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
B. `mg sin theta sqrt(1+mu^(2))`
C. `mg sin theta`
D. `sqrt((mg sin theta)^(2)+(mumg cos theta)^(2))`

Answer

To solve the problem of finding the contact force between a block of mass \( m \) sliding down an inclined plane with an inclination \( \theta \) and a coefficient of friction \( \mu \), we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
The forces acting on the block are:
1. The gravitational force \( \vec{F_g} = mg \) acting vertically downward.
2. The normal force \( \vec{N} \) acting perpendicular to the inclined plane.
3. The frictional force \( \vec{F_f} \) acting parallel to the inclined plane, opposing the motion.

### Step 2: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- Along the inclined plane:
\[ F_{\text{parallel}} = mg \sin \theta \]
- Perpendicular to the inclined plane:
\[ F_{\text{perpendicular}} = mg \cos \theta \]

### Step 3: Analyze the Motion of the Block
Since the block is sliding down the incline with uniform speed, the net force acting along the incline must be zero (i.e., no acceleration). Therefore, the frictional force must balance the component of gravitational force acting down the incline:
\[ F_f = mg \sin \theta \]

### Step 4: Determine the Normal Force
The normal force \( N \) acting on the block can be determined from the component of the gravitational force acting perpendicular to the incline:
\[ N = mg \cos \theta \]

### Step 5: Calculate the Contact Force
The contact force \( R \) is the resultant 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:
\[ R = \sqrt{N^2 + F_f^2} \]
Substituting the expressions for \( N \) and \( F_f \):
\[ R = \sqrt{(mg \cos \theta)^2 + (mg \sin \theta)^2} \]

### Step 6: Simplify the Expression
Factoring out \( m^2g^2 \):
\[ R = \sqrt{m^2g^2 (\cos^2 \theta + \sin^2 \theta)} \]
Using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ R = \sqrt{m^2g^2 \cdot 1} = mg \]

### Final Answer
Thus, the contact force between the block and the inclined plane is:
\[ R = mg \]