When a wire of lenght 5m & radius 0.5mm is stretched by a load of 49N,the elongation produced in the wire is 0.1cm. Find the energy stored per unit volume of the wire.

Answer 1

σ=F/A=F/πr²
ε=ΔL/L

Hook’s law for the wire:
k ε= σ
k= σ/ ε= F L /πr²ΔL

E=k•ΔL²/2= F• L•ΔL²/2πr²ΔL=
=F•L•ΔL/2πr²=49•5•10⁻³/2•π•25•10⁻⁸=1.56•10⁵ J
E₀=E/V= E/πr²L = 1.56•10⁵/π•25•10⁻⁸•5=3.97•10¹⁰ J/m³