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. `(a)mg`
B. `(b)mu mg cos theta`
C. ` (c)mg sin theta`
D. `(d)mu mg tan theta`

Explanation

To find the total force that the inclined plane exerts on the block, we need to analyze the forces acting on the block placed on the inclined plane. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block are:
1. **Weight (mg)**: This force acts vertically downward.
2. **Normal Force (N)**: This force acts perpendicular to the surface of the inclined plane.
3. **Frictional Force (F)**: This force acts parallel to the surface of the inclined plane and opposes the motion.

### Step 2: Resolve the Weight into Components
The weight of the block can be resolved into two components:
- **Perpendicular to the incline**: \( mg \cos(\theta) \)
- **Parallel to the incline**: \( mg \sin(\theta) \)

### Step 3: Apply Newton's Second Law
Since the block is at rest, the net force acting on it must be zero. Therefore, we can set up the following equations based on the forces acting on the block:

1. In the direction perpendicular to the incline:
\[ N = mg \cos(\theta) \]

2. In the direction parallel to the incline (considering friction):
The frictional force \( F \) must balance the component of the weight acting down the incline:
\[ F = mg \sin(\theta) \]

### Step 4: 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. However, since the frictional force acts parallel to the incline and the normal force acts perpendicular to it, we can find the resultant force \( R \) using the Pythagorean theorem:

\[ R = \sqrt{N^2 + F^2} \]

Substituting the expressions for \( N \) and \( F \):
\[ R = \sqrt{(mg \cos(\theta))^2 + (mg \sin(\theta))^2} \]

### Step 5: Simplify the Expression
Using the Pythagorean identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \):
\[ R = \sqrt{m^2g^2(\cos^2(\theta) + \sin^2(\theta))} = \sqrt{m^2g^2} = mg \]

### Conclusion
Thus, the total force that the inclined plane exerts on the block is:
\[ \text{Total Force} = mg \]