Block of mass 10 kg is moving on inclined plane with constant velocity 10 m/s. The coedfficient of kinetic friction between incline plane and block is :-

Options

A. 0.57
B. 0.75
C. 0.5
D. None of these

Answer

To solve the problem of finding the coefficient of kinetic friction (\( \mu_k \)) between the inclined plane and the block, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
The block of mass \( m = 10 \, \text{kg} \) is moving with a constant velocity of \( 10 \, \text{m/s} \) on an inclined plane. Since it is moving with constant velocity, the net force acting on the block along the incline must be zero.

### Step 2: Identify the Forces
1. **Weight of the block (\( W \))**: The weight acts vertically downward and is given by \( W = mg \), where \( g \approx 9.8 \, \text{m/s}^2 \).
2. **Normal force (\( N \))**: This acts perpendicular to the inclined surface.
3. **Frictional force (\( F_f \))**: This opposes the motion of the block and is given by \( F_f = \mu_k N \).
4. **Components of Weight**:
- Parallel to the incline: \( W_{\parallel} = mg \sin \theta \)
- Perpendicular to the incline: \( W_{\perpendicular} = mg \cos \theta \)

### Step 3: Set Up the Equations
Since the block is moving with constant velocity, the net force along the incline is zero:
\[ mg \sin \theta - F_f = 0 \]
Substituting the frictional force:
\[ mg \sin \theta - \mu_k N = 0 \]

### Step 4: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

### Step 5: Substitute Normal Force into the Equation
Substituting \( N \) into the friction equation:
\[ mg \sin \theta - \mu_k (mg \cos \theta) = 0 \]
This simplifies to:
\[ mg \sin \theta = \mu_k (mg \cos \theta) \]

### Step 6: Cancel Out Common Terms
Since \( mg \) is common on both sides, we can cancel it out (assuming \( m \neq 0 \)):
\[ \sin \theta = \mu_k \cos \theta \]

### Step 7: Solve for Coefficient of Kinetic Friction
Rearranging gives:
\[ \mu_k = \frac{\sin \theta}{\cos \theta} = \tan \theta \]

### Step 8: Calculate \( \tan \theta \)
Given that \( \theta = 37^\circ \):
\[ \tan 37^\circ = \frac{3}{4} = 0.75 \]

### Final Answer
Thus, the coefficient of kinetic friction (\( \mu_k \)) between the inclined plane and the block is:
\[ \mu_k = 0.75 \]