A horizontal force F pulls a 20 kg box at a constant speed along a rough horizotal floor. The coefficient of friction between the box and the floor is 0.25. The work done by force F on the block in displacing it by 2 m is

Options

A. 49 J
B. 98 J
C. 147 J
D. 196 J

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the frictional force
The frictional force (f) can be calculated using the formula:
\[ f = \mu \cdot m \cdot g \]
where:
- \( \mu \) is the coefficient of friction (0.25),
- \( m \) is the mass of the box (20 kg),
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values:
\[ f = 0.25 \cdot 20 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \]
\[ f = 0.25 \cdot 196 \, \text{N} \]
\[ f = 49 \, \text{N} \]

### Step 2: Determine the applied force
Since the box is moving at a constant speed, the applied force (F) must equal the frictional force to maintain equilibrium. Therefore:
\[ F = f = 49 \, \text{N} \]

### Step 3: Calculate the work done by the force
The work done (W) by the force is given by the formula:
\[ W = F \cdot d \]
where:
- \( F \) is the applied force (49 N),
- \( d \) is the displacement (2 m).

Substituting the values:
\[ W = 49 \, \text{N} \cdot 2 \, \text{m} \]
\[ W = 98 \, \text{J} \]

### Conclusion
The work done by the force F on the block in displacing it by 2 m is **98 joules**.

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