When a force of 80 N is applied on a uniform steel wire, its free end elongates by 1.2 mm. Find the energy stored in the wire.

Answer

To find the energy stored in the steel wire when a force is applied, we can use the formula for elastic potential energy stored in a stretched wire. The energy (U) stored in the wire can be calculated using the formula:

\[ U = \frac{1}{2} F \Delta L \]

Where:
- \( U \) is the energy stored in the wire,
- \( F \) is the force applied,
- \( \Delta L \) is the elongation (change in length) of the wire.

**Step 1: Identify the values given in the problem.**
- Force \( F = 80 \, \text{N} \)
- Elongation \( \Delta L = 1.2 \, \text{mm} = 1.2 \times 10^{-3} \, \text{m} \)

**Step 2: Substitute the values into the energy formula.**
Using the formula for the energy stored:

\[ U = \frac{1}{2} \times 80 \, \text{N} \times (1.2 \times 10^{-3} \, \text{m}) \]

**Step 3: Calculate the energy stored.**
Now, perform the calculation:

\[ U = \frac{1}{2} \times 80 \times 1.2 \times 10^{-3} \]

\[ U = 40 \times 1.2 \times 10^{-3} \]

\[ U = 48 \times 10^{-3} \, \text{J} \]

**Step 4: Finalize the answer.**
Thus, the energy stored in the wire is:

\[ U = 0.048 \, \text{J} \]

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