A block of mass 2 kg is placed on the floor . The coefficient of static friction is `0.4` . If a force of `2.8` N is applied on the block parallel to floor , the force of friction between the block and floor is : (Taking g = `10 m//s^(2)`)

Options

A. 2.8N
B. 8N
C. 2.0N
D. zero

Explanation

To solve the problem, we need to determine the force of friction acting on the block when a force is applied. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the weight of the block
The weight (W) of the block can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 2 \, \text{kg} \) (mass of the block)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ W = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 2: Calculate the normal force (N)
Since the block is on a horizontal surface and there are no vertical forces acting other than its weight and the normal force, the normal force (N) is equal to the weight of the block:
\[ N = W = 20 \, \text{N} \]

### Step 3: Calculate the limiting friction (F_friction)
The limiting friction can be calculated using the formula:
\[ F_{\text{friction}} = \mu \cdot N \]
where:
- \( \mu = 0.4 \) (coefficient of static friction)

Substituting the values:
\[ F_{\text{friction}} = 0.4 \cdot 20 \, \text{N} = 8 \, \text{N} \]

### Step 4: Compare applied force with limiting friction
The applied force (F_applied) is given as:
\[ F_{\text{applied}} = 2.8 \, \text{N} \]

Now, we compare the applied force with the limiting friction:
- Limiting friction = 8 N
- Applied force = 2.8 N

Since the applied force (2.8 N) is less than the limiting friction (8 N), the block will not move.

### Step 5: Determine the force of friction
When the block is not moving, the force of friction will be equal to the applied force (as it opposes the motion). Therefore, the force of friction is:
\[ F_{\text{friction}} = F_{\text{applied}} = 2.8 \, \text{N} \]

### Final Answer
The force of friction between the block and the floor is **2.8 N**.

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