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The problem involves a system of three connected masses on a table with kinetic friction. The masses are 4.00 kg, 1.00 kg, and 2.00 kg, connected by cords passing over frictionless pulleys. The kinetic friction coefficient between the 1.00 kg mass and the table is 0.350. The task is to find the acceleration of each object and the tension in the two cords.

### Solution Steps:

1. **Define the forces acting on each mass:**
   - **Mass 1 (1.00 kg on the table):**
     - Weight: \( W_1 = m_1 \cdot g = 1.00 \times 9.8 = 9.8 \, \text{N} \).
     - Normal force: \( N = W_1 = 9.8 \, \text{N} \).
     - Friction force: \( f_k = \mu_k \cdot N = 0.350 \times 9.8 = 3.43 \, \text{N} \).
     - Tensions: \( T_1 \) pulling to the left (from 4.00 kg) and \( T_2 \) pulling to the right (from 2.00 kg).

   - **Mass 2 (4.00 kg hanging on the left):**
     - Weight: \( W_2 = m_2 \cdot g = 4.00 \times 9.8 = 39.2 \, \text{N} \).
     - Tension: \( T_1 \) acts upward.

   - **Mass 3 (2.00 kg hanging on the right):**
     - Weight: \( W_3 = m_3 \cdot g = 2.00 \times 9.8 = 19.6 \, \text{N} \).
     - Tension: \( T_2 \) acts upward.

2. **Set up the equations of motion:**
   - For the 1.00 kg mass on the table:
     \[
     m_1 a = T_2 - T_1 - f_k.
     \]
   - For the 4.00 kg mass:
     \[
     m_2 a = m_2 g - T_1.
     \]
   - For the 2.00 kg mass:
     \[
     m_3 a = T_2 - m_3 g.
     \]

3. **Combine the equations:**
   Substitute the values into the equations to solve for acceleration \( a \) and tensions \( T_1 \) and \( T_2 \).

Let's calculate the values:

### Calculations:
1. From the equations:
   \[
   1.00a = T_2 - T_1 - 3.43,
   \]
   \[
   4.00a = 39.2 - T_1,
   \]
   \[
   2.00a = T_2 - 19.6.
   \]

2. Solving the system of equations by substituting and eliminating variables.

Let's compute these values.The calculated values are:

- **Acceleration (\(a\)) of the system:** \( 4.606 \, \text{m/s}^2 \).
- **Tension in the first cord (\(T_1\)):** \( 20.776 \, \text{N} \).
- **Tension in the second cord (\(T_2\)):** \( 28.812 \, \text{N} \).

### Summary:
- The acceleration of all the masses is \( 4.606 \, \text{m/s}^2 \).
- The tension in the cord connecting the 4.00 kg mass is \( 20.776 \, \text{N} \).
- The tension in the cord connecting the 2.00 kg mass is \( 28.812 \, \text{N} \).

Would you like a step-by-step breakdown of the calculations or have any questions?

### Related Questions:
1. How would the system change if the coefficient of kinetic friction was different?
2. What would happen to the tensions if the mass of the block on the table changed?
3. How does the acceleration of each object compare to the overall system acceleration?
4. What forces are acting on each mass, and how do they contribute to the system's motion?
5. How would introducing friction to the pulleys affect the results?

**Tip:** Always ensure to account for all forces acting on each mass, including friction, tension, and gravity, when setting up equations of motion in similar problems.

Three objects are connected on the table as shown. The table is rough and has a coefficient of kinetic friction of 0.350. The objects have masses of 4.00 kg, 1.00 kg, and 2.00 kg, and the pulleys are frictionless. Determine the acceleration of each object and their directions. Determine the tensions in the two cords.

Solve a system of three masses (4.00 kg, 1.00 kg, 2.00 kg) connected by frictionless pulleys on a table with kinetic friction. Learn how to calculate the acceleration of each object and the tensions in the cords. This problem involves Newton's Second Law and kinetic friction, suitable for Grades 10-12.

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