A steel wire 4.0m in length is stretched through 2.0mm.The cross -sectional area of the wire is 2.0 mm^(2).If young's modulus of steel is 2.0xx10^(11) N//m^(2) (a) the enrgy density of wire, (b) the elastic potential energy stored in the wire.
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the energy density = 2.5 × 10⁴ J/m³ and energy stored in the wire = 0.2
(a) given, length of steel wire, l = 4m
cross sectional area of the wire, A = 2 mm² = 2 × 10^-6 m²
wire stretched , ∆l = 2mm = 2 × 10^-3 m
Young's modulus, Y = 2 × 10¹¹ N/m²
energy density = 1/2 × stress × strain
= 1/2 × (Y × strain) × strain
= 1/2 × Y × strain²
= 1/2 × 2 × 10¹¹ × (2 × 10^-3/4)²
= 1/2 × 2 × 10¹¹ × 1/4 × 10^-6
= 0.25 × 10^5 J/m³
= 2.5 × 10⁴ J/m³
(b) elastic potential energy stored in the wire, U = energy density × volume of wire.
= 2.5 × 10⁴ J/m³ × (Al)
= 2.5 × 10⁴ × 2 × 10^-6 × 4
= 20 × 10^-2
= 0.2 J
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