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 step by step, we will follow these steps:

### Step 1: Identify the given values
- Mass of the box (m) = 20 kg
- Coefficient of friction (μ) = 0.25
- Displacement (d) = 2 m
- Acceleration due to gravity (g) = 9.8 m/s² (standard value)

### Step 2: Calculate the normal force (N)
Since the box is on a horizontal surface, the normal force (N) is equal to the weight of the box.
\[ N = m \cdot g \]
Substituting the values:
\[ N = 20 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 196 \, \text{N} \]

### Step 3: Calculate the frictional force (f)
The frictional force can be calculated using the formula:
\[ f = \mu \cdot N \]
Substituting the values:
\[ f = 0.25 \cdot 196 \, \text{N} = 49 \, \text{N} \]

### Step 4: Determine the force F
Since the box is moving at a constant speed, the pulling force F must be equal to the frictional force. Therefore:
\[ F = f = 49 \, \text{N} \]

### Step 5: Calculate the work done by force F
The work done (W) by the force F when it displaces the box by 2 m can be calculated using the formula:
\[ W = F \cdot d \]
Substituting the values:
\[ W = 49 \, \text{N} \cdot 2 \, \text{m} = 98 \, \text{J} \]

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

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