To determine the extension produced in the wire and the energy stored in the extended wire, we will proceed with the following steps:
### Given Data:
- Length of the wire (L): [tex]\( 2.5 \)[/tex] m
- Area of cross-section (A): [tex]\( 1 \times 10^{-6} \ \text{m}^2 \)[/tex]
- Mass hanging on the wire (m): [tex]\( 16 \)[/tex] kg
- Acceleration due to gravity (g): [tex]\( 9.81 \ \text{m/s}^2 \)[/tex]
- Young's modulus (Y): [tex]\( 2 \times 10^{11} \ \text{N/m}^2 \)[/tex]
### Step-by-Step Solution:
1. Calculate the force applied by the mass:
The force ([tex]\( F \)[/tex]) applied by the mass can be determined using:
[tex]\[
F = m \times g
\][/tex]
where [tex]\( m \)[/tex] is the mass and [tex]\( g \)[/tex] is the acceleration due to gravity.
Plugging in the values:
[tex]\[
F = 16 \ \text{kg} \times 9.81 \ \text{m/s}^2
\][/tex]
So, the force is:
[tex]\[
F = 156.96 \ \text{N}
\][/tex]
2. Calculate the extension produced in the wire:
The extension ([tex]\( \Delta L \)[/tex]) can be determined using Hooke's law for materials in the elastic range:
[tex]\[
\Delta L = \frac{F \times L}{Y \times A}
\][/tex]
where [tex]\( F \)[/tex] is the force, [tex]\( L \)[/tex] is the original length of the wire, [tex]\( Y \)[/tex] is Young's modulus, and [tex]\( A \)[/tex] is the cross-sectional area.
Plugging in the values:
[tex]\[
\Delta L = \frac{156.96 \ \text{N} \times 2.5 \ \text{m}}{2 \times 10^{11} \ \text{N/m}^2 \times 1 \times 10^{-6} \ \text{m}^2}
\][/tex]
So, the extension produced is:
[tex]\[
\Delta L = 0.001962 \ \text{m}
\][/tex]
3. Calculate the energy stored in the extended wire:
The energy stored ([tex]\( U \)[/tex]) in the wire can be determined using the formula for elastic potential energy:
[tex]\[
U = \frac{1}{2} \times F \times \Delta L
\][/tex]
where [tex]\( F \)[/tex] is the force and [tex]\( \Delta L \)[/tex] is the extension.
Plugging in the values:
[tex]\[
U = \frac{1}{2} \times 156.96 \ \text{N} \times 0.001962 \ \text{m}
\][/tex]
So, the energy stored is:
[tex]\[
U = 0.15397776 \ \text{J}
\][/tex]
### Summary:
- The extension produced in the wire is [tex]\( 0.001962 \ \text{m} \)[/tex] or [tex]\( 1.962 \ \text{mm} \)[/tex].
- The energy stored in the extended wire is [tex]\( 0.15397776 \ \text{J} \)[/tex].
