A given object takes n times as much time to slide down a `45^(@)` rough incline as it takes to slide down a perfectly smooth `45^(@)` incline. The coefficeint of kinetic friction between the object and the incline is given by:

Options

A. `(1-(1)/(n^(2)))`
B. `((1)/(1-n^(2)))`
C. `sqrt((1-(1)/(n^(2))))`
D. `sqrt(((1)/(1-n^(2))))`

Explanation

To solve the problem of finding the coefficient of kinetic friction (μ) between an object and a rough incline, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have an object sliding down a 45-degree incline.
- The time taken to slide down the rough incline is n times the time taken to slide down a smooth incline.

2. **Forces on the Smooth Incline**:
- On a smooth incline, the only force acting along the incline is the component of gravitational force.
- The gravitational force acting on the object is \( mg \), and its component along the incline is \( mg \sin(45^\circ) = mg \frac{\sqrt{2}}{2} \).
- Using Newton's second law, the acceleration \( a_1 \) of the object on the smooth incline is given by:
\[ a_1 = g \sin(45^\circ) = g \frac{\sqrt{2}}{2} \]

3. **Forces on the Rough Incline**:
- On the rough incline, the forces acting on the object include the gravitational force and the frictional force.
- The frictional force can be expressed as \( f = \mu N \), where \( N = mg \cos(45^\circ) = mg \frac{\sqrt{2}}{2} \).
- The net force acting down the incline is:
\[ F_{\text{net}} = mg \sin(45^\circ) - f = mg \sin(45^\circ) - \mu mg \cos(45^\circ) \]
- This gives us the equation:
\[ mg \sin(45^\circ) - \mu mg \cos(45^\circ) = ma_2 \]
- Dividing through by \( m \) and substituting \( \sin(45^\circ) \) and \( \cos(45^\circ) \):
\[ g \frac{\sqrt{2}}{2} - \mu g \frac{\sqrt{2}}{2} = a_2 \]
- Simplifying, we find:
\[ a_2 = g \frac{\sqrt{2}}{2} (1 - \mu) \]

4. **Relating Time and Distance**:
- The distance \( L \) covered by the object is the same for both inclines.
- Using the equation of motion \( s = ut + \frac{1}{2} a t^2 \) and noting that the initial velocity \( u = 0 \):
- For the smooth incline:
\[ L = \frac{1}{2} a_1 t_1^2 = \frac{1}{2} g \frac{\sqrt{2}}{2} t_1^2 \]
- For the rough incline:
\[ L = \frac{1}{2} a_2 t_2^2 = \frac{1}{2} g \frac{\sqrt{2}}{2} (1 - \mu) t_2^2 \]

5. **Using the Time Relationship**:
- We know \( t_2 = n t_1 \). Substituting this into the equation for the rough incline:
\[ L = \frac{1}{2} g \frac{\sqrt{2}}{2} (1 - \mu) (n t_1)^2 \]
- Setting the two expressions for \( L \) equal to each other:
\[ \frac{1}{2} g \frac{\sqrt{2}}{2} t_1^2 = \frac{1}{2} g \frac{\sqrt{2}}{2} (1 - \mu) n^2 t_1^2 \]

6. **Cancelling Common Terms**:
- Cancelling \( \frac{1}{2} g \frac{\sqrt{2}}{2} t_1^2 \) from both sides (assuming \( t_1 \neq 0 \)):
\[ 1 = (1 - \mu) n^2 \]
- Rearranging gives:
\[ 1 - n^2 = -\mu n^2 \implies \mu = 1 - \frac{1}{n^2} \]

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