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 finding the work done against friction when a body of mass M slides down an inclined plane of inclination θ with a coefficient of friction μ through a distance s, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Body**:
- The weight of the body (mg) acts vertically downward.
- The component of the weight acting parallel to the incline is \( mg \sin \theta \).
- The component of the weight acting perpendicular to the incline is \( mg \cos \theta \).

2. **Determine the Normal Force**:
- The normal force (N) acting on the body is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

3. **Calculate the Frictional Force**:
- The frictional force (f) opposing the motion is given by:
\[ f = \mu N = \mu (mg \cos \theta) \]

4. **Work Done Against Friction**:
- Work done against friction (W) is calculated using the formula:
\[ W = -f \cdot d \cdot \cos(180^\circ) \]
- Since \(\cos(180^\circ) = -1\), the equation simplifies to:
\[ W = f \cdot s \]
- Substituting the expression for the frictional force:
\[ W = \mu (mg \cos \theta) \cdot s \]

5. **Final Expression for Work Done Against Friction**:
- Thus, the work done against friction when the body slides down the incline is:
\[ W = \mu m g \cos \theta \cdot s \]

### Final Answer:
The work done against friction is:
\[ W = \mu M g \cos \theta \cdot s \]

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