Elastic potential energy in stretched wire
Stress, strain and Hooke's law · Physics
Study notes
Problem: A wire of length 2m and cross-sectional area 10^-6 m^2 is stretched by a force of 1000N. The Young's Modulus of the material is 2x10^11 Pa. Calculate the elastic potential energy stored in the wire. Steps: 1. First, find the extension (x) using the formula: Stress = Young's Modulus * Strain. Stress = Force / Area = 1000 / 10^-6 = 10^9 Pa. Strain = Extension / Original Length = x / 2. So, 10^9 = 2x10^11 * (x / 2). 10^9 = 10^11 * x. x = 10^9 / 10^11 = 0.01 m. 2. Now, calculate the Elastic Potential Energy (U) using the formula: U = (1/2) * F * x. U = 0.5 * 1000 N * 0.01 m. U = 5 J. Answer: The elastic potential energy stored is 5 Joules.