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What is the magnitude and direction of the acceleration of each block?

To solve for the magnitude and direction of the acceleration of each block in this pulley system, we need to:

1. Apply Newton's second law to each mass, considering forces acting on them.
2. Consider the system's constraints (acceleration and forces across pulleys and tension in the strings).
3. Set up a system of equations to solve for acceleration \(a\).

### Step-by-Step Analysis:

We are given:
- \( m_A = 1.25 \, \text{kg} \)
- \( m_B = 2.50 \, \text{kg} \)
- \( m_C = 3.75 \, \text{kg} \)
- Coefficient of kinetic friction \( \mu_k = 0.602 \) for block \( m_B \).
- All blocks accelerate with the same magnitude of acceleration \( a \).

#### 1. Forces on Each Block:

- **Block A**: Since it's hanging, the forces acting are:
  - Gravitational force: \( F_{gA} = m_A g \)
  - Tension in the string: \( T_A \)
  
  Applying Newton's second law:
  \[
  m_A g - T_A = m_A a \tag{1}
  \]
  
- **Block B**: This block is on the ground, with friction acting on it. Forces are:
  - Gravitational force: \( F_{gB} = m_B g \) (normal force)
  - Frictional force: \( F_{\text{friction}} = \mu_k F_{gB} = \mu_k m_B g \)
  - Tension in the string pulling: \( T_B \)

  Newton's second law for block \( B \):
  \[
  T_B - \mu_k m_B g = m_B a \tag{2}
  \]

- **Block C**: Another hanging mass, with forces:
  - Gravitational force: \( F_{gC} = m_C g \)
  - Tension in the string: \( T_C \)

  Newton's second law for block \( C \):
  \[
  T_C - m_C g = -m_C a \tag{3}
  \]
  (Note that \( C \) accelerates downward, hence the negative sign).

#### 2. System Constraints:
For an ideal pulley system, we can assume that the tension in the string is the same across the pulleys (due to massless and frictionless assumptions). Therefore, we have \( T_A = T_B = T_C \).

#### 3. Solving the System:

We now have three equations:

1. \( m_A g - T = m_A a \)
2. \( T - \mu_k m_B g = m_B a \)
3. \( T - m_C g = -m_C a \)

Solving these equations step by step:

- From equation (3):
  \[
  T = m_C g - m_C a
  \]
  
- Substituting \( T \) into equation (1):
  \[
  m_A g - (m_C g - m_C a) = m_A a
  \]
  Simplifying:
  \[
  m_A g - m_C g + m_C a = m_A a
  \]
  \[
  m_A g - m_C g = (m_A + m_C) a
  \]
  \[
  a = \frac{m_A g - m_C g}{m_A + m_C}
  \]
  Substituting values \( m_A = 1.25 \, \text{kg} \), \( m_C = 3.75 \, \text{kg} \), and \( g = 9.8 \, \text{m/s}^2 \):
  \[
  a = \frac{(1.25)(9.8) - (3.75)(9.8)}{1.25 + 3.75}
  \]
  \[
  a = \frac{12.25 - 36.75}{5} = \frac{-24.5}{5} = -4.9 \, \text{m/s}^2
  \]

This value implies that block \( C \) moves downward, and block \( A \) moves upward with an acceleration of \( 4.9 \, \text{m/s}^2 \).

Would you like further details or explanations on any part of this solution? Here are some related questions that might help clarify:

1. How does tension in the string affect the system?
2. Why is the acceleration the same for all blocks?
3. How would the result change if the coefficient of friction were different?
4. What happens if the pulleys were not frictionless?
5. How does Newton's second law apply to each block individually?

**Tip:** When analyzing pulley systems, always write out the force equations for each object involved and consider the direction of forces and accelerations carefully.

Learn how to calculate the acceleration of a system involving masses and a pulley. This problem involves Newton's Second Law, friction, and pulley dynamics. The system consists of three masses (1.25 kg, 2.50 kg, and 3.75 kg), with a coefficient of kinetic friction of 0.602, and all blocks accelerate with the same magnitude.

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