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. (a) `mu_k=sqrt(1-1/n^2)`
B. (b) `mu_k=1-1/n^2`
C. (c) `mu_s=sqrt(1-1/n^2)`
D. (d) `mu_s=1-1/n^2`

Explanation

To solve the problem, we will analyze the motion of the block on both the smooth incline and the rough incline. We will derive the expressions for time taken on both surfaces and then find the coefficient of friction.

### Step 1: Analyze the motion on the smooth incline
1. The block is released from rest on a smooth incline at an angle of \(45^\circ\).
2. The only force acting on the block along the incline is the component of gravitational force, which can be calculated as:
\[ F = mg \sin(45^\circ) = mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}} \]
3. The acceleration \(a\) of the block down the incline is given by Newton's second law:
\[ a = \frac{F}{m} = \frac{mg/\sqrt{2}}{m} = \frac{g}{\sqrt{2}} \]

### Step 2: Calculate the time taken on the smooth incline
1. Using the equation of motion \(d = \frac{1}{2} a t^2\), we can solve for time \(t\):
\[ d = \frac{1}{2} \cdot \frac{g}{\sqrt{2}} \cdot t^2 \]
Rearranging gives:
\[ t^2 = \frac{2d\sqrt{2}}{g} \]
Therefore, the time \(t_s\) taken on the smooth incline is:
\[ t_s = \sqrt{\frac{2d\sqrt{2}}{g}} \]

### Step 3: Analyze the motion on the rough incline
1. On the rough incline, the frictional force will oppose the motion. The frictional force \(F_f\) is given by:
\[ F_f = \mu mg \cos(45^\circ) = \mu mg \cdot \frac{1}{\sqrt{2}} = \frac{\mu mg}{\sqrt{2}} \]
2. The net force \(F_{net}\) acting on the block down the incline is:
\[ F_{net} = mg \sin(45^\circ) - F_f = \frac{mg}{\sqrt{2}} - \frac{\mu mg}{\sqrt{2}} = \frac{mg(1 - \mu)}{\sqrt{2}} \]
3. The acceleration \(a_r\) on the rough incline is:
\[ a_r = \frac{F_{net}}{m} = \frac{g(1 - \mu)}{\sqrt{2}} \]

### Step 4: Calculate the time taken on the rough incline
1. Using the same equation of motion \(d = \frac{1}{2} a_r t^2\):
\[ d = \frac{1}{2} \cdot \frac{g(1 - \mu)}{\sqrt{2}} \cdot t_r^2 \]
Rearranging gives:
\[ t_r^2 = \frac{2d\sqrt{2}}{g(1 - \mu)} \]
Therefore, the time \(t_r\) taken on the rough incline is:
\[ t_r = \sqrt{\frac{2d\sqrt{2}}{g(1 - \mu)}} \]

### Step 5: Relate the times taken on both inclines
1. According to the problem, the time taken on the rough incline is \(n\) times the time taken on the smooth incline:
\[ t_r = n t_s \]
Substituting the expressions for \(t_r\) and \(t_s\):
\[ \sqrt{\frac{2d\sqrt{2}}{g(1 - \mu)}} = n \sqrt{\frac{2d\sqrt{2}}{g}} \]

### Step 6: Simplify and solve for the coefficient of friction \(\mu\)
1. Squaring both sides:
\[ \frac{2d\sqrt{2}}{g(1 - \mu)} = n^2 \cdot \frac{2d\sqrt{2}}{g} \]
2. Canceling \(2d\sqrt{2}/g\) from both sides:
\[ \frac{1}{1 - \mu} = n^2 \]
3. Rearranging gives:
\[ 1 - \mu = \frac{1}{n^2} \]
Therefore:
\[ \mu = 1 - \frac{1}{n^2} \]

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