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Let's go through the problem step by step to solve it.

### Given:
- Mass of the wooden block \( m_1 = 8 \, \text{kg} \)
- Mass of the second particle \( m_2 = 6 \, \text{kg} \)
- The block is released from rest and travels \( 0.6 \, \text{m} \) in \( 0.8 \, \text{s} \)
- The string passes over a smooth, light pulley
- There is friction with coefficient \( \mu \) between the block and the table.

We are tasked with finding:
- (a) The acceleration of the block
- (b) The tension in the string
- (c) The coefficient of friction \( \mu \)

### (a) Finding the acceleration of the block:

Using the equation of motion, the displacement \( s \), initial velocity \( u = 0 \), time \( t = 0.8 \, \text{s} \), and displacement \( s = 0.6 \, \text{m} \), we use the kinematic equation:
\[
s = ut + \frac{1}{2} a t^2
\]
Substituting the known values:
\[
0.6 = 0 \cdot 0.8 + \frac{1}{2} a (0.8)^2
\]
\[
0.6 = \frac{1}{2} a \cdot 0.64
\]
\[
a = \frac{0.6 \times 2}{0.64} = \frac{1.2}{0.64} = 1.875 \, \text{m/s}^2
\]

Thus, the acceleration of the block is \( 1.875 \, \text{m/s}^2 \).

### (b) Finding the tension in the string:

For this, we need to consider the forces acting on both the wooden block and the second particle.

- For the block:
  The block experiences tension \( T \) opposing its motion and friction \( f = \mu m_1 g \). Using Newton’s second law, we can write the equation for the block:
  \[
  T - \mu m_1 g = m_1 a
  \]
  
- For the second particle:
  The second particle experiences the force of gravity pulling it down, which is \( m_2 g \), and the tension in the string \( T \) opposing the motion. The equation of motion is:
  \[
  m_2 g - T = m_2 a
  \]
  
We have two equations now:

1. \( T - \mu m_1 g = m_1 a \)
2. \( m_2 g - T = m_2 a \)

Solving these two equations will help us find the tension \( T \) and the coefficient of friction \( \mu \).

Let me calculate these next.### (b) The tension in the string:
The tension in the string is \( T = 47.61 \, \text{N} \).

### (c) The coefficient of friction \( \mu \):
The coefficient of friction between the block and the table is \( \mu = 0.416 \) (approximately).

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Let me know if you would like further clarification on any part, and feel free to ask if you have more questions!

Here are some related questions to expand on this:
1. What happens if the pulley is not smooth? How does this affect the system?
2. How would increasing the mass of the second particle affect the acceleration?
3. What is the relationship between tension and friction in this setup?
4. If the coefficient of friction \( \mu \) were zero, how would the problem change?
5. How would the results change if the block were on an inclined plane?

**Tip:** When dealing with connected systems involving pulleys, always write separate equations for each object to keep track of forces acting on them individually.

The diagram shows a wooden block, of mass 8 kg, which is attached to a second particle, of mass 6 kg, by a light string. The string passes over a smooth, light pulley. The coefficient of friction between the block and the table is μ. The block is released from rest and travels 0.6 meters in 0.8 seconds.
(a) Find the acceleration of the block.
(b) Find the tension in the string.
(c) Find μ.

Solve a block and pulley system problem involving an 8 kg block and a 6 kg particle. Learn how to calculate the acceleration of the block, the tension in the string, and the coefficient of friction. Suitable for students in Grades 10-12, this solution uses kinematic equations, Newton's second law, and friction formulas.

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