A block of mass 50 kg slides over a horizontal distance of 1m. If the coefficient of friction between their surface is 0.2, then work done against friction is (take `g = 9.8 m//s^(2)`) :

Options

A. 98 J
B. 72 J
C. 56 J
D. 34 J

Explanation

To solve the problem of calculating the work done against friction when a block of mass 50 kg slides over a horizontal distance of 1 meter with a coefficient of friction of 0.2, we can follow these steps:

### Step 1: Calculate the Normal Force
The normal force (N) acting on the block can be calculated using the formula:
\[ N = mg \]
where:
- \( m = 50 \, \text{kg} \) (mass of the block)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ N = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N} \]

### Step 2: Calculate the Frictional Force
The frictional force (F_f) can be calculated using the formula:
\[ F_f = \mu N \]
where:
- \( \mu = 0.2 \) (coefficient of friction)

Substituting the values:
\[ F_f = 0.2 \times 490 \, \text{N} = 98 \, \text{N} \]

### Step 3: Calculate the Work Done Against Friction
The work done (W) against friction can be calculated using the formula:
\[ W = F_f \times d \]
where:
- \( d = 1 \, \text{m} \) (distance moved)

Substituting the values:
\[ W = 98 \, \text{N} \times 1 \, \text{m} = 98 \, \text{J} \]

### Step 4: Conclusion
The work done against friction is:
\[ W = 98 \, \text{J} \]

### Final Answer
The work done against friction is **98 joules**.

---