0)$ is:
$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times (2)^2 = 4 \ m$
The work done by gravity on the $3 \ kg$ block is the product of the gravitational force $(Mg)$ and the displacement $(s)$:
$W = Mgs = 3 \times 10 \times 4 = 120 \ J$
Thus,the work done is $120 \ J$.

(A) Let the masses be $m_1$ and $m_2$. The pulley is accelerating upwards with $a_0 = g$. In the frame of the pulley,each block experiences a pseudo-force $m_i g$ downwards.
For mass $m_1$ (assuming it moves downwards relative to the pulley with acceleration $a$):
$m_1 g + m_1 g - T = m_1 a \implies 2m_1 g - T = m_1 a$ $(i)$
For mass $m_2$ (moving upwards relative to the pulley with acceleration $a$):
$T - m_2 g - m_2 g = m_2 a \implies T - 2m_2 g = m_2 a$ (ii)
Adding $(i)$ and (ii):
$2m_1 g - 2m_2 g = (m_1 + m_2) a \implies a = \frac{2g(m_1 - m_2)}{m_1 + m_2}$
Substitute $a$ into (ii):
$T = m_2(a + 2g) = m_2 \left( \frac{2g(m_1 - m_2)}{m_1 + m_2} + 2g \right)$
$T = 2m_2 g \left( \frac{m_1 - m_2 + m_1 + m_2}{m_1 + m_2} \right) = 2m_2 g \left( \frac{2m_1}{m_1 + m_2} \right) = \frac{4 m_1 m_2 g}{m_1 + m_2}$

(B) Let the mass of the rod be $M$ and the mass of the ball be $m = 1.2M$. Let $T$ be the tension in the string attached to the rod. The movable pulley is supported by two segments of the string,each with tension $T$,so the upward force on the movable pulley is $2T$. This force is transmitted to the ball,so the tension in the string attached to the ball is $2T$.
For the ball (mass $m$),the equation of motion is: $m a_1 = 2T - mg$ $(i)$
For the rod (mass $M$),the equation of motion is: $M a_2 = Mg - T$ (ii)
Since the string length is constant,the constraint relation between accelerations is $a_2 = 2a_1$,or $a_1 = a_2/2$.
Substituting $a_1$ into $(i)$: $m(a_2/2) = 2T - mg \Rightarrow T = \frac{m a_2}{4} + \frac{mg}{2}$.
Substitute $T$ into (ii): $M a_2 = Mg - (\frac{m a_2}{4} + \frac{mg}{2})$.
Rearranging for $a_2$: $a_2(M + m/4) = g(M - m/2)$.
$a_2 = g \frac{M - m/2}{M + m/4} = g \frac{1 - (m/M)/2}{1 + (m/M)/4}$.
Given $m/M = 1.2$,$a_2 = 10 \times \frac{1 - 0.6}{1 + 0.3} = 10 \times \frac{0.4}{1.3} = \frac{4}{1.3} \approx 3.07 \,m/s^2$. The closest option is $3 \,m/s^2$.

(D) Let the masses be $m_1 = 4 \ kg$ (left side),$m_2 = 3 \ kg$ (upper right),and $m_3 = 3 \ kg$ (lower right).
The total mass on the right side is $M_R = m_2 + m_3 = 3 \ kg + 3 \ kg = 6 \ kg$.
The mass on the left side is $M_L = 4 \ kg$.
Since $M_R > M_L$,the system accelerates towards the right with acceleration $a$.
The acceleration $a$ is given by $a = \frac{(M_R - M_L)g}{M_R + M_L} = \frac{(6 - 4)g}{6 + 4} = \frac{2g}{10} = 0.2g$.
For the lower $3 \ kg$ block,the equation of motion is $T_1 - m_3g = m_3a$.
$T_1 = m_3(g + a) = 3(g + 0.2g) = 3(1.2g) = 3.6g$.
For the entire system,the tension $T_2$ in the string passing over the pulley is $T_2 = \frac{2 M_L M_R g}{M_L + M_R} = \frac{2(4)(6)g}{4 + 6} = \frac{48g}{10} = 4.8g$.
The ratio of tensions is $\frac{T_1}{T_2} = \frac{3.6g}{4.8g} = \frac{36}{48} = \frac{3}{4}$.

(D) Let $T$ be the tension in the string and $a$ be the magnitude of the acceleration of the masses.
Since $m_{2} > m_{1}$,the mass $m_{2}$ moves downwards with acceleration $a$,and the mass $m_{1}$ moves upwards with acceleration $a$.
For mass $m_{1}$ (moving upwards): $T - m_{1}g = m_{1}a$ ....$(i)$
For mass $m_{2}$ (moving downwards): $m_{2}g - T = m_{2}a$ ....(ii)
Adding equations $(i)$ and (ii),we get:
$(T - m_{1}g) + (m_{2}g - T) = m_{1}a + m_{2}a$
$(m_{2} - m_{1})g = (m_{1} + m_{2})a$
$a = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} g$

(B) For a system of two masses connected by a string over a pulley,the acceleration of each mass is given by $a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g$.
Since the two masses move in opposite directions,the acceleration of the center of mass $(a_{CM})$ is calculated as:
$a_{CM} = \frac{m_{1}a_{1} + m_{2}a_{2}}{m_{1} + m_{2}}$.
Taking the direction of $m_{1}$ as positive,$a_{1} = a$ and $a_{2} = -a$.
$a_{CM} = \frac{m_{1}a - m_{2}a}{m_{1} + m_{2}} = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right) a$.
Substituting the value of $a$:
$a_{CM} = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right) \times \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right) g = \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right)^{2} g$.
The total external force acting on the system is $F_{ext} = (m_{1} + m_{2}) a_{CM}$.
$F_{ext} = (m_{1} + m_{2}) \times \left(\frac{m_{1} - m_{2}}{m_{1} + m_{2}}\right)^{2} g = \frac{(m_{1} - m_{2})^{2}}{m_{1} + m_{2}} g$.

(B) Let the acceleration of the system be $a$. The system moves such that $m_3$ moves downwards and $m_1$ moves upwards.
For mass $m_3$: $m_3g - T_2 = m_3a \implies 60 - T_2 = 6a$ ---$(1)$
For the system $(m_1 + m_2)$: $T_2 - (m_1 + m_2)g = (m_1 + m_2)a \implies T_2 - 80 = 8a$ ---$(2)$
Adding $(1)$ and $(2)$: $60 - 80 = 14a \implies -20 = 14a \implies a = -20/14 = -10/7 \text{ m/s}^2$.
The negative sign indicates that the system moves in the opposite direction (i.e.,$m_1$ moves down and $m_3$ moves up).
Using $a = 10/7 \text{ m/s}^2$ (magnitude),$m_3$ moves up: $T_2 - m_3g = m_3a \implies T_2 = m_3(g + a) = 6(10 + 10/7) = 6(80/7) = 480/7 \text{ N}$.
For the whole system hanging from $T_1$: $2T_1 = (m_1 + m_2 + m_3)g + (m_1 + m_2 + m_3)a_{\text{cm}}$ is not required here. The tension $T_1$ supports the entire system: $2T_1 = (m_1 + m_2 + m_3)g + \text{net force}$.
Actually,$T_1$ is the tension in the string passing over the pulley. Since the pulley is fixed and frictionless,$T_1$ balances the weights. The total downward force is $(m_1 + m_2 + m_3)g = 140 \text{ N}$. Thus,$2T_1 = 140 \implies T_1 = 70 \text{ N}$.
Ratio $T_1/T_2 = 70 / (480/7) = 490 / 480 = 49/48 \approx 1$. Given the options,let's re-evaluate the setup: $T_1$ is the tension in the string connected to $m_1$ and the $(m_2+m_3)$ system. $T_1 = m_1(g+a) = 4(10 + 10/7) = 4(80/7) = 320/7 \text{ N}$.
Ratio $T_1/T_2 = (320/7) / (480/7) = 320/480 = 2/3$.