Starting from rest a body slides down a `45^(@)` inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of the body and the inclined plane is :

Options

A. 0.75
B. 0.33
C. 0.25
D. `0.80`

Explanation

To solve the problem, we need to analyze the motion of a body sliding down a 45-degree inclined plane with and without friction. Let's break it down step by step.

### Step 1: Understand the Forces Acting on the Body
When the body is sliding down the incline, two main forces act on it:
1. The gravitational force component acting down the incline: \( F_{\text{gravity}} = mg \sin \theta \)
2. The frictional force acting up the incline: \( F_{\text{friction}} = \mu N = \mu mg \cos \theta \)

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

### Step 2: Write the Equation of Motion
The net force acting on the body when it is sliding down the incline with friction is given by:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
Thus, the net acceleration \( a \) of the body can be expressed as:
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
Dividing through by \( m \):
\[ a = g \sin \theta - \mu g \cos \theta \]

### Step 3: Calculate the Acceleration with and without Friction
1. **Without Friction (Case 1)**:
\[ a_1 = g \sin 45^\circ = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

2. **With Friction (Case 2)**:
\[ a_2 = g \sin 45^\circ - \mu g \cos 45^\circ = \frac{g}{\sqrt{2}} - \mu \cdot \frac{g}{\sqrt{2}} = \frac{g}{\sqrt{2}}(1 - \mu) \]

### Step 4: Relate the Distances and Times
The distance \( d \) traveled down the incline can be expressed in terms of time \( t \) using the equation:
\[ d = \frac{1}{2} a t^2 \]

For the case without friction, the distance is:
\[ d = \frac{1}{2} a_1 t^2 = \frac{1}{2} \cdot \frac{g}{\sqrt{2}} t^2 \]

For the case with friction, since it takes twice the time \( 2t \):
\[ d = \frac{1}{2} a_2 (2t)^2 = \frac{1}{2} a_2 \cdot 4t^2 = 2 a_2 t^2 \]

### Step 5: Set the Distances Equal
Since the distance \( d \) is the same in both cases:
\[ \frac{1}{2} \cdot \frac{g}{\sqrt{2}} t^2 = 2 a_2 t^2 \]
Cancelling \( t^2 \) from both sides (assuming \( t \neq 0 \)):
\[ \frac{g}{2\sqrt{2}} = 2 a_2 \]
Substituting \( a_2 \):
\[ \frac{g}{2\sqrt{2}} = 2 \cdot \frac{g}{\sqrt{2}}(1 - \mu) \]
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{1}{2\sqrt{2}} = 2 \cdot \frac{1}{\sqrt{2}}(1 - \mu) \]

### Step 6: Simplify and Solve for \( \mu \)
\[ \frac{1}{2\sqrt{2}} = \frac{2(1 - \mu)}{\sqrt{2}} \]
Multiplying both sides by \( 2\sqrt{2} \):
\[ 1 = 4(1 - \mu) \]
Solving for \( \mu \):
\[ 1 = 4 - 4\mu \implies 4\mu = 4 - 1 \implies 4\mu = 3 \implies \mu = \frac{3}{4} = 0.75 \]

### Final Answer
The coefficient of friction \( \mu \) between the body and the inclined plane is \( 0.75 \).

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