Acceleration of block parallel to plane (if `M = 2m, theta = 45, mu = 0.5)`

Options

A. `(5g)/(9sqrt2)`
B. `(5g)/(18sqrt2)`
C. `(5g)/(3sqrt2)`
D. `(5g)/(8sqrt2)`

Answer

To solve the problem of finding the acceleration of a block on an inclined plane, we can follow these steps:

### Step 1: Understand the Problem
We have a block of mass \( M = 2m \) on an inclined plane with an angle \( \theta = 45^\circ \) and a coefficient of friction \( \mu = 0.5 \). We need to find the acceleration of the block parallel to the plane.

### Step 2: Draw the Free Body Diagram
1. Draw the inclined plane and the block on it.
2. Identify the forces acting on the block:
- Weight acting downwards: \( W = 2mg \)
- Normal force \( N \) acting perpendicular to the inclined plane
- Frictional force \( f \) acting parallel to the plane (upwards, opposing motion)

### Step 3: Resolve Forces
1. Resolve the weight into components:
- Parallel to the plane: \( W_{\parallel} = 2mg \sin(45^\circ) = 2mg \cdot \frac{1}{\sqrt{2}} = \frac{2mg}{\sqrt{2}} \)
- Perpendicular to the plane: \( W_{\perpendicular} = 2mg \cos(45^\circ) = 2mg \cdot \frac{1}{\sqrt{2}} = \frac{2mg}{\sqrt{2}} \)

### Step 4: Calculate Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = W_{\perpendicular} = \frac{2mg}{\sqrt{2}} \]

### Step 5: Calculate Maximum Static Friction
The maximum static friction \( f_s \) can be calculated using:
\[ f_s = \mu N = \mu \left(\frac{2mg}{\sqrt{2}}\right) = 0.5 \cdot \frac{2mg}{\sqrt{2}} = \frac{mg}{\sqrt{2}} \]

### Step 6: Calculate Net Force
The net force \( F_{\text{net}} \) acting on the block parallel to the incline is given by:
\[ F_{\text{net}} = W_{\parallel} - f_s \]
Substituting the values:
\[ F_{\text{net}} = \frac{2mg}{\sqrt{2}} - \frac{mg}{\sqrt{2}} = \frac{mg}{\sqrt{2}} \]

### Step 7: Calculate Acceleration
Using Newton's second law, \( F = ma \):
\[ a = \frac{F_{\text{net}}}{\text{mass}} = \frac{\frac{mg}{\sqrt{2}}}{2m} \]
This simplifies to:
\[ a = \frac{g}{2\sqrt{2}} \]

### Final Answer
The acceleration of the block parallel to the plane is:
\[ a = \frac{g}{2\sqrt{2}} \]

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