A smooth block is released at rest on a `45^(@)` incline and then slides a distance `d`. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

Options

A. `mu_(k) = 1- ( 1)/( n^(2))`
B. `mu_(k) = sqrt( 1 - ( 1)/ ( n^(2)))`
C. `mu_(s) = 1- ( 1)/( n^(2))`
D. `mu_(s) = sqrt( 1 - ( 1)/ ( n^(2)))`

Answer

To solve the problem, we need to analyze the motion of a block sliding down a 45-degree incline under two conditions: one is a smooth incline (no friction), and the other is a rough incline (with friction).

### Step-by-Step Solution:

1. **Identify the Forces on the Smooth Incline:**
- When the block is on a smooth incline, the only force acting along the incline is the component of the gravitational force.
- The gravitational force acting on the block is \( mg \).
- The component of this force acting down the incline is \( mg \sin(45^\circ) \).
- Thus, the acceleration \( a_1 \) of the block on the smooth incline is given by:
\[ a_1 = g \sin(45^\circ) = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

2. **Identify the Forces on the Rough Incline:**
- On the rough incline, there is also a frictional force acting against the motion of the block.
- The frictional force \( F_f \) is given by \( F_f = \mu N \), where \( N = mg \cos(45^\circ) \).
- Thus, \( F_f = \mu mg \cdot \frac{1}{\sqrt{2}} \).
- The net force acting down the incline is:
\[ F_{\text{net}} = mg \sin(45^\circ) - F_f = mg \sin(45^\circ) - \mu mg \cos(45^\circ) \]
- Therefore, the acceleration \( a_2 \) on the rough incline is:
\[ a_2 = g \sin(45^\circ) - \mu g \cos(45^\circ) = g \cdot \frac{1}{\sqrt{2}} - \mu g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1 - \mu) \]

3. **Relate the Distances and Times:**
- The distance \( d \) covered by the block is the same in both cases.
- The time taken to slide down the smooth incline is \( t \), and the time taken to slide down the rough incline is \( nt \).
- Using the equation of motion \( d = \frac{1}{2} a t^2 \), we can write:
\[ d = \frac{1}{2} a_1 t^2 \quad \text{(for smooth incline)} \]
\[ d = \frac{1}{2} a_2 (nt)^2 \quad \text{(for rough incline)} \]

4. **Set the Distances Equal:**
- Since both expressions equal \( d \), we can set them equal to each other:
\[ \frac{1}{2} a_1 t^2 = \frac{1}{2} a_2 (nt)^2 \]
- Simplifying gives:
\[ a_1 t^2 = a_2 n^2 t^2 \]
- Dividing both sides by \( t^2 \) (assuming \( t \neq 0 \)):
\[ a_1 = a_2 n^2 \]

5. **Substitute the Values of \( a_1 \) and \( a_2 \):**
- Substituting the expressions for \( a_1 \) and \( a_2 \):
\[ \frac{g}{\sqrt{2}} = \left(\frac{g}{\sqrt{2}}(1 - \mu)\right) n^2 \]
- Cancelling \( \frac{g}{\sqrt{2}} \) from both sides (assuming \( g \neq 0 \)):
\[ 1 = (1 - \mu) n^2 \]

6. **Solve for the Coefficient of Friction \( \mu \):**
- Rearranging gives:
\[ 1 - \mu = \frac{1}{n^2} \]
\[ \mu = 1 - \frac{1}{n^2} \]

### Final Answer:
The coefficient of friction \( \mu \) is:
\[ \mu = 1 - \frac{1}{n^2} \]