When a body of mass M slides down an inclined plane of inclination `theta`, having coefficient of friction `mu` through a distance s, the work done against friction is :

Options

A. `mu Mg cos theta s`
B. `mu Mg sin theta s`
C. `Mg(mu cos theta-sin theta)s`
D. none of the above

Answer

To solve the problem of calculating the work done against friction when a body of mass \( M \) slides down an inclined plane, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Work Done**:
The work done against friction can be calculated using the formula:
\[ \text{Work Done} = \text{Frictional Force} \times \text{Displacement} \]

2. **Identify the Frictional Force**:
The frictional force \( F_f \) acting on the body can be expressed as:
\[ F_f = \mu \cdot N \]
where \( \mu \) is the coefficient of friction and \( N \) is the normal force.

3. **Calculate the Normal Force**:
For a body on an inclined plane, the normal force \( N \) can be calculated as:
\[ N = M g \cos \theta \]
where \( M \) is the mass of the body, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of inclination.

4. **Substitute the Normal Force into the Frictional Force Equation**:
Now substituting the expression for \( N \) into the equation for frictional force:
\[ F_f = \mu \cdot (M g \cos \theta) = \mu M g \cos \theta \]

5. **Calculate the Work Done Against Friction**:
Now, substituting the expression for \( F_f \) into the work done equation:
\[ \text{Work Done} = F_f \cdot s = (\mu M g \cos \theta) \cdot s \]
Thus, the work done against friction when the body slides down the inclined plane through a distance \( s \) is:
\[ \text{Work Done} = \mu M g \cos \theta \cdot s \]

### Final Answer:
The work done against friction is:
\[ \text{Work Done} = \mu M g \cos \theta \cdot s \]