A block of mass `m` is placed at rest on an inclination `theta` to the horizontal.If the coefficient of friction between the block and the plane is `mu`, then the total force the inclined plane exerts on the block is

Options

A. `mg`
B. `mu mg cos theta`
C. ` mg sin theta`
D. `mu mg tan theta`

Explanation

To solve the problem of finding the total force that the inclined plane exerts on the block, we will analyze the forces acting on the block placed on the inclined plane.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The block experiences three main forces:
- **Gravitational Force (Weight)**: This acts vertically downward and is given by \( W = mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.
- **Normal Force (N)**: This acts perpendicular to the surface of the inclined plane.
- **Frictional Force (f)**: This opposes the motion of the block and acts parallel to the surface of the inclined plane.

2. **Resolve the Gravitational Force**:
- The gravitational force can be resolved into two components:
- **Parallel to the Incline**: \( W_{\parallel} = mg \sin(\theta) \)
- **Perpendicular to the Incline**: \( W_{\perpendicular} = mg \cos(\theta) \)

3. **Determine the Normal Force (N)**:
- The normal force balances the perpendicular component of the weight. Thus, we have:
\[ N = W_{\perpendicular} = mg \cos(\theta) \]

4. **Determine the Frictional Force (f)**:
- The frictional force can be calculated using the coefficient of friction (\( \mu \)):
\[ f = \mu N = \mu (mg \cos(\theta)) \]

5. **Calculate the Total Force Exerted by the Inclined Plane**:
- The total force exerted by the inclined plane on the block is the vector sum of the normal force and the frictional force. Since these two forces are perpendicular to each other, we can use the Pythagorean theorem:
\[ R = \sqrt{N^2 + f^2} \]
- Substituting the expressions for \( N \) and \( f \):
\[ R = \sqrt{(mg \cos(\theta))^2 + (\mu mg \cos(\theta))^2} \]
- Factor out \( (mg \cos(\theta))^2 \):
\[ R = mg \cos(\theta) \sqrt{1 + \mu^2} \]

### Final Result:
The total force that the inclined plane exerts on the block is:
\[ R = mg \cos(\theta) \sqrt{1 + \mu^2} \]