A block slides down an inclined surface of inclination `30^0` with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficietn of kinetic friction between the two.

Explanation

To solve the problem of a block sliding down an inclined plane with a given angle and distance, we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
- The forces acting on the block include:
1. Gravitational force acting downwards: \( mg \)
2. Normal force acting perpendicular to the inclined surface: \( N \)
3. Frictional force acting opposite to the direction of motion: \( f_k = \mu_k N \)

### Step 2: Resolve the Gravitational Force
- The gravitational force can be resolved into two components:
- Parallel to the incline: \( F_{\parallel} = mg \sin \theta \)
- Perpendicular to the incline: \( F_{\perpendicular} = mg \cos \theta \)

Given that \( \theta = 30^\circ \):
- \( F_{\parallel} = mg \sin 30^\circ = mg \cdot \frac{1}{2} = \frac{mg}{2} \)
- \( F_{\perpendicular} = mg \cos 30^\circ = mg \cdot \frac{\sqrt{3}}{2} = \frac{mg\sqrt{3}}{2} \)

### Step 3: Calculate the Normal Force
- The normal force \( N \) balances the perpendicular component of the weight:
\[ N = F_{\perpendicular} = mg \cos 30^\circ = \frac{mg\sqrt{3}}{2} \]

### Step 4: Write the Equation of Motion
- According to Newton's second law, the net force along the incline is equal to mass times acceleration:
\[ F_{\text{net}} = F_{\parallel} - f_k = ma \]
Substituting the expressions for \( F_{\parallel} \) and \( f_k \):
\[ mg \sin 30^\circ - \mu_k N = ma \]
\[ \frac{mg}{2} - \mu_k \left(\frac{mg\sqrt{3}}{2}\right) = ma \]

### Step 5: Simplify the Equation
- Since mass \( m \) appears in all terms, we can cancel it out:
\[ \frac{g}{2} - \mu_k \left(\frac{g\sqrt{3}}{2}\right) = a \]

### Step 6: Calculate Acceleration
- We know the block covers a distance of 8 m in 2 seconds starting from rest. Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \), \( s = 8 \, \text{m} \), and \( t = 2 \, \text{s} \):
\[ 8 = 0 + \frac{1}{2} a (2^2) \]
\[ 8 = 2a \implies a = 4 \, \text{m/s}^2 \]

### Step 7: Substitute Acceleration into the Equation
- Substitute \( a = 4 \, \text{m/s}^2 \) into the equation:
\[ \frac{g}{2} - \mu_k \left(\frac{g\sqrt{3}}{2}\right) = 4 \]
Using \( g \approx 10 \, \text{m/s}^2 \):
\[ \frac{10}{2} - \mu_k \left(\frac{10\sqrt{3}}{2}\right) = 4 \]
\[ 5 - 5\sqrt{3} \mu_k = 4 \]
\[ 5\sqrt{3} \mu_k = 1 \implies \mu_k = \frac{1}{5\sqrt{3}} \]

### Step 8: Calculate the Coefficient of Kinetic Friction
- Evaluating \( \mu_k \):
\[ \mu_k \approx \frac{1}{5 \cdot 1.732} \approx \frac{1}{8.66} \approx 0.115 \]

### Final Answer
The coefficient of kinetic friction \( \mu_k \) is approximately \( 0.11 \).

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