A body is projected up along a rough inclined plane of inclination ` 45 ^@ ` . The coefficient of friction is 0.5. Then the retardation of the block is

Options

A. ` ( g ) /( 2 sqrt 2 ) `
B. ` ( g ) /(2 ) `
C. ` ( 3g )/( 2sqrt 2 ) `
D. ` ( g ) /( sqrt 2 ) `

Explanation

To find the retardation of the block projected up along a rough inclined plane with an inclination of \( 45^\circ \) and a coefficient of friction of \( 0.5 \), we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
When the block is projected up the incline, two main forces act on it:
1. The gravitational force (\( mg \)) acting downwards.
2. The frictional force, which opposes the motion of the block.

### Step 2: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

For \( \theta = 45^\circ \):
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

Thus, the components become:
- \( mg \sin 45^\circ = mg \cdot \frac{1}{\sqrt{2}} \)
- \( mg \cos 45^\circ = mg \cdot \frac{1}{\sqrt{2}} \)

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

### Step 4: Calculate the Frictional Force
The frictional force (\( f \)) can be calculated using the coefficient of friction (\( \mu \)):
\[ f = \mu N = \mu (mg \cos \theta) = 0.5 \cdot (mg \cdot \frac{1}{\sqrt{2}}) = \frac{0.5mg}{\sqrt{2}} \]

### Step 5: Determine the Net Force Acting on the Block
The net force (\( F_{\text{net}} \)) acting on the block as it moves up the incline is given by:
\[ F_{\text{net}} = -mg \sin \theta - f \]
Substituting the values:
\[ F_{\text{net}} = -mg \cdot \frac{1}{\sqrt{2}} - \frac{0.5mg}{\sqrt{2}} \]
\[ F_{\text{net}} = -mg \cdot \frac{1 + 0.5}{\sqrt{2}} = -mg \cdot \frac{1.5}{\sqrt{2}} \]

### Step 6: Calculate the Retardation
Using Newton's second law, \( F = ma \), we can find the retardation (\( a \)):
\[ -mg \cdot \frac{1.5}{\sqrt{2}} = ma \]
Dividing both sides by \( m \):
\[ a = -g \cdot \frac{1.5}{\sqrt{2}} \]

### Step 7: Substitute the Value of \( g \)
Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ a = -9.8 \cdot \frac{1.5}{\sqrt{2}} \]
Calculating this gives:
\[ a \approx -9.8 \cdot \frac{1.5}{1.414} \approx -10.4 \, \text{m/s}^2 \]

Thus, the retardation of the block is approximately \( 10.4 \, \text{m/s}^2 \).

### Final Answer
The retardation of the block is \( 10.4 \, \text{m/s}^2 \).

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