To solve for the tension in the rope on the right side of the pulley, we need to apply Newton's second law to both the block and the pulley.

Given:

• Mass of the block $m_b = 1.4 \, \text{kg}$
• Acceleration of the block $a_b = 1.2 \, \text{m/s}^2$ (upward)
• Mass of the pulley $m_p = 2.0 \, \text{kg}$
• Radius of the pulley $R = 0.50 \, \text{m}$
• Moment of inertia of the pulley (solid cylinder) $I = \frac{1}{2} m_p R^2$
• Gravitational acceleration $g = 9.81 \, \text{m/s}^2$

Step 1: Block's equation of motion

For the block, we use Newton’s second law: $F_{\text{net}} = m_b \cdot a_b$ The net force on the block is the difference between the downward gravitational force and the upward tension in the rope on the left side ($T_1$): $m_b \cdot g - T_1 = m_b \cdot a_b$ Solving for $T_1$: $T_1 = m_b \cdot g - m_b \cdot a_b$ $T_1 = (1.4 \, \text{kg}) \cdot (9.81 \, \text{m/s}^2) - (1.4 \, \text{kg}) \cdot (1.2 \, \text{m/s}^2)$ $T_1 = 13.734 \, \text{N} - 1.68 \, \text{N}$ $T_1 = 12.054 \, \text{N}$

Step 2: Pulley’s equation of motion

Now, let's apply Newton's second law for rotation to the pulley. The torque on the pulley comes from the tension difference on the left and right sides of the rope. This torque causes angular acceleration of the pulley.

The moment of inertia for a solid cylinder is: $I = \frac{1}{2} m_p R^2 = \frac{1}{2} \cdot (2.0 \, \text{kg}) \cdot (0.50 \, \text{m})^2 = 0.25 \, \text{kg} \cdot \text{m}^2$

The net torque is: $\tau = (T_1 - T_2) \cdot R$ The angular acceleration $\alpha$ of the pulley is related to the linear acceleration $a_b$ of the block by $\alpha = \frac{a_b}{R}$. The rotational form of Newton's second law gives: $\tau = I \cdot \alpha$ Substitute for $\tau$ and $\alpha$: $(T_1 - T_2) \cdot R = I \cdot \frac{a_b}{R}$ $(T_1 - T_2) = \frac{I \cdot a_b}{R^2}$ $(T_1 - T_2) = \frac{(0.25 \, \text{kg} \cdot \text{m}^2) \cdot (1.2 \, \text{m/s}^2)}{(0.50 \, \text{m})^2}$ $(T_1 - T_2) = \frac{0.30 \, \text{N} \cdot \text{m}}{0.25 \, \text{m}^2}$ $T_1 - T_2 = 1.2 \, \text{N}$

Step 3: Solve for $T_2$

Now that we know $T_1 - T_2 = 1.2 \, \text{N}$, we can solve for $T_2$: $T_2 = T_1 - 1.2 \, \text{N}$ $T_2 = 12.054 \, \text{N} - 1.2 \, \text{N}$ $T_2 = 10.854 \, \text{N}$

Final Answer:

The tension on the right side of the rope is $\boxed{10.9 \, \text{N}}$.

Would you like to go over any details or explore another aspect of the problem? Here are some related questions:

1. How would the tension change if the block’s acceleration were greater?
2. How would the moment of inertia of the pulley affect the tension in the rope?
3. What happens if the pulley had a different shape, like a hollow cylinder?
4. How would friction between the rope and the pulley affect the solution?
5. How can we calculate the angular velocity of the pulley after a certain time?

Tip: When dealing with rotating objects, remember that torque $\tau$ is the rotational equivalent of force, and angular acceleration $\alpha$ corresponds to linear acceleration through $\alpha = a / R$.