A block of mass m is at rest on a rough inclined plane, whose friction coefficient is `mu.` Net force exerted by surface on block is :-

Options

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

Explanation

To solve the problem of finding the net force exerted by the surface on a block of mass \( m \) resting on a rough inclined plane with a friction coefficient \( \mu \), we can follow these steps:

### Step 1: Identify Forces Acting on the Block
The forces acting on the block are:
1. The gravitational force (\( \vec{W} \)) acting downward, which can be expressed as \( W = mg \), where \( g \) is the acceleration due to gravity.
2. The normal force (\( \vec{N} \)) exerted by the inclined plane on the block, acting perpendicular to the surface.
3. The frictional force (\( \vec{F}_{\text{friction}} \)) acting parallel to the surface, opposing the motion.

### Step 2: Resolve the Weight into Components
The weight of the block can be resolved into two components:
- A component parallel to the incline: \( W_{\parallel} = mg \sin \theta \)
- A component perpendicular to the incline: \( W_{\perpendicular} = mg \cos \theta \)

### Step 3: Apply Equilibrium Conditions
Since the block is at rest, the net force acting on it must be zero. Therefore, we can set up the following equilibrium conditions:

1. **In the direction perpendicular to the incline**:
\[ N = mg \cos \theta \]
(The normal force balances the perpendicular component of the weight.)

2. **In the direction parallel to the incline**:
\[ F_{\text{friction}} = mg \sin \theta \]
(The frictional force balances the parallel component of the weight.)

### Step 4: Calculate the Net Force Exerted by the Surface
The net force exerted by the surface 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 to find the resultant force:

\[ F_{\text{net}} = \sqrt{N^2 + F_{\text{friction}}^2} \]

Substituting the expressions we found:
\[ F_{\text{net}} = \sqrt{(mg \cos \theta)^2 + (mg \sin \theta)^2} \]

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

### Conclusion
The net force exerted by the surface on the block is:
\[ F_{\text{net}} = mg \]

### Final Answer
The net force exerted by the surface on the block is \( mg \).

---