The coefficient of friction between a body of mass 1kg and the surface of an inclined plane at `45^(@)` is 0.5. if `g=9.8m//s^(2)`, the acceleration of the body downwards in `m//s^(2)` is

Options

A. `(4.9)/sqrt(2)`
B. `4.9sqrt(2)`
C. `19.6sqrt(2)`
D. `4.9`

Explanation

To solve the problem, we need to determine the acceleration of a body of mass 1 kg sliding down an inclined plane at an angle of 45° with a coefficient of friction of 0.5. We will use Newton's laws of motion and the concepts of forces acting on the body.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- The weight of the body (W) acts vertically downwards:
\[ W = mg = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]
- This weight can be resolved into two components:
- Perpendicular to the incline: \( W_{\perp} = mg \cos(45^\circ) = \frac{9.8}{\sqrt{2}} \, \text{N} \)
- Parallel to the incline: \( W_{\parallel} = mg \sin(45^\circ) = \frac{9.8}{\sqrt{2}} \, \text{N} \)

2. **Calculate the Normal Force (N)**:
- The normal force (N) is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = mg \cos(45^\circ) = \frac{9.8}{\sqrt{2}} \, \text{N} \]

3. **Calculate the Frictional Force (F_f)**:
- The frictional force (F_f) opposing the motion is given by:
\[ F_f = \mu N = 0.5 \times N = 0.5 \times \frac{9.8}{\sqrt{2}} = \frac{9.8}{2\sqrt{2}} \, \text{N} \]

4. **Determine the Net Force (F_net)**:
- The net force acting on the body down the incline is the difference between the parallel component of the weight and the frictional force:
\[ F_{\text{net}} = W_{\parallel} - F_f = \frac{9.8}{\sqrt{2}} - \frac{9.8}{2\sqrt{2}} \]
- Simplifying this:
\[ F_{\text{net}} = \frac{9.8}{\sqrt{2}} \left(1 - \frac{1}{2}\right) = \frac{9.8}{\sqrt{2}} \times \frac{1}{2} = \frac{9.8}{2\sqrt{2}} \, \text{N} \]

5. **Calculate the Acceleration (a)**:
- According to Newton's second law, \( F_{\text{net}} = ma \):
\[ a = \frac{F_{\text{net}}}{m} = \frac{\frac{9.8}{2\sqrt{2}}}{1} = \frac{9.8}{2\sqrt{2}} \, \text{m/s}^2 \]
- Evaluating this gives:
\[ a = \frac{9.8}{2 \times 1.414} \approx \frac{9.8}{2.828} \approx 3.464 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the body downwards is approximately \( 3.464 \, \text{m/s}^2 \).

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