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, we need to analyze the motion of the block on the inclined plane and then on the horizontal surface.

### Step-by-Step Solution:

1. **Identify Forces on the Inclined Plane:**
- The weight of the block is \( mg \).
- The component of gravitational force acting down the incline is \( mg \sin \theta \).
- The normal force acting on the block is \( N = mg \cos \theta \).
- The frictional force opposing the motion is \( F_{\text{friction}} = \mu N = \mu mg \cos \theta \).

2. **Net Force and Acceleration on the Inclined Plane:**
- The net force acting on the block along the incline is:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
- According to Newton's second law, this net force equals mass times acceleration:
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
- Dividing through by \( m \):
\[ a = g \sin \theta - \mu g \cos \theta \]

3. **Calculate the Velocity at the Bottom of the Incline:**
- Using the kinematic equation \( v^2 = u^2 + 2as \) where \( u = 0 \) (initial velocity), \( a = g \sin \theta - \mu g \cos \theta \), and \( s = l \):
\[ v^2 = 0 + 2(g \sin \theta - \mu g \cos \theta) l \]
- Therefore, the velocity at the bottom of the incline is:
\[ v^2 = 2l(g \sin \theta - \mu g \cos \theta) \]

4. **Motion on the Horizontal Surface:**
- On the horizontal surface, the only force acting against the motion is the frictional force:
\[ F_{\text{friction}} = \mu mg \]
- The deceleration (retardation) due to friction is:
\[ a' = -\mu g \]

5. **Calculate the Distance \( d \) on the Horizontal Surface:**
- Using the kinematic equation again \( v'^2 = u'^2 + 2a's' \) where \( v' = 0 \) (final velocity), \( u' = v \) (initial velocity from the incline), and \( a' = -\mu g \):
\[ 0 = v^2 + 2(-\mu g)d \]
- Rearranging gives:
\[ d = \frac{v^2}{2\mu g} \]
- Substituting \( v^2 \) from the previous step:
\[ d = \frac{2l(g \sin \theta - \mu g \cos \theta)}{2\mu g} \]
- Simplifying:
\[ d = \frac{l(g \sin \theta - \mu g \cos \theta)}{\mu g} \]
- This can be further simplified to:
\[ d = \frac{l(\sin \theta - \mu \cos \theta)}{\mu} \]

### Final Answer:
\[ d = \frac{l(\sin \theta - \mu \cos \theta)}{\mu} \]