A block has been placed on an inclined plane . The slope angle of `theta` of the plane is such that the block slides down the plane at a constant speed . The coefficient of kinetic friction is equal to :

Options

A. `sin theta`
B. `cos theta`
C. g
D. `tan theta`

Explanation

To solve the problem of finding the coefficient of kinetic friction (\( \mu_k \)) for a block sliding down an inclined plane at a constant speed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Block**:
- The weight of the block (\( mg \)) acts vertically downward.
- This weight can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

2. **Determine the Normal Force**:
- The normal force (\( N \)) acting on the block is equal to the component of the weight acting perpendicular to the incline. Therefore, we have:
\[ N = mg \cos \theta \]

3. **Identify the Frictional Force**:
- The kinetic friction force (\( F_k \)) acting against the motion of the block is given by:
\[ F_k = \mu_k N \]
- Substituting the expression for the normal force, we get:
\[ F_k = \mu_k (mg \cos \theta) \]

4. **Apply the Condition of Constant Speed**:
- Since the block is sliding down the incline at a constant speed, the net force acting along the incline is zero. This means that the force due to gravity down the incline is balanced by the frictional force:
\[ mg \sin \theta = F_k \]
- Substituting the expression for the frictional force, we have:
\[ mg \sin \theta = \mu_k (mg \cos \theta) \]

5. **Simplify the Equation**:
- We can cancel \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \sin \theta = \mu_k \cos \theta \]

6. **Solve for the Coefficient of Kinetic Friction**:
- Rearranging the equation gives:
\[ \mu_k = \frac{\sin \theta}{\cos \theta} \]
- This simplifies to:
\[ \mu_k = \tan \theta \]

### Final Answer:
The coefficient of kinetic friction (\( \mu_k \)) is equal to \( \tan \theta \).

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