A block can slide on a smooth inclined plane of inclination `theta` kept on the floor of a lift. When the lift is descending with a retardation a , the acceleration of the block relative to the incline is

Options

A. `(g+a)sintheta`
B. `(g-a)`
C. `gsintheta`
D. `(g-a)sintheta`

Answer

To find the acceleration of the block relative to the inclined plane when the lift is descending with a retardation \( a \), we will follow these steps:

### Step 1: Understand the Forces Acting on the Block
When the lift is descending with retardation \( a \), we can consider the effective acceleration acting on the block due to gravity and the retardation of the lift. The gravitational force acting on the block is \( mg \) (where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity).

### Step 2: Define the Effective Acceleration
Since the lift is descending with retardation \( a \), we can consider this as an upward acceleration of \( a \) for the block relative to the lift. Therefore, the effective acceleration acting on the block can be expressed as:
\[ g_{\text{effective}} = g + a \]
This is because the block experiences an additional upward pseudo force due to the retardation of the lift.

### Step 3: Resolve Forces Along the Incline
The block is on an inclined plane with an angle \( \theta \). The component of the effective gravitational force acting down the incline can be calculated using:
\[ F_{\text{down incline}} = (g + a) \cdot \sin(\theta) \]
This is the force that causes the block to accelerate down the incline.

### Step 4: Apply Newton's Second Law
According to Newton's second law, the net force acting on the block along the incline is equal to the mass of the block multiplied by its acceleration relative to the incline:
\[ F = m \cdot a_{\text{relative}} \]
Substituting the expression for the force down the incline, we have:
\[ m \cdot (g + a) \cdot \sin(\theta) = m \cdot a_{\text{relative}} \]

### Step 5: Solve for the Acceleration of the Block Relative to the Incline
We can cancel the mass \( m \) from both sides of the equation (assuming \( m \neq 0 \)):
\[ (g + a) \cdot \sin(\theta) = a_{\text{relative}} \]
Thus, the acceleration of the block relative to the incline is given by:
\[ a_{\text{relative}} = (g + a) \cdot \sin(\theta) \]

### Final Answer
The acceleration of the block relative to the incline is:
\[ a_{\text{relative}} = (g + a) \cdot \sin(\theta) \]