A load of 10 kg is suspended by a metal wire that is 3 m long and possesses a cross-sectional area of 4 mm ^2 . Determine the elongation. The Young modulus of the metal is 2.0 10¹¹ N/m ^2 .

Step-by-step solution

Given: m = 10 kg L = 3 m A = 4 mm ^2 = 4 10⁻⁶ m ^2 Y = 2.0 10¹¹ N/m ^2 g = 9.8 m/s ^2 The force applied on the wire is F = mg = 10 9.8 = 98 N . The elongation L is given by the formula: L = FL AY Substituting the values: L = 98 3 4 10⁻⁶ 2.0 10¹¹ L = 294 8 10^5 L = 36.75 10⁻⁵ m = 3.675 10⁻⁴ m