https://file.aiquickdraw.com/mathbot/file/1726775290798ygkfz555.jpg

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

Let's go step by step to determine the magnitude and direction of the acceleration of each block, taking into account the system setup from the image.

### Step 1: Understanding the system and forces involved

We have three masses connected by a pulley system:
- \( m_A = 1.25 \, \text{kg} \)
- \( m_B = 2.50 \, \text{kg} \)
- \( m_C = 3.75 \, \text{kg} \)
- Coefficient of kinetic friction between \( m_B \) and the ground: \( \mu_k = 0.602 \)

The blocks are connected with light strings, so the tension is the same on both sides. The system is massless and frictionless except for \( m_B \), which is subject to kinetic friction as it moves on the ground.

Let:
- \( T \) be the tension in the strings.
- \( a \) be the acceleration (same for all masses due to the string connection).
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 2: Free-body diagrams

For each mass, we can write equations based on Newton's second law.

#### For \( m_A \):
\( m_A \) is moving vertically, so:
\[
T - m_A g = m_A a
\]

#### For \( m_B \):
\( m_B \) is moving horizontally and subject to kinetic friction, so:
\[
T - f_k = m_B a
\]
Where \( f_k \) is the kinetic friction force, given by:
\[
f_k = \mu_k m_B g
\]
Substituting this:
\[
T - \mu_k m_B g = m_B a
\]

#### For \( m_C \):
\( m_C \) is also moving vertically, so:
\[
m_C g - T = m_C a
\]

### Step 3: Set up the equations

Now, we have the following three equations:

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

We will solve this system to find \( a \).

### Step 4: Solving for acceleration \( a \)

First, solve equation (1) for \( T \):
\[
T = m_A g + m_A a
\]

Substitute this into equation (3):
\[
m_C g - (m_A g + m_A a) = m_C a
\]
Simplifying:
\[
m_C g - m_A g - m_A a = m_C a
\]
\[
m_C g - m_A g = (m_C + m_A) a \quad \text{(Equation 4)}
\]

Next, substitute \( T = m_A g + m_A a \) into equation (2):
\[
m_A g + m_A a - \mu_k m_B g = m_B a
\]
Simplifying:
\[
m_A g - \mu_k m_B g = (m_B - m_A) a \quad \text{(Equation 5)}
\]

### Step 5: Solve for \( a \)

Now, we have two equations to solve for \( a \):

**From equation 4:**
\[
a = \frac{m_C g - m_A g}{m_C + m_A}
\]

Substituting the values:
\[
a = \frac{(3.75)(9.8) - (1.25)(9.8)}{3.75 + 1.25}
\]
\[
a = \frac{36.75 - 12.25}{5} = \frac{24.5}{5} = 4.9 \, \text{m/s}^2
\]

Thus, the acceleration is \( a = 4.9 \, \text{m/s}^2 \).

### Step 6: Determine the direction

- Block \( m_A \) accelerates **upwards**.
- Block \( m_B \) moves **horizontally to the right**.
- Block \( m_C \) accelerates **downwards**.

Would you like more details on any of these steps or have further questions? 

### Related Questions:
1. How does the coefficient of friction impact the acceleration of \( m_B \)?
2. What happens to the system if we change the mass of \( m_A \)?
3. How would the situation change if the pulley had mass and friction?
4. Can we solve for the tension in the string after finding the acceleration?
5. How does the system behave if one of the masses reaches the pulley or ground first?

**Tip:** When analyzing pulley systems, always identify the forces acting on each block separately to create a clear picture for solving the system.

Learn how to calculate the acceleration of three blocks in a pulley system, including kinetic friction for one block. This problem uses Newton's Second Law and involves calculating tension and acceleration. Suitable for college-level physics.

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