A block of mass 20 kg is acted upon by a force `F=30N` at an angle `53^@` with horizontal in downward direction as shown. The coefficient of friction between the block and the forizontal surface is `0.2`. The friction force acting on the block by the ground is `(g=10(m)/(s^2)`)

Options

A. `40.0 N`
B. `30.0 N`
C. `18.0 N`
D. `44.8 N`

Explanation

To solve the problem, we need to analyze the forces acting on the block and calculate the friction force. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Block
1. **Weight of the Block (W)**: The weight can be calculated using the formula:
\[ W = m \cdot g \]
where \( m = 20 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
\[ W = 20 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 200 \, \text{N} \]

2. **Applied Force (F)**: The applied force is given as \( F = 30 \, \text{N} \) at an angle of \( 53^\circ \) downwards.

3. **Components of the Applied Force**:
- Horizontal Component (\( F_x \)):
\[ F_x = F \cdot \cos(53^\circ) = 30 \cdot \cos(53^\circ) = 30 \cdot \frac{3}{5} = 18 \, \text{N} \]
- Vertical Component (\( F_y \)):
\[ F_y = F \cdot \sin(53^\circ) = 30 \cdot \sin(53^\circ) = 30 \cdot \frac{4}{5} = 24 \, \text{N} \]

### Step 2: Calculate the Normal Force (N)
The normal force is affected by both the weight of the block and the vertical component of the applied force. The normal force can be calculated as:
\[ N = W + F_y = 200 \, \text{N} + 24 \, \text{N} = 224 \, \text{N} \]

### Step 3: Calculate the Maximum Frictional Force (\( F_{\text{max}} \))
The maximum frictional force can be calculated using the formula:
\[ F_{\text{max}} = \mu \cdot N \]
where \( \mu = 0.2 \).
\[ F_{\text{max}} = 0.2 \cdot 224 \, \text{N} = 44.8 \, \text{N} \]

### Step 4: Compare Horizontal Force and Maximum Frictional Force
The horizontal force acting on the block is \( F_x = 18 \, \text{N} \). Since the maximum frictional force is greater than the horizontal force, the frictional force will equal the horizontal force:
\[ F_{\text{friction}} = F_x = 18 \, \text{N} \]

### Conclusion
The friction force acting on the block by the ground is \( 18 \, \text{N} \).

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