A wooden box of mass `8kg` slides down an inclined plane of inclination `30^(@)` to the horizontal with a constant acceleration of `0.4ms^(-2)` What is the force of friction between the box and inclined plane ? `(g = 10m//s^(2))` .

Options

A. 12.2 BN
B. 24.4 N
C. 36.8 N
D. 48.8 N

Explanation

To solve the problem step by step, we will analyze the forces acting on the wooden box as it slides down the inclined plane.

### Step 1: Identify the forces acting on the box
The forces acting on the box are:
1. The gravitational force (weight) acting downwards, \( W = mg \).
2. The normal force (\( N \)) acting perpendicular to the inclined plane.
3. The frictional force (\( F_f \)) acting opposite to the direction of motion (up the incline).

Given:
- Mass of the box, \( m = 8 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Inclination angle, \( \theta = 30^\circ \)
- Acceleration of the box, \( a = 0.4 \, \text{m/s}^2 \)

### Step 2: Calculate the weight of the box
The weight of the box is calculated as:
\[ W = mg = 8 \, \text{kg} \times 10 \, \text{m/s}^2 = 80 \, \text{N} \]

### Step 3: Resolve the weight into components
The weight can be resolved into two components:
1. The component parallel to the incline:
\[ W_{\parallel} = W \sin \theta = 80 \sin 30^\circ = 80 \times \frac{1}{2} = 40 \, \text{N} \]
2. The component perpendicular to the incline:
\[ W_{\perpendicular} = W \cos \theta = 80 \cos 30^\circ = 80 \times \frac{\sqrt{3}}{2} = 40\sqrt{3} \, \text{N} \]

### Step 4: Apply Newton's second law along the incline
According to Newton's second law, the net force acting on the box along the incline is equal to the mass times its acceleration:
\[ \text{Net force} = m \cdot a \]
The net force acting down the incline is given by:
\[ W_{\parallel} - F_f = m \cdot a \]
Substituting the known values:
\[ 40 \, \text{N} - F_f = 8 \, \text{kg} \times 0.4 \, \text{m/s}^2 \]
\[ 40 \, \text{N} - F_f = 3.2 \, \text{N} \]

### Step 5: Solve for the frictional force
Rearranging the equation to find \( F_f \):
\[ F_f = 40 \, \text{N} - 3.2 \, \text{N} = 36.8 \, \text{N} \]

### Conclusion
The force of friction between the box and the inclined plane is \( 36.8 \, \text{N} \).

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