A vertical steel wire of length 4 m and cross-sectional area 2 mm ^2 has its upper end fixed. A block of mass 5 kg is attached to the unstretched lower end and released from rest. The maximum elastic potential energy stored in the wire during its subsequent motion is _____ mJ . (Given: Y = 1.0 10¹¹ N m ⁻² and g = 10 m s ⁻² )

Step-by-step solution

The wire acts like a spring with an equivalent spring constant k = YA L . Substituting the given values: k = 1.0 10¹¹ 2 10⁻⁶ 4 = 5 10^4 N m ⁻¹ Let the maximum extension of the wire be x_m . At this point, the block momentarily comes to rest. By the conservation of mechanical energy, the loss in gravitational potential energy equals the gain in elastic potential energy: Mgx_m = 1 2 kx_m^2 x_m = 2Mg k = 2 5 10 5 10^4 = 2 10⁻³ m The maximum elastic potential energy stored in the wire is: U_ max = 1 2 kx_m^2 = Mgx_m U_