A given object taken n time more time to slide down `45^(@)` rough inclined plane as it taken to slide down a perfectly smooth `45^(@)` incline The coefficient of kintic friction between the object and the incline is .

Options

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

Answer

To solve the problem, we need to analyze the motion of an object sliding down two different inclined planes: one smooth and one rough. We'll derive the coefficient of kinetic friction based on the given information.

### Step-by-Step Solution:

1. **Define Variables**:
- Let the length of the incline be \( s \).
- Let the time taken to slide down the smooth incline be \( t \).
- The time taken to slide down the rough incline is \( n \cdot t \).
- The angle of inclination \( \theta = 45^\circ \).

2. **Acceleration on Smooth Incline**:
- For the smooth incline, the only force acting along the incline is the component of gravitational force.
- The acceleration \( a \) on the smooth incline is given by:
\[ a = g \sin \theta = g \sin 45^\circ = \frac{g}{\sqrt{2}} \]
- Using the second equation of motion \( s = ut + \frac{1}{2} a t^2 \) (where \( u = 0 \)):
\[ s = 0 + \frac{1}{2} \left(\frac{g}{\sqrt{2}}\right) t^2 \]
\[ s = \frac{g}{2\sqrt{2}} t^2 \quad \text{(Equation 1)} \]

3. **Acceleration on Rough Incline**:
- For the rough incline, the forces acting on the object include the gravitational component and the frictional force.
- The net acceleration \( a' \) is given by:
\[ a' = g \sin \theta - \mu g \cos \theta = g \sin 45^\circ - \mu g \cos 45^\circ = \frac{g}{\sqrt{2}} - \mu \frac{g}{\sqrt{2}} = \frac{g(1 - \mu)}{\sqrt{2}} \]
- The time taken to slide down the rough incline is \( n \cdot t \):
\[ s = 0 + \frac{1}{2} a' (n t)^2 \]
\[ s = \frac{1}{2} \left(\frac{g(1 - \mu)}{\sqrt{2}}\right) (n t)^2 \]
\[ s = \frac{g(1 - \mu)}{2\sqrt{2}} n^2 t^2 \quad \text{(Equation 2)} \]

4. **Equating the Two Equations**:
- Since both expressions represent the same distance \( s \), we can set Equation 1 equal to Equation 2:
\[ \frac{g}{2\sqrt{2}} t^2 = \frac{g(1 - \mu)}{2\sqrt{2}} n^2 t^2 \]
- We can cancel \( \frac{g}{2\sqrt{2}} t^2 \) from both sides (assuming \( t \neq 0 \)):
\[ 1 = (1 - \mu) n^2 \]
- Rearranging gives:
\[ 1 - \mu = \frac{1}{n^2} \]
\[ \mu = 1 - \frac{1}{n^2} \]

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