What is meant by elastic potential energy? Derive an expression for the elastic potential energy of stretched wire. Prove that its elastic energy density is equal to 1/2 stress x strain.

Definition of Elastic Potential Energy:
Elastic potential energy is the energy stored in an elastic material when it is deformed, such as when it is stretched or compressed. This energy is released when the material returns to its original shape.

Derivation of Elastic Potential Energy:
1. Consider a wire of length L and cross-sectional area A with Young’s modulus Y.
2. The tensile force F stretches the wire by an amount x.
3. Stress in the wire is given by
Stress (σ) = F / A
4. Strain in the wire is given by
Strain (ε) = x / L
5. Young’s modulus is defined as
Y = σ / ε = (F/A) / (x/L) => F = (Y * A * x) / L
6. The work done (W) in stretching the wire is given by the area under the stress-strain curve:
W = ∫(from 0 to x) F dx = ∫(from 0 to x) (Y * A * (x/L)) dx
7. Integrating:
W = (Y * A / L) * ∫(from 0 to x) x dx
= (Y * A / L) * [x²/2] (from 0 to x) = (Y * A / L) * (x²/2)
8. Hence, the elastic potential energy (U) stored in the wire is:
U = (Y * A * x²) / (2L)

Proof of Elastic Energy Density is 1/2 Stress x Strain:
1. The density of elastic energy:
is energy per volume
u = U / (A * L)
= [(Y * A * x²) / (2L)] / (A * L)
= (Y * x²) / (2L²)
2. If one starts by assuming the presence of Young’s modulus:
Y = (F / A) / (x / L)
3. Thus, the stress is:
Stress (σ) = F / A = (Y * x) / L
4. And strain is:
Strain (ε) = x / L
5. Substituting for stress and strain:
u = (1/2) * σ * ε

Hence, the elastic energy density is equal to 1/2 stress x strain.

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