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. 36.8N
B. 76.8 N
C. 65.6 N
D. none of these

Explanation

To find the force of friction between the wooden box and the inclined plane, we can follow these steps:

### Step 1: Identify the given data
- Mass of the box (m) = 8 kg
- Acceleration (a) = 0.4 m/s²
- Angle of inclination (θ) = 30°
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the gravitational force acting on the box
The gravitational force (weight) acting on the box can be calculated using the formula:
\[ F_g = m \cdot g \]
Substituting the values:
\[ F_g = 8 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 80 \, \text{N} \]

### Step 3: Resolve the gravitational force into components
The gravitational force can be resolved into two components:
1. Perpendicular to the inclined plane:
\[ F_{\perp} = F_g \cdot \cos(\theta) \]
2. Parallel to the inclined plane:
\[ F_{\parallel} = F_g \cdot \sin(\theta) \]

Calculating \( F_{\parallel} \):
\[ F_{\parallel} = 80 \, \text{N} \cdot \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ F_{\parallel} = 80 \, \text{N} \cdot \frac{1}{2} = 40 \, \text{N} \]

### Step 4: Apply Newton's second law
According to Newton's second law, the net force acting on the box along the inclined plane is given by:
\[ F_{\text{net}} = m \cdot a \]

The net force can also be expressed as:
\[ F_{\text{net}} = F_{\parallel} - F_f \]
Where \( F_f \) is the force of friction.

Setting the two expressions for net force equal gives:
\[ F_{\parallel} - F_f = m \cdot a \]

### Step 5: Solve for the force of friction
Rearranging the equation to solve for \( F_f \):
\[ F_f = F_{\parallel} - m \cdot a \]

Substituting the known values:
\[ F_f = 40 \, \text{N} - (8 \, \text{kg} \cdot 0.4 \, \text{m/s}^2) \]
\[ F_f = 40 \, \text{N} - 3.2 \, \text{N} \]
\[ F_f = 36.8 \, \text{N} \]

### Final Answer
The force of friction between the box and the inclined plane is **36.8 N**.

---