A block is kept on a inclined plane of inclination `theta` of length l. The velocity of particle at the bottom of inclined is (the coefficient of friction is `mu`

Options

A. `sqrt(2 gl (mu cos theta - sin theta))`
B. `sqrt(2 gl (sin theta - mu cos theta))`
C. `sqrt(2 gl (sin theta + mu cos theta))`
D. `sqrt(2 gl (cos theta - mu sin theta))`

Explanation

To find the velocity of a block sliding down an inclined plane with a given angle of inclination \( \theta \), length \( l \), and coefficient of friction \( \mu \), we can follow these steps:

### Step 1: Identify Forces Acting on the Block
The forces acting on the block include:
- Gravitational force \( mg \) acting downward.
- Normal force \( N \) acting perpendicular to the inclined plane.
- Frictional force \( f \) acting opposite to the direction of motion along the incline.

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

### Step 3: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Step 4: Calculate the Frictional Force
The frictional force \( f \) can be calculated using the coefficient of friction:
\[ f = \mu N = \mu (mg \cos \theta) \]

### Step 5: Apply Newton's Second Law
Using Newton's second law along the incline, we can set up the equation:
\[ F_{\text{net}} = ma \]
Where \( F_{\text{net}} = F_{\parallel} - f \):
\[ mg \sin \theta - \mu mg \cos \theta = ma \]
This simplifies to:
\[ mg (\sin \theta - \mu \cos \theta) = ma \]

### Step 6: Solve for Acceleration
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ a = g (\sin \theta - \mu \cos \theta) \]

### Step 7: Use Kinematics to Find Final Velocity
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( u = 0 \) (initial velocity)
- \( a = g (\sin \theta - \mu \cos \theta) \)
- \( s = l \) (length of the incline)

Substituting these values in:
\[ v^2 = 0 + 2 \cdot g (\sin \theta - \mu \cos \theta) \cdot l \]
Thus,
\[ v^2 = 2gl (\sin \theta - \mu \cos \theta) \]

### Step 8: Solve for Final Velocity
Taking the square root gives us the final velocity:
\[ v = \sqrt{2gl (\sin \theta - \mu \cos \theta)} \]

### Final Answer
The velocity of the particle at the bottom of the inclined plane is:
\[ v = \sqrt{2gl (\sin \theta - \mu \cos \theta)} \]

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