A block A can slide on a frictionless incline of angle `theta` and length l, kept inside an elevator going up with uniform velocity v. Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.

Explanation

To solve the problem of finding the time taken by the block to slide down the incline, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
The block A is sliding down a frictionless incline of angle θ. The forces acting on the block are:
- The gravitational force (mg) acting downward.
- The normal force (N) acting perpendicular to the incline.

Since the elevator is moving upwards with a uniform velocity v, there is no pseudo force acting on the block.

### Step 2: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( mg \cos(\theta) \)
- Parallel to the incline: \( mg \sin(\theta) \)

Since the incline is frictionless, the only force causing the block to slide down the incline is the component of gravity acting parallel to the incline, which is \( mg \sin(\theta) \).

### Step 3: Apply Newton's Second Law
Using Newton's second law, we can write:
\[ F_{\text{net}} = m a \]
where \( F_{\text{net}} \) is the net force acting on the block along the incline. Here, the net force is \( mg \sin(\theta) \).

Thus, we have:
\[ mg \sin(\theta) = m a \]
Cancelling m from both sides (since m ≠ 0), we get:
\[ a = g \sin(\theta) \]

### Step 4: Use the Kinematic Equation
We will use the second equation of motion to find the time taken to slide down the incline. The equation is:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the distance along the incline (which is L),
- \( u \) is the initial velocity (which is 0, since the block is released),
- \( a \) is the acceleration along the incline (which we found to be \( g \sin(\theta) \)),
- \( t \) is the time taken to slide down.

Substituting the known values into the equation:
\[ L = 0 \cdot t + \frac{1}{2} (g \sin(\theta)) t^2 \]
This simplifies to:
\[ L = \frac{1}{2} g \sin(\theta) t^2 \]

### Step 5: Solve for Time (t)
Rearranging the equation to solve for \( t^2 \):
\[ t^2 = \frac{2L}{g \sin(\theta)} \]
Taking the square root gives:
\[ t = \sqrt{\frac{2L}{g \sin(\theta)}} \]

### Final Answer
The time taken by the block to slide down the length of the incline is:
\[ t = \sqrt{\frac{2L}{g \sin(\theta)}} \]

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