Let's go through each part of this problem.

Part (a): Free-Body Diagrams

For each mass, we need to consider the forces acting on it:

1. Brick Load (Mass $m_1 = 22.5 \, \text{kg}$):

   • Gravitational Force (Weight): $F_{g1} = m_1 g$, directed downward.
   • Tension (T) in the rope, directed upward.

2. Counterweight (Mass $m_2 = 17.0 \, \text{kg}$):

   • Gravitational Force (Weight): $F_{g2} = m_2 g$, directed downward.
   • Tension (T) in the rope, directed upward.

Since the rope passes over a frictionless pulley, both masses will experience the same tension $T$ and acceleration $a$.

Part (b): Calculating the Acceleration of the Brick Load

To find the acceleration, we'll set up Newton's second law for each mass. Let $g = 9.8 \, \text{m/s}^2$.

For the brick load ($m_1$): $m_1 g - T = m_1 a$

For the counterweight ($m_2$): $T - m_2 g = m_2 a$

Adding these two equations eliminates $T$ and gives us: $m_1 g - m_2 g = (m_1 + m_2)a$ $a = \frac{(m_1 - m_2) g}{m_1 + m_2}$

Substitute $m_1 = 22.5 \, \text{kg}$, $m_2 = 17.0 \, \text{kg}$, and $g = 9.8 \, \text{m/s}^2$: $a = \frac{(22.5 - 17.0) \times 9.8}{22.5 + 17.0}$ $a = \frac{5.5 \times 9.8}{39.5}$ $a \approx 1.36 \, \text{m/s}^2$

Part (c): Calculating the Tension in the Rope

Now that we know $a$, we can substitute it back into either equation to solve for $T$. Let's use the equation for the counterweight:

$T = m_2 g + m_2 a$ $T = 17.0 \times 9.8 + 17.0 \times 1.36$ $T = 166.6 + 23.12$ $T \approx 189.72 \, \text{N}$

Part (d): Comparing the Tension with the Weight of the Brick Load and the Counterweight

1. Tension vs. Weight of Brick Load:

   • The weight of the brick load is $m_1 g = 22.5 \times 9.8 = 220.5 \, \text{N}$.
   • Since $T \approx 189.72 \, \text{N}$ and $189.72 \, \text{N} < 220.5 \, \text{N}$, the tension in the rope is less than the weight of the brick load.

2. Tension vs. Weight of Counterweight:

   • The weight of the counterweight is $m_2 g = 17.0 \times 9.8 = 166.6 \, \text{N}$.
   • Since $T \approx 189.72 \, \text{N}$ and $189.72 \, \text{N} > 166.6 \, \text{N}$, the tension in the rope is greater than the weight of the counterweight.

Summary of Results

1. Free-body diagrams should be drawn as described for each mass.
2. The acceleration of the brick load is approximately $1.36 \, \text{m/s}^2$.
3. The tension in the rope as the brick load moves is approximately $189.72 \, \text{N}$.
4. The tension in the rope is less than the weight of the brick load but greater than the weight of the counterweight.

Would you like more details on any of these calculations, or have any other questions?

Here are five related questions to explore further:

1. How would the results change if there was friction in the pulley?
2. What would happen if the masses were equal?
3. How would we solve for tension if the rope had mass?
4. How would the setup change if the flywheel had rotational inertia?
5. How would the system behave if released from a non-rest initial velocity?

Tip: Always check if forces like tension change across segments if a pulley has friction.