A block is kept on an inclined plane of inclination `theta` of length l. the velocity of particle at the bottom of inclined is (the coefficient of friciton is `mu`)

Options

A. `sqrt(2gl(mucostheta-sintheta))`
B. `sqrt(2gl(sintheta-mucostheta))`
C. `sqrt(2gl(sintheta+mucostheta))`
D. `sqrt(2gl(costheta+musintheta))`

Answer

To solve the problem of finding the velocity of a block at the bottom of an inclined plane with a given angle of inclination \( \theta \), length \( l \), and coefficient of friction \( \mu \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The weight of the block \( mg \) can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)
- The normal force \( N \) acting on the block is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]
- The frictional force \( f \) acting against the motion of the block is given by:
\[ f = \mu N = \mu (mg \cos \theta) \]

2. **Apply Newton's Second Law**:
- Along the incline, the net force \( F \) acting on the block is given by:
\[ F = mg \sin \theta - f \]
- Substituting for \( f \):
\[ F = mg \sin \theta - \mu (mg \cos \theta) \]
- According to Newton's second law, this net force is equal to the mass \( m \) times the acceleration \( a \):
\[ ma = mg \sin \theta - \mu (mg \cos \theta) \]
- Dividing through by \( m \):
\[ a = g \sin \theta - \mu g \cos \theta \]

3. **Use Kinematic Equations**:
- We can use the kinematic equation to find the final velocity \( v \) of the block at the bottom of the incline:
\[ v^2 = u^2 + 2as \]
- Here, the initial velocity \( u = 0 \), the distance \( s = l \), and the acceleration \( a \) we found in the previous step:
\[ v^2 = 0 + 2 \left(g \sin \theta - \mu g \cos \theta\right) l \]
- Simplifying, we get:
\[ v^2 = 2l \left(g \sin \theta - \mu g \cos \theta\right) \]

4. **Calculate the Final Velocity**:
- Taking the square root to find \( v \):
\[ v = \sqrt{2l \left(g \sin \theta - \mu g \cos \theta\right)} \]
- This can be further simplified to:
\[ v = \sqrt{2g l \left(\sin \theta - \mu \cos \theta\right)} \]

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