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

Explanation

To solve the problem step by step, we will analyze the motion of a block sliding down a smooth incline and then a rough incline, and relate the times taken for both cases.

### Step 1: Analyze the smooth incline
When the block is released on a smooth incline at an angle of \(45^\circ\), the only force acting on it along the incline is the component of gravitational force.

The gravitational force acting on the block can be resolved into two components:
- Parallel to the incline: \(F_{\parallel} = mg \sin \theta\)
- Perpendicular to the incline: \(F_{\perpendicular} = mg \cos \theta\)

For a \(45^\circ\) incline, \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\).

Thus, the acceleration \(A_1\) down the incline is given by:
\[ A_1 = g \sin 45^\circ = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

### Step 2: Analyze the rough incline
On a rough incline, there is also a frictional force acting on the block. The frictional force can be expressed as:
\[ F_{\text{friction}} = \mu mg \cos \theta \]

The net force acting on the block down the incline is now:
\[ F_{\text{net}} = mg \sin \theta - F_{\text{friction}} = mg \sin \theta - \mu mg \cos \theta \]

Substituting for \(\sin 45^\circ\) and \(\cos 45^\circ\):
\[ F_{\text{net}} = mg \cdot \frac{1}{\sqrt{2}} - \mu mg \cdot \frac{1}{\sqrt{2}} = mg \cdot \frac{1 - \mu}{\sqrt{2}} \]

The acceleration \(A_2\) down the rough incline is given by:
\[ A_2 = \frac{F_{\text{net}}}{m} = \frac{g(1 - \mu)}{\sqrt{2}} \]

### Step 3: Relate the times taken to slide down both inclines
Let \(t\) be the time taken to slide down the smooth incline and \(nt\) be the time taken to slide down the rough incline. The distance \(d\) is the same for both cases.

Using the equation of motion:
\[ d = \frac{1}{2} A_1 t^2 \quad \text{(for smooth incline)} \]
\[ d = \frac{1}{2} A_2 (nt)^2 \quad \text{(for rough incline)} \]

Substituting the expressions for \(A_1\) and \(A_2\):
\[ d = \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 \]
\[ d = \frac{1}{2} \left(\frac{g(1 - \mu)}{\sqrt{2}}\right) (nt)^2 \]

### Step 4: Set the equations equal to each other
Since both expressions equal \(d\), we can set them equal:
\[ \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 = \frac{1}{2} \left(\frac{g(1 - \mu)}{\sqrt{2}}\right) (nt)^2 \]

Canceling common terms:
\[ \frac{g}{\sqrt{2}} t^2 = \frac{g(1 - \mu)}{\sqrt{2}} n^2 t^2 \]

### Step 5: Simplify and solve for \(\mu\)
Dividing both sides by \(\frac{g}{\sqrt{2}} t^2\) (assuming \(t \neq 0\)):
\[ 1 = (1 - \mu) n^2 \]

Rearranging gives:
\[ 1 - \mu = \frac{1}{n^2} \]
\[ \mu = 1 - \frac{1}{n^2} \]

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