A block is placed on a smooth inclined plane at an angle `'theta'` to the horizontal. The acceleration must the plane be moved horizontally so that the block to fall freely is 

Options

A. `g tan q`
B. `g cot q`
C. `g sin q`
D. `g cos q`

Answer

To solve the problem of determining the acceleration that the inclined plane must have in order for the block to fall freely, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
- The block experiences gravitational force acting downward, which is equal to \( mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.
- The inclined plane is at an angle \( \theta \) to the horizontal.

2. **Condition for Free Fall**:
- For the block to fall freely, the normal force \( N \) between the block and the inclined plane must be zero. This means that the only force acting on the block in the vertical direction is its weight.

3. **Setting Up the Equations**:
- When the inclined plane accelerates horizontally with acceleration \( a \), we can analyze the forces acting on the block.
- The component of the gravitational force acting down the incline is \( mg \sin \theta \).
- The component of the gravitational force acting perpendicular to the incline is \( mg \cos \theta \).

4. **Applying Newton's Second Law**:
- In the direction perpendicular to the incline, the block is in equilibrium when the normal force is zero:
\[ N - mg \cos \theta = 0 \quad \Rightarrow \quad N = mg \cos \theta \]
- Since \( N = 0 \) for free fall, we have:
\[ 0 = mg \cos \theta \]

5. **Analyzing the Incline's Acceleration**:
- The block will also experience a pseudo force due to the acceleration of the inclined plane. This pseudo force acts horizontally opposite to the direction of the incline’s acceleration.
- The effective force acting on the block along the incline due to this pseudo force is \( ma \sin \theta \).
- For the block to fall freely, the net force along the incline must equal the gravitational force component down the incline:
\[ ma \sin \theta = mg \cos \theta \]

6. **Solving for Acceleration \( a \)**:
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a \sin \theta = g \cos \theta \]
- Rearranging gives:
\[ a = \frac{g \cos \theta}{\sin \theta} \]
- This simplifies to:
\[ a = g \cot \theta \]

### Final Answer:
The acceleration that the inclined plane must have for the block to fall freely is:
\[ a = g \cot \theta \]