A block of mass m is placed on a smooth inclined plane of inclination `theta` with the horizontal. The force exerted by the plane on the block has a magnitude

Options

A. mg
B. `mg/costheta`
C. `mg costheta`
D. `mg tantheta`

Answer

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

### Step 1: Identify the Forces Acting on the Block
The block of mass \( m \) experiences two main forces:
1. The gravitational force \( \vec{F_g} = m\vec{g} \) acting vertically downward.
2. The normal force \( \vec{N} \) exerted by the inclined plane acting perpendicular to the surface of the plane.

### Step 2: Resolve the Gravitational Force
The gravitational force can be resolved into two components relative to the inclined plane:
- A component acting perpendicular to the inclined plane: \( F_{\perp} = mg \cos \theta \)
- A component acting parallel to the inclined plane: \( F_{\parallel} = mg \sin \theta \)

### Step 3: Analyze the Motion of the Block
Since the block is placed on a smooth inclined plane, it is not accelerating. This implies that the net force acting on the block is zero. Therefore, the normal force must balance the perpendicular component of the gravitational force.

### Step 4: Set Up the Equation for the Normal Force
The normal force \( N \) exerted by the inclined plane must equal the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Conclusion
Thus, the magnitude of the force exerted by the inclined plane on the block is:
\[ N = mg \cos \theta \]