A mass of 0.5 kg is suspended from wiere, then length of wire increase by 3 mm then work done

Options

A. `4.5xx10^(-3)` Joule
B. `7.3xx10^(-3)` Joule
C. `9.3xx10^(-2)` Joule
D. `2.5` Joule

Answer

To solve the problem of calculating the work done when a mass of 0.5 kg is suspended from a wire that stretches by 3 mm, we can follow these steps:

### Step 1: Calculate the Weight of the Mass
The weight (W) of the mass can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 0.5 \, \text{kg} \) (mass)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 0.5 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 4.9 \, \text{N} \]

### Step 2: Convert the Change in Length to Meters
The change in length (ΔL) is given as 3 mm. We need to convert this to meters:
\[ \Delta L = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \]

### Step 3: Calculate the Work Done
The work done (W_d) when the wire stretches can be calculated using the formula:
\[ W_d = \frac{1}{2} \cdot F \cdot \Delta L \]
where:
- \( F \) is the force (weight of the mass) and \( \Delta L \) is the change in length.

Substituting the values:
\[ W_d = \frac{1}{2} \cdot 4.9 \, \text{N} \cdot (3 \times 10^{-3} \, \text{m}) \]

Calculating the work done:
\[ W_d = \frac{1}{2} \cdot 4.9 \cdot 3 \times 10^{-3} \]
\[ W_d = \frac{14.7 \times 10^{-3}}{2} \]
\[ W_d = 7.35 \times 10^{-3} \, \text{J} \]

### Final Answer
The work done in stretching the wire is:
\[ W_d = 7.35 \times 10^{-3} \, \text{J} \]

---