Let's break down this problem, focusing on both parts (a) and (b).

Given Information:

• Mass of block 1: $m_1 = 9.4 \, \text{kg}$
• Angle of incline: $\theta = 12.0^\circ$
• The pulley is frictionless, and the incline is smooth (no friction).

General Approach:

1. Draw Free-Body Diagrams (FBDs) for both masses $m_1$ and $m_2$.
2. Identify the forces acting on both blocks.
3. Use Newton's second law ($F = ma$) to set up the equations for equilibrium in part (a) and acceleration in part (b).

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Part (a): System in Equilibrium

For the system to be in equilibrium, the net force on each block must be zero. Let's consider the forces acting on each block.

Forces on $m_1$ (on the incline):

• Gravitational force: $m_1 g$, where $g = 9.8 \, \text{m/s}^2$
• Component of gravitational force parallel to the incline: $m_1 g \sin(\theta)$
• Tension in the string: $T$ (acting up the incline)

Since there is no motion, the forces must balance: $T = m_1 g \sin(\theta)$

Forces on $m_2$ (hanging vertically):

• Gravitational force: $m_2 g$
• Tension in the string: $T$

For equilibrium: $T = m_2 g$

Set up the equilibrium condition:

Since the tension is the same in both cases (no acceleration), we can equate the two expressions for $T$:

$m_1 g \sin(\theta) = m_2 g$

Cancel $g$ from both sides: $m_1 \sin(\theta) = m_2$

Solve for $m_2$: $m_2 = m_1 \sin(\theta)$

Substitute the given values: $m_2 = 9.4 \times \sin(12.0^\circ)$ $m_2 \approx 9.4 \times 0.2079 = 1.954 \, \text{kg}$

Thus, the correct value of $m_2$ for equilibrium is approximately 1.95 kg.

Part (b): System Accelerating Down the Incline

If the block on the incline is accelerating down with an acceleration of $a = 1.1 \, \text{m/s}^2$, we need to consider the forces in a non-equilibrium situation. Newton's second law will help here.

Forces on $m_1$:

The net force on $m_1$ down the incline is: $F_{\text{net}, m_1} = m_1 g \sin(\theta) - T = m_1 a$ Thus, $T = m_1 g \sin(\theta) - m_1 a$

Forces on $m_2$:

The net force on $m_2$ is: $F_{\text{net}, m_2} = m_2 g - T = m_2 a$ Thus, $T = m_2 g - m_2 a$

Set the two expressions for $T$ equal:

$m_1 g \sin(\theta) - m_1 a = m_2 g - m_2 a$

Rearrange to isolate $m_2$: $m_2 (g - a) = m_1 (g \sin(\theta) - a)$

Solve for $m_2$: $m_2 = \frac{m_1 (g \sin(\theta) - a)}{g - a}$

Substitute the values: $m_2 = \frac{9.4 \times (9.8 \times \sin(12^\circ) - 1.1)}{9.8 - 1.1}$ $m_2 = \frac{9.4 \times (9.8 \times 0.2079 - 1.1)}{8.7}$ $m_2 = \frac{9.4 \times (2.038 - 1.1)}{8.7} = \frac{9.4 \times 0.938}{8.7} \approx \frac{8.817}{8.7} \approx 1.01 \, \text{kg}$

Thus, the correct value of $m_2$ for an acceleration of $1.1 \, \text{m/s}^2$ is approximately 1.01 kg.

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Would you like more details on any step or help with a related concept?

Here are five related questions:

1. What would happen if the incline had friction? How would the equations change?
2. How do you derive the components of the gravitational force for an inclined plane?
3. What if the angle of the incline increased? How would $m_2$ change for equilibrium?
4. What is the relationship between mass and acceleration in a pulley system with unequal masses?
5. How can you experimentally determine the coefficient of friction on an incline?

Tip: Always double-check the signs of forces when applying Newton's second law to ensure consistent direction.