A block of mass 6 kg resting on a
horizontal surface is connected by a cord
passing over a light, frictionless pulley to
a hanging block of mass 4 kg. The
coefficient of kinetic friction between the
block and the horizontal surface is 0.5.
After the blocks are released find: (a) the
acceleration of each block (b) The
tension on the cord

To solve this problem, let’s denote the following variables for convenience:

- \( m_1 = 6 \, \text{kg} \): mass of the block on the horizontal surface.
- \( m_2 = 4 \, \text{kg} \): mass of the hanging block.
- \( \mu_k = 0.5 \): coefficient of kinetic friction between \( m_1 \) and the horizontal surface.
- \( g = 9.8 \, \text{m/s}^2 \): acceleration due to gravity.

### Step 1: Identify the Forces on Each Block

#### For Block \( m_1 \) on the Horizontal Surface:
- Gravitational force: \( F_{g1} = m_1 \cdot g = 6 \cdot 9.8 = 58.8 \, \text{N} \).
- Normal force (equal to gravitational force since it's on a horizontal surface): \( N = 58.8 \, \text{N} \).
- Frictional force opposing motion: \( F_f = \mu_k \cdot N = 0.5 \cdot 58.8 = 29.4 \, \text{N} \).
- Tension in the cord: \( T \) (this is unknown and is the same on both blocks since the pulley is light and frictionless).

#### For Block \( m_2 \) (the Hanging Block):
- Gravitational force: \( F_{g2} = m_2 \cdot g = 4 \cdot 9.8 = 39.2 \, \text{N} \).
- Tension in the cord: \( T \).

### Step 2: Set Up the Equations of Motion

The system accelerates in the direction of the hanging block, so we can apply Newton’s Second Law for each block.

#### For \( m_1 \):
\[
T - F_f = m_1 \cdot a
\]
Substituting \( F_f = 29.4 \, \text{N} \):
\[
T - 29.4 = 6a \quad \text{(Equation 1)}
\]

#### For \( m_2 \):
\[
m_2 \cdot g - T = m_2 \cdot a
\]
Substituting \( F_{g2} = 39.2 \, \text{N} \):
\[
39.2 - T = 4a \quad \text{(Equation 2)}
\]

### Step 3: Solve for Acceleration and Tension

Add Equations 1 and 2 to eliminate \( T \):
\[
(T - 29.4) + (39.2 - T) = 6a + 4a
\]
\[
9.8 = 10a
\]
\[
a = \frac{9.8}{10} = 0.98 \, \text{m/s}^2
\]

Now, substitute \( a = 0.98 \, \text{m/s}^2 \) into Equation 1 to find \( T \):
\[
T - 29.4 = 6 \times 0.98
\]
\[
T - 29.4 = 5.88
\]
\[
T = 5.88 + 29.4 = 35.28 \, \text{N}
\]

### Final Answers:
- (a) The acceleration of each block is \( 0.98 \, \text{m/s}^2 \).
- (b) The tension on the cord is \( 35.28 \, \text{N} \).

Would you like more details, or do you have any questions about this problem?

### Related Questions
1. How would changing the coefficient of kinetic friction affect the acceleration?
2. What would happen if the mass of \( m_2 \) were increased?
3. How does the tension in the cord compare to the weight of \( m_1 \) and \( m_2 \)?
4. If the pulley were not frictionless, how would it affect the system?
5. What would the acceleration be if the surface were frictionless?

#### Tip:
In pulley problems, always start by analyzing forces on each mass separately and set up equations for each block using Newton's Second Law.

A block of mass 6 kg resting on a horizontal surface is connected by a cord passing over a light, frictionless pulley to a hanging block of mass 4 kg. The coefficient of kinetic friction between the block and the horizontal surface is 0.5. After the blocks are released, find: (a) the acceleration of each block (b) the tension on the cord.

This physics problem explores calculating the acceleration and tension in a system where a 6 kg block on a horizontal surface is connected to a 4 kg hanging block. With a friction coefficient of 0.5 and a light, frictionless pulley, learn to apply Newton's laws and frictional force formulas. Ideal for students in Grades 11-12.

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