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

Options

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

Answer

To solve the problem, we need to determine the acceleration of a body sliding down an inclined plane at an angle of \(45^\circ\) with a coefficient of friction of \(0.5\). We will use the following steps:

### Step 1: Identify the forces acting on the body
The forces acting on the body are:
- The gravitational force acting downwards, \(mg\).
- The component of gravitational force acting parallel to the incline, \(mg \sin \theta\).
- The component of gravitational force acting perpendicular to the incline, \(mg \cos \theta\).
- The frictional force acting opposite to the direction of motion, which is given by \(f_k = \mu N\), where \(N\) is the normal force.

### Step 2: Calculate the components of the gravitational force
For an incline of \(45^\circ\):
- \(mg \sin 45^\circ = mg \cdot \frac{\sqrt{2}}{2}\)
- \(mg \cos 45^\circ = mg \cdot \frac{\sqrt{2}}{2}\)

### Step 3: Calculate the normal force
The normal force \(N\) is equal to the component of the gravitational force acting perpendicular to the incline:
\[ N = mg \cos 45^\circ = mg \cdot \frac{\sqrt{2}}{2} \]

### Step 4: Calculate the frictional force
The frictional force \(f_k\) can be calculated using the coefficient of friction:
\[ f_k = \mu N = 0.5 \cdot \left(mg \cdot \frac{\sqrt{2}}{2}\right) = \frac{0.5mg\sqrt{2}}{2} = \frac{0.5mg\sqrt{2}}{2} = \frac{mg\sqrt{2}}{4} \]

### Step 5: Set up the equation of motion
The net force \(F_{net}\) acting on the body in the direction of motion is given by:
\[ F_{net} = mg \sin 45^\circ - f_k \]
Substituting the values we calculated:
\[ F_{net} = mg \cdot \frac{\sqrt{2}}{2} - \frac{mg\sqrt{2}}{4} \]

### Step 6: Simplify the equation
Factoring out \(mg\):
\[ F_{net} = mg\left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{4}\right) \]
To combine the terms:
\[ F_{net} = mg\left(\frac{2\sqrt{2}}{4} - \frac{\sqrt{2}}{4}\right) = mg\left(\frac{2\sqrt{2} - \sqrt{2}}{4}\right) = mg\left(\frac{\sqrt{2}}{4}\right) \]

### Step 7: Apply Newton's second law
According to Newton's second law:
\[ F_{net} = ma \]
So we can set up the equation:
\[ mg\left(\frac{\sqrt{2}}{4}\right) = ma \]

### Step 8: Solve for acceleration \(a\)
Dividing both sides by \(m\):
\[ g\left(\frac{\sqrt{2}}{4}\right) = a \]
Substituting \(g = 9.8 \, \text{m/s}^2\):
\[ a = 9.8 \cdot \frac{\sqrt{2}}{4} \]

### Step 9: Calculate the final value
Calculating the numerical value:
\[ a = \frac{9.8\sqrt{2}}{4} \]
To simplify:
\[ a = \frac{9.8 \cdot 1.414}{4} \approx \frac{13.86}{4} \approx 3.465 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the body downwards is approximately \(3.465 \, \text{m/s}^2\).