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 is a step-by-step solution:

### Step 1: Identify the Forces Acting on the Block
1. **Weight of the Block (W)**: This acts downward and is given by:
\[ W = mg = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N} \]

2. **Applied Force (F)**: The force of 30 N is applied at an angle of 53° downward from the horizontal. We can resolve this force into horizontal and vertical components:
- Horizontal component (\(F_x\)):
\[ F_x = F \cos(53°) = 30 \cos(53°) \]
- Vertical component (\(F_y\)):
\[ F_y = F \sin(53°) = 30 \sin(53°) \]

### Step 2: Calculate the Components of the Applied Force
Using the trigonometric values:
- \(\cos(53°) = \frac{3}{5}\)
- \(\sin(53°) = \frac{4}{5}\)

Calculating the components:
\[ F_x = 30 \times \frac{3}{5} = 18 \, \text{N} \]
\[ F_y = 30 \times \frac{4}{5} = 24 \, \text{N} \]

### Step 3: 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 4: Calculate the Maximum Friction Force (F_friction_max)
The maximum static friction force can be calculated using the coefficient of friction (\(\mu\)):
\[ F_{\text{friction max}} = \mu N = 0.2 \times 224 \, \text{N} = 44.8 \, \text{N} \]

### Step 5: Determine the Required Friction Force
The required friction force is equal to the horizontal component of the applied force:
\[ F_{\text{friction required}} = F_x = 18 \, \text{N} \]

### Step 6: Compare the Required Friction Force with Maximum Friction Force
Since the required friction force (18 N) is less than the maximum friction force (44.8 N), the actual friction force will equal the required friction force:
\[ F_{\text{friction}} = 18 \, \text{N} \]

### Final Answer
The friction force acting on the block by the ground is **18 N**.

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