Starting from rest, the time taken by a body sliding down on a rough inclined plane at `45^(@)` with the horizontal is twice the time taken to travel on a smooth plane of same inclination and same distance. Then, the coefficient of kinetic friction is

Options

A. 0.25
B. 0.33
C. 0.5
D. 0.75

Explanation

To solve the problem, we need to analyze the motion of a body sliding down a rough inclined plane and compare it to a smooth inclined plane. Let's break down the solution step by step:

### Step 1: Understand the motion on the smooth inclined plane

1. **Given Information**: The body starts from rest and slides down a smooth inclined plane at an angle of \(45^\circ\).
2. **Acceleration on Smooth Plane**: The only force acting along the incline is the component of gravitational force:
\[ a = g \sin \theta \]
For \( \theta = 45^\circ \), \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), so:
\[ a = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

3. **Distance and Time Relation**: Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \):
\[ s = \frac{1}{2} a t^2 = \frac{1}{2} \cdot \frac{g}{\sqrt{2}} \cdot t^2 \]
Rearranging gives:
\[ t^2 = \frac{2s \sqrt{2}}{g} \]

### Step 2: Understand the motion on the rough inclined plane

1. **Forces Acting on the Body**: On the rough inclined plane, the forces acting are:
- Gravitational force component down the incline: \( mg \sin \theta \)
- Normal force: \( mg \cos \theta \)
- Frictional force: \( f = \mu mg \cos \theta \)

2. **Net Force and Acceleration**: The net force acting down the incline is:
\[ F_{\text{net}} = mg \sin \theta - f = mg \sin \theta - \mu mg \cos \theta \]
Thus, the net acceleration \( a' \) is:
\[ ma' = mg \sin \theta - \mu mg \cos \theta \]
Simplifying gives:
\[ a' = g \sin \theta - \mu g \cos \theta \]
For \( \theta = 45^\circ \):
\[ a' = g \cdot \frac{1}{\sqrt{2}} - \mu g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1 - \mu) \]

3. **Distance and Time Relation**: The time taken to slide down the rough plane is twice that of the smooth plane, so:
\[ 2t = \frac{1}{2} a' (2t)^2 \]
This gives:
\[ s = \frac{1}{2} a' (4t^2) = 2 a' t^2 \]
Substituting \( a' \):
\[ s = 2 \cdot \frac{g}{\sqrt{2}}(1 - \mu) t^2 \]

### Step 3: Equating the two distances

1. **Equating Distances**:
\[ \frac{2s \sqrt{2}}{g} = 2 \cdot \frac{g}{\sqrt{2}}(1 - \mu) t^2 \]
Canceling \( s \) and rearranging gives:
\[ 1 = (1 - \mu) \cdot \frac{4}{2} = 2(1 - \mu) \]
Thus:
\[ 1 = 2 - 2\mu \implies 2\mu = 1 \implies \mu = 0.5 \]

### Final Answer
The coefficient of kinetic friction \( \mu \) is \( 0.5 \).

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