A body just slides a rough plane inclined at an angle of `30^(@)` with the horizontal. Calculate the acceleration with which the body will slide down when the inclination in the plane is changed to `45^(@)`. Take `g=9.8ms^(-2)`

Answer

To solve the problem step by step, we will first determine the coefficient of friction using the angle of repose and then calculate the acceleration when the angle of inclination is changed to 45 degrees.

### Step 1: Determine the Coefficient of Friction (μ)
The angle of repose is given as \(30^\circ\). The coefficient of friction (μ) can be calculated using the tangent of the angle of repose:

\[ \mu = \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \]

### Step 2: Set Up the Forces Acting on the Body at \(45^\circ\)
When the body is on the inclined plane at \(45^\circ\), the forces acting on it are:
- Gravitational force acting down the incline: \(F_{\text{gravity}} = mg \sin(45^\circ)\)
- Normal force acting perpendicular to the incline: \(N = mg \cos(45^\circ)\)
- Frictional force acting up the incline: \(F_{\text{friction}} = \mu N = \mu mg \cos(45^\circ)\)

### Step 3: Write the Equation of Motion
The net force acting on the body along the incline can be expressed as:

\[ F_{\text{net}} = F_{\text{gravity}} - F_{\text{friction}} \]

Substituting the expressions for the forces:

\[ F_{\text{net}} = mg \sin(45^\circ) - \mu mg \cos(45^\circ) \]

### Step 4: Simplify the Equation
Since \(m\) appears in every term, we can cancel it out:

\[ F_{\text{net}} = g \sin(45^\circ) - \mu g \cos(45^\circ) \]

Now substituting the values of \(\sin(45^\circ)\) and \(\cos(45^\circ)\):

\[ F_{\text{net}} = g \left(\frac{1}{\sqrt{2}}\right) - \mu g \left(\frac{1}{\sqrt{2}}\right) \]

Factoring out \(g\):

\[ F_{\text{net}} = g \left(\frac{1}{\sqrt{2}} - \mu \frac{1}{\sqrt{2}}\right) = g \frac{1 - \mu}{\sqrt{2}} \]

### Step 5: Substitute the Values
Now substituting \(g = 9.8 \, \text{m/s}^2\) and \(\mu = \frac{1}{\sqrt{3}}\):

\[ F_{\text{net}} = 9.8 \left(\frac{1 - \frac{1}{\sqrt{3}}}{\sqrt{2}}\right) \]

Calculating \(1 - \frac{1}{\sqrt{3}}\):

\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]

Now substituting this back into the equation:

\[ F_{\text{net}} = 9.8 \cdot \frac{\sqrt{3} - 1}{\sqrt{3} \sqrt{2}} \]

### Step 6: Calculate the Acceleration
The acceleration \(a\) can be calculated as:

\[ a = \frac{F_{\text{net}}}{m} = \frac{g \left(1 - \mu\right)}{\sqrt{2}} \]

Substituting the values:

\[ a = 9.8 \cdot \frac{1 - \frac{1}{\sqrt{3}}}{\sqrt{2}} \approx 9.8 \cdot \frac{0.577}{1.414} \approx 9.8 \cdot 0.408 \approx 4.0 \, \text{m/s}^2 \]

### Final Answer
The acceleration with which the body will slide down the plane when the inclination is changed to \(45^\circ\) is approximately:

\[ \boxed{4.0 \, \text{m/s}^2} \]