A body of mass m slides down a rough plane of inclination `alpha` . If `mu` is the coefficient of friction, then acceleration of the body will be

Explanation

To solve the problem of a body of mass \( m \) sliding down a rough inclined plane at an angle \( \alpha \) with a coefficient of friction \( \mu \), we will follow these steps:

### Step 1: Identify the forces acting on the body
The forces acting on the body are:
1. The gravitational force \( mg \) acting downwards.
2. The normal force \( N \) acting perpendicular to the inclined plane.
3. The frictional force \( F \) acting opposite to the direction of motion (up the incline).

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

### Step 3: Write the equation for the normal force
Since there is no motion in the direction perpendicular to the incline, the sum of forces in the y-direction (perpendicular to the incline) is zero:
\[ N - mg \cos \alpha = 0 \]
Thus, the normal force \( N \) is given by:
\[ N = mg \cos \alpha \]

### Step 4: Write the equation for the frictional force
The frictional force \( F \) can be expressed in terms of the normal force:
\[ F = \mu N = \mu (mg \cos \alpha) \]

### Step 5: Write the equation of motion along the incline
In the direction parallel to the incline (x-direction), we can write the equation of motion:
\[ \text{Net force} = m a \]
The net force acting down the incline is the component of gravity minus the frictional force:
\[ mg \sin \alpha - F = ma \]
Substituting the expression for the frictional force:
\[ mg \sin \alpha - \mu (mg \cos \alpha) = ma \]

### Step 6: Simplify the equation
Factoring out \( m \) from the left side:
\[ m(g \sin \alpha - \mu g \cos \alpha) = ma \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ g \sin \alpha - \mu g \cos \alpha = a \]

### Step 7: Final expression for acceleration
Thus, the acceleration \( a \) of the body sliding down the rough inclined plane is given by:
\[ a = g \sin \alpha - \mu g \cos \alpha \]

### Summary of the solution:
The acceleration of the body sliding down a rough inclined plane of inclination \( \alpha \) with coefficient of friction \( \mu \) is:
\[ a = g \sin \alpha - \mu g \cos \alpha \]

---