A block of mass m slides from the top of an inclined plane of angle of inclination `theta` & length `l`. The coefficient of friction between the plane and the block is `mu`. Then it is observed over a distance d along the horizontal surface having the same coefficient of friction `mu`, before it comes to a stop. Find the value of d.

Answer

To solve the problem step by step, we will apply the principles of physics, particularly the work-energy theorem, to find the distance \( d \) that the block travels on the horizontal surface before coming to a stop.

### Step 1: Analyze the motion on the inclined plane

1. **Identify the forces acting on the block on the inclined plane**:
- The gravitational force acting downwards: \( mg \)
- The normal force acting perpendicular to the surface: \( N = mg \cos \theta \)
- The frictional force acting opposite to the direction of motion: \( f = \mu N = \mu mg \cos \theta \)

2. **Calculate the net force acting on the block**:
- The component of gravitational force acting down the incline: \( F_{\text{gravity}} = mg \sin \theta \)
- The net force acting on the block:
\[ F_{\text{net}} = mg \sin \theta - f = mg \sin \theta - \mu mg \cos \theta \]

### Step 2: Apply the work-energy theorem

3. **Set up the work-energy theorem**:
- The work done by all forces is equal to the change in kinetic energy:
\[ W_{\text{gravity}} + W_{\text{friction}} = \Delta KE \]
- Since the block starts from rest and comes to rest, the change in kinetic energy is zero:
\[ W_{\text{gravity}} + W_{\text{friction}} = 0 \]

4. **Calculate the work done by gravity**:
- The work done by gravity as the block moves down the incline is:
\[ W_{\text{gravity}} = (mg \sin \theta) \cdot L \]

5. **Calculate the work done by friction**:
- The work done by friction while the block moves down the incline is:
\[ W_{\text{friction}} = -(\mu mg \cos \theta) \cdot L \]

### Step 3: Set up the equation

6. **Combine the work done expressions**:
\[ mg \sin \theta \cdot L - \mu mg \cos \theta \cdot L = 0 \]

7. **Factor out common terms**:
\[ mgL (\sin \theta - \mu \cos \theta) = 0 \]

### Step 4: Analyze motion on the horizontal surface

8. **Now consider the block's motion on the horizontal surface**:
- The frictional force acting on the block is the same:
\[ f = \mu mg \]
- The block will decelerate due to this frictional force until it comes to rest.

9. **Use the work-energy principle again**:
- The work done by friction on the horizontal surface is equal to the initial kinetic energy of the block when it reaches the horizontal surface:
\[ W_{\text{friction}} = -f \cdot d = -\mu mg \cdot d \]

### Step 5: Relate the two parts of the motion

10. **Set the work done by friction equal to the kinetic energy at the bottom of the incline**:
- The kinetic energy at the bottom of the incline can be calculated using the work done by gravity:
\[ \frac{1}{2} mv^2 = mgL (\sin \theta - \mu \cos \theta) \]

11. **Equate the two expressions**:
\[ \mu mg d = \frac{1}{2} mv^2 \]

### Step 6: Solve for \( d \)

12. **Substituting the expression for \( v^2 \)**:
\[ d = \frac{v^2}{2\mu g} \]

### Step 7: Final expression for \( d \)

13. **Substituting \( v^2 \) from the work done**:
\[ d = \frac{L(\sin \theta - \mu \cos \theta)}{2\mu} \]

### Final Result
Thus, the distance \( d \) that the block travels on the horizontal surface before coming to a stop is given by:
\[ d = \frac{L(\sin \theta - \mu \cos \theta)}{2\mu} \]