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This is a classic problem of a system of two masses connected by a pulley, with the masses on inclined planes and experiencing friction. The aim is to calculate the acceleration of block \( M_2 \). Here's the strategy for solving the problem:

### Given:
- \( M_1 = 2.25 \, \text{kg} \)
- \( M_2 = 7.35 \, \text{kg} \)
- \( \theta_1 = 47.5^\circ \), \( \theta_2 = 33.5^\circ \)
- Coefficient of kinetic friction for \( M_1 \): \( \mu_1 = 0.205 \)
- Coefficient of kinetic friction for \( M_2 \): \( \mu_2 = 0.105 \)

### Steps:

1. **Forces on Block \( M_1 \):**
   - The force due to gravity along the incline: \( F_{g1} = M_1 g \sin(\theta_1) \)
   - The friction force on \( M_1 \): \( F_{\text{friction}_1} = \mu_1 M_1 g \cos(\theta_1) \)
   - Total net force on \( M_1 \): 
     \[
     F_{M_1} = M_1 g \sin(\theta_1) - \mu_1 M_1 g \cos(\theta_1)
     \]

2. **Forces on Block \( M_2 \):**
   - The force due to gravity along the incline: \( F_{g2} = M_2 g \sin(\theta_2) \)
   - The friction force on \( M_2 \): \( F_{\text{friction}_2} = \mu_2 M_2 g \cos(\theta_2) \)
   - Total net force on \( M_2 \):
     \[
     F_{M_2} = M_2 g \sin(\theta_2) - \mu_2 M_2 g \cos(\theta_2)
     \]

3. **Applying Newton's Second Law:**
   The net forces on both blocks will result in their acceleration, and since they are connected by a pulley, they will have the same magnitude of acceleration \( a \).

   For block \( M_1 \):
   \[
   M_1 a = M_1 g \sin(\theta_1) - \mu_1 M_1 g \cos(\theta_1)
   \]

   For block \( M_2 \):
   \[
   M_2 a = M_2 g \sin(\theta_2) - \mu_2 M_2 g \cos(\theta_2)
   \]

4. **Equating the forces:**
   Since the tension in the rope is the same for both blocks and they have the same acceleration \( a \), we solve these two equations simultaneously for \( a \).

### Calculation:
Substitute the given values into the equations:

For block \( M_1 \):
\[
M_1 a = M_1 g \sin(47.5^\circ) - \mu_1 M_1 g \cos(47.5^\circ)
\]
\[
2.25 a = 2.25 \times 9.8 \times \sin(47.5^\circ) - 0.205 \times 2.25 \times 9.8 \times \cos(47.5^\circ)
\]

For block \( M_2 \):
\[
M_2 a = M_2 g \sin(33.5^\circ) - \mu_2 M_2 g \cos(33.5^\circ)
\]
\[
7.35 a = 7.35 \times 9.8 \times \sin(33.5^\circ) - 0.105 \times 7.35 \times 9.8 \times \cos(33.5^\circ)
\]

Let me calculate the values now.The magnitude of the acceleration \( a_2 \) of block \( M_2 \) is approximately \( 2.11 \, \text{m/s}^2 \).

Would you like more details or have any questions?

Here are five related questions to expand your understanding:
1. How does the coefficient of kinetic friction affect the motion of the blocks?
2. What would happen to the acceleration if the incline angles were increased or decreased?
3. How would adding mass to block \( M_1 \) affect the system’s acceleration?
4. What would the acceleration be if the pulley had mass or friction?
5. How can we calculate the tension in the string?

**Tip:** Always check if the forces acting parallel and perpendicular to the incline are correctly decomposed into their respective components when solving similar problems.

Two blocks, M_1 and M_2, are connected by a massless string that passes over a massless pulley. M_1 has a mass of 2.25 kg and is on an incline of angle θ_1 = 47.5° that has a coefficient of kinetic friction μ_1 = 0.205. M_2 has a mass of 7.35 kg and is on an incline of angle θ_2 = 33.5° that has a coefficient of kinetic friction μ_2 = 0.105. The two-block system is in motion, with block M_2 sliding down the ramp. Find the magnitude a_2 of the acceleration of block M_2.

Solve for the acceleration of two blocks connected by a massless string on inclined planes with friction. M_1 (2.25 kg) is on a 47.5° incline with μ_1 = 0.205, and M_2 (7.35 kg) is on a 33.5° incline with μ_2 = 0.105. Using Newton's Second Law, determine the system's acceleration.

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