A block of mass m slides down an inclined plane of inclination `theta` with uniform speed The coefficient of friction between the block and the plane is `mu`. The contact force between the block and the plane is .

Options

A. `mg sin thetasqrt(1+mu^(2))`
B. `sqrt((mg sin theta)^(2)+(mumgcostheta)^(2))`
C. `mg sin theta`
D. `mg`

Explanation

To solve the problem of a block of mass \( m \) sliding down an inclined plane with an angle \( \theta \) and a coefficient of friction \( \mu \), we need to analyze the forces acting on the block. Given that the block is sliding down with uniform speed, we can conclude that the net force acting on the block is zero.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The gravitational force acting downwards: \( F_g = mg \)
- The normal force acting perpendicular to the inclined plane: \( N \)
- The frictional force acting opposite to the direction of motion: \( F_f = \mu N \)

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: \( F_{\perpendicular} = mg \cos \theta \)

3. **Apply Newton's Second Law:**
- Since the block is moving with uniform speed, the net force along the incline is zero. Therefore:
\[ F_{\parallel} - F_f = 0 \]
- Substituting the expressions for the forces:
\[ mg \sin \theta - \mu N = 0 \]

4. **Relate Normal Force to Weight:**
- The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

5. **Substitute Normal Force into the Equation:**
- Now substitute \( N \) into the friction force equation:
\[ mg \sin \theta - \mu (mg \cos \theta) = 0 \]

6. **Simplify the Equation:**
- Factor out \( mg \):
\[ mg (\sin \theta - \mu \cos \theta) = 0 \]
- Since \( mg \neq 0 \), we can simplify to:
\[ \sin \theta - \mu \cos \theta = 0 \]

7. **Solve for the Coefficient of Friction:**
- Rearranging gives:
\[ \mu = \frac{\sin \theta}{\cos \theta} = \tan \theta \]

8. **Calculate the Contact Force:**
- Now, substituting back to find the contact force \( N \):
\[ N = mg \cos \theta \]

### Final Result:
The contact force between the block and the inclined plane is:
\[ N = mg \cos \theta \]