When load of 5 kg is hung on a wire than extension of 3 m takes place, then work done will be

Options

A. `75 "joule"`
B. `60 "joule"`
C. `50 "joule" `
D. `100 "joule" `

Explanation

To find the work done when a load of 5 kg is hung on a wire that extends by 3 meters, we can follow these steps:

### Step 1: Understand the relationship between force and extension
When a load is hung on a wire, it creates a force due to gravity. This force can be calculated using the formula:
\[ F = mg \]
where:
- \( m \) is the mass (5 kg),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 2: Calculate the force
Substituting the values into the equation:
\[ F = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \]

### Step 3: Use the work done formula for elastic materials
The work done on the wire (which behaves like a spring) when it is extended can be calculated using the formula:
\[ W = \frac{1}{2} F x \]
where:
- \( W \) is the work done,
- \( F \) is the force applied (50 N),
- \( x \) is the extension (3 m).

### Step 4: Substitute the values into the work done formula
Now, substituting the values into the work done formula:
\[ W = \frac{1}{2} \times 50 \, \text{N} \times 3 \, \text{m} \]

### Step 5: Calculate the work done
Calculating the above expression:
\[ W = \frac{1}{2} \times 50 \times 3 = 25 \times 3 = 75 \, \text{J} \]

### Final Answer
The work done when a load of 5 kg is hung on the wire, causing an extension of 3 meters, is:
\[ \boxed{75 \, \text{J}} \]

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