A block of mass 4kg rests on an an inclined plane. The inclination of the plane is gradually increased. It is found that when the inclination is 3 in 5 `(sin=theta(3)/(5))` the block just begins to slide down the plane. The coefficient of friction between the block and the plane is.

Options

A. `0.4`
B. `0.6`
C. `0.8`
D. `0.75`

Explanation

To find the coefficient of friction between the block and the inclined plane, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
When the block is on the inclined plane, the forces acting on it include:
- The gravitational force acting downwards, \( F_g = mg \), where \( m = 4 \, \text{kg} \) and \( g \approx 9.8 \, \text{m/s}^2 \).
- The normal force \( N \) acting perpendicular to the surface of the incline.
- The frictional force \( F_f \) acting up the incline, which opposes the motion.

### Step 2: Calculate the Gravitational Force
The gravitational force can be calculated as:
\[ F_g = mg = 4 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 39.2 \, \text{N} \]

### Step 3: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( F_{\perpendicular} = mg \cos(\theta) \)
- Parallel to the incline: \( F_{\parallel} = mg \sin(\theta) \)

Given that \( \sin(\theta) = \frac{3}{5} \), we can find \( \cos(\theta) \) using the Pythagorean identity:
\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Step 4: Calculate the Components of the Gravitational Force
Now calculate the components:
- \( F_{\perpendicular} = mg \cos(\theta) = 39.2 \, \text{N} \times \frac{4}{5} = 31.36 \, \text{N} \)
- \( F_{\parallel} = mg \sin(\theta) = 39.2 \, \text{N} \times \frac{3}{5} = 23.52 \, \text{N} \)

### Step 5: Set Up the Equation for Motion
At the point just before the block begins to slide, the frictional force is equal to the component of the gravitational force acting down the incline:
\[ F_f = F_{\parallel} \]
The frictional force can also be expressed as:
\[ F_f = \mu N \]
Where \( \mu \) is the coefficient of friction and \( N = F_{\perpendicular} \).

### Step 6: Substitute and Solve for Coefficient of Friction
Substituting the values we have:
\[ \mu N = F_{\parallel} \]
\[ \mu (31.36 \, \text{N}) = 23.52 \, \text{N} \]
Now solve for \( \mu \):
\[ \mu = \frac{23.52 \, \text{N}}{31.36 \, \text{N}} \approx 0.75 \]

### Final Answer
The coefficient of friction between the block and the inclined plane is approximately \( \mu \approx 0.75 \).

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