A body of 5 kg weight kept on a rough inclined plane of angle `30^(@)` starts sliding with a constant velocity. Then the coefficient of friction is (assume `g=10m//s^(2)`)

Options

A. `1//sqrt(3)`
B. `2//sqrt(3)`
C. `sqrt(3)`
D. `2sqrt(3)`

Answer

To solve the problem, we need to find the coefficient of friction (μ) for a body of weight 5 kg that is sliding down a rough inclined plane at a constant velocity. Given that the angle of inclination (θ) is 30 degrees and the acceleration due to gravity (g) is 10 m/s², we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body:**
- The weight of the body (W) can be calculated as:
\[ W = mg = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \]
- The weight can be resolved into two components:
- Parallel to the incline: \( W_{\parallel} = mg \sin \theta \)
- Perpendicular to the incline: \( W_{\perpendicular} = mg \cos \theta \)

2. **Calculate the Components of Weight:**
- For θ = 30 degrees:
\[ W_{\parallel} = 50 \sin(30^\circ) = 50 \times \frac{1}{2} = 25 \, \text{N} \]
\[ W_{\perpendicular} = 50 \cos(30^\circ) = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \, \text{N} \]

3. **Determine the Normal Force (N):**
- The normal force (N) acting on the body is equal to the perpendicular component of the weight:
\[ N = W_{\perpendicular} = 25\sqrt{3} \, \text{N} \]

4. **Apply the Condition for Constant Velocity:**
- Since the body is sliding down with constant velocity, the net force acting along the incline is zero:
\[ W_{\parallel} - F_k = 0 \]
- Where \( F_k \) is the kinetic friction force, which can be expressed as:
\[ F_k = \mu N \]
- Therefore, we have:
\[ mg \sin \theta = \mu N \]

5. **Substituting Known Values:**
- Substitute \( W_{\parallel} \) and \( N \):
\[ 25 = \mu (25\sqrt{3}) \]

6. **Solving for the Coefficient of Friction (μ):**
- Rearranging the equation gives:
\[ \mu = \frac{25}{25\sqrt{3}} = \frac{1}{\sqrt{3}} \]
- This can be simplified to:
\[ \mu = \frac{\sqrt{3}}{3} \]

### Final Answer:
The coefficient of friction (μ) is:
\[ \mu = \frac{\sqrt{3}}{3} \]