A body of mass 2kg is placed on a horizontal surface having kinetic friction `0.4` and static friction `0.5` . If the force applied on the body is `2.5N` , then the frictional force acting on the body will be

Options

A. 8N
B. 10N
C. 20N
D. `2.5N`

Explanation

To solve the problem, we need to determine the frictional force acting on a body of mass 2 kg placed on a horizontal surface with given coefficients of static and kinetic friction. The applied force on the body is 2.5 N.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the body, \( m = 2 \, \text{kg} \)
- Coefficient of static friction, \( \mu_s = 0.5 \)
- Coefficient of kinetic friction, \( \mu_k = 0.4 \)
- Applied force, \( F = 2.5 \, \text{N} \)

2. **Calculate the Normal Force (N):**
The normal force \( N \) acting on the body can be calculated using the formula:
\[ N = mg \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).
\[ N = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \]

3. **Calculate the Limiting Static Friction (F_s):**
The limiting static friction can be calculated using the formula:
\[ F_s = \mu_s \times N \]
Substituting the values:
\[ F_s = 0.5 \times 20 \, \text{N} = 10 \, \text{N} \]

4. **Compare the Applied Force with Limiting Friction:**
The applied force \( F = 2.5 \, \text{N} \) is less than the limiting static friction \( F_s = 10 \, \text{N} \). Since the applied force is less than the maximum static friction, the body will not move.

5. **Determine the Frictional Force:**
Since the body does not move, the frictional force will adjust to balance the applied force. Therefore, the frictional force \( F_f \) will be equal to the applied force:
\[ F_f = F = 2.5 \, \text{N} \]

### Final Answer:
The frictional force acting on the body is \( 2.5 \, \text{N} \).

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