The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross-sectional area of the wire is mm². (Given Y=2.0×10¹¹ N m⁻²)

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40. Given: L=20 m, Δ L=2 cm=2×10⁻² m, U=80 J, Y=2.0×10¹¹ N m⁻². U=1/2Y((Δ L)/L)²AL, so A=(2UL)/(Y(Δ L)²). ε=(2×10⁻²)/(20)=10⁻³, so 1/2Yε²=1/2×2.0×10¹¹×10⁻⁶=10⁵ J m⁻³. 80=10⁵× A×20, giving A=4×10⁻⁵ m². A=4×10⁻⁵×10⁶=40 mm².

The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross-sectional area of the wire is ______ mm². (Given Y=2.0×10¹¹ N m⁻²)

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