A block sliding down on a rough `45^(@)` inclined planes has half the velocity it would have been, the inclined plane is smooth. The coefficient of sliding friction between the the block and the inclined plane is

Options

A. `1/4`
B. `3/4`
C. `1/(2sqrt2)`
D. `1/sqrt2`

Explanation

To solve the problem step by step, we need to analyze the forces acting on the block sliding down the inclined plane and derive the coefficient of friction.

### Step 1: Identify the forces acting on the block
When the block is on a rough inclined plane, the forces acting on it are:
- The gravitational force \( mg \) acting vertically downward.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( F_f \) acting opposite to the direction of motion.

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- Parallel to the incline: \( F_{\parallel} = mg \sin \theta \)
- Perpendicular to the incline: \( F_{\perpendicular} = mg \cos \theta \)

Given that the angle \( \theta = 45^\circ \), we have:
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

### Step 3: Write the normal force equation
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta = mg \cdot \frac{1}{\sqrt{2}} \]

### Step 4: Write the frictional force equation
The frictional force \( F_f \) can be expressed as:
\[ F_f = \mu N = \mu \cdot mg \cos \theta = \mu \cdot mg \cdot \frac{1}{\sqrt{2}} \]

### Step 5: Write the net force equation
The net force acting on the block when it is sliding down is given by:
\[ F_{\text{net}} = F_{\parallel} - F_f = mg \sin \theta - F_f \]
Substituting the expressions we derived:
\[ F_{\text{net}} = mg \cdot \frac{1}{\sqrt{2}} - \mu \cdot mg \cdot \frac{1}{\sqrt{2}} \]
Factoring out \( mg \cdot \frac{1}{\sqrt{2}} \):
\[ F_{\text{net}} = mg \cdot \frac{1}{\sqrt{2}} (1 - \mu) \]

### Step 6: Write the acceleration equation
Using Newton's second law, we can relate the net force to the acceleration \( a \):
\[ ma = mg \cdot \frac{1}{\sqrt{2}} (1 - \mu) \]
Dividing both sides by \( m \):
\[ a = g \cdot \frac{1}{\sqrt{2}} (1 - \mu) \]

### Step 7: Calculate the velocity for smooth and rough surfaces
1. **For the smooth surface** (no friction):
- The acceleration is \( a = g \cdot \frac{1}{\sqrt{2}} \).
- Using the kinematic equation \( v^2 = u^2 + 2as \) (where \( u = 0 \)):
\[ v^2 = 2g \cdot \frac{1}{\sqrt{2}} \cdot s = \sqrt{2}gs \]

2. **For the rough surface**:
- The acceleration is \( a = g \cdot \frac{1}{\sqrt{2}} (1 - \mu) \).
- The final velocity is given as \( \frac{v}{2} \):
\[ \left(\frac{v}{2}\right)^2 = 0 + 2 \cdot g \cdot \frac{1}{\sqrt{2}} (1 - \mu) \cdot s \]
\[ \frac{v^2}{4} = 2g \cdot \frac{1}{\sqrt{2}} (1 - \mu) \cdot s \]

### Step 8: Set the equations equal and solve for \( \mu \)
From the two equations:
\[ \sqrt{2}gs = \frac{v^2}{2} \]
Substituting this into the rough surface equation:
\[ \frac{v^2}{4} = 2g \cdot \frac{1}{\sqrt{2}} (1 - \mu) \cdot \frac{v^2}{2\sqrt{2}g} \]
Cancelling \( v^2 \) and simplifying:
\[ \frac{1}{4} = \frac{(1 - \mu)}{2} \]
Multiplying through by 2:
\[ \frac{1}{2} = 1 - \mu \]
Thus, solving for \( \mu \):
\[ \mu = \frac{1}{2} \]

### Final Answer
The coefficient of sliding friction \( \mu \) between the block and the inclined plane is \( \frac{1}{2} \).