Q.Calculate the work done in stretching a steel wire of length 2 m and cross sectional area 0.0225 mm² when a load of 100 N is slowly applied to its free end. [Young's modulus of steel = 2×10¹¹ N/m²]

Because the restoring force in a wire grows linearly with extension, the work done in stretching it to extension l is W = (1/2) F l = Y A l^2 / (2L).

Consider a wire of natural length L and cross-sectional area A, with Young's modulus Y. When it is stretched by a small extension x, the restoring/applied force required is, by the definition of Young's modulus (Y = stress/strain = (F/A)/(x/L)):

F(x) = (Y A / L) x

This force increases linearly with the extension x (like a spring, F ∝ x), starting from 0 and reaching a maximum value F = YAl/L when the total extension is l. The small work done in stretching the wire by a further dx is dW = F(x) dx. The total work done in stretching the wire from x = 0 to x = l is:

W = ∫(0 to l) (Y A / L) x dx = (Y A / L) × (l^2 / 2) = (Y A l^2) / (2L)

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