A body of mass 0.2 kg is pressed into a vertical wall normally with a horizontal force 20 N. Coefficient of friction is 0.9. Another horizontal force of 2N is applied on the body in the vertical plane. The frictional force acting on the body is `K sqrt2N.` The value of K is

Options

A. 2
B. 4
C. 8
D. 6

Answer

To solve the problem step by step, we will analyze the forces acting on the body pressed against the wall and calculate the frictional force.

### Step 1: Identify the Given Information
- Mass of the body (m) = 0.2 kg
- Horizontal force pressing the body into the wall (N) = 20 N
- Coefficient of friction (μ) = 0.9
- Additional horizontal force applied in the vertical plane (F) = 2 N

### Step 2: Calculate the Weight of the Body
The weight (W) of the body can be calculated using the formula:
\[ W = m \cdot g \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).
\[ W = 0.2 \, \text{kg} \times 10 \, \text{m/s}^2 = 2 \, \text{N} \]

### Step 3: Calculate the Maximum Static Friction Force
The maximum static friction force (\( F_s^{max} \)) can be calculated using the formula:
\[ F_s^{max} = \mu \cdot N \]
where \( N \) is the normal force (which is the horizontal force pressing the body into the wall).
\[ F_s^{max} = 0.9 \cdot 20 \, \text{N} = 18 \, \text{N} \]

### Step 4: Determine the Resultant Force Acting on the Body
The body experiences two horizontal forces:
1. The normal force (20 N) directed into the wall.
2. The additional horizontal force (2 N) acting in the vertical plane.

Since the additional force is horizontal, we can consider it as a force acting outward from the wall. The resultant force (\( F_{net} \)) can be calculated using the Pythagorean theorem:
\[ F_{net} = \sqrt{(20 \, \text{N})^2 + (2 \, \text{N})^2} \]
\[ F_{net} = \sqrt{400 + 4} = \sqrt{404} \approx 20.1 \, \text{N} \]

### Step 5: Compare the Resultant Force with Maximum Static Friction
Since \( F_{net} \) (approximately 20.1 N) is less than \( F_s^{max} \) (18 N), the frictional force will equal the resultant force to prevent the body from sliding.

### Step 6: Express the Frictional Force in the Required Form
The problem states that the frictional force acting on the body is \( K \sqrt{2} \, \text{N} \). We need to express the frictional force in this form.

Since we found that the frictional force equals the resultant force, we can set:
\[ F_{friction} = K \sqrt{2} \]
From our calculations, we have:
\[ F_{friction} = 2 \sqrt{2} \, \text{N} \]

### Step 7: Solve for K
To find \( K \):
\[ K \sqrt{2} = 2 \sqrt{2} \]
Dividing both sides by \( \sqrt{2} \):
\[ K = 2 \]

### Final Answer
The value of \( K \) is \( 2 \).

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