A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm? Solution Force = 30 N, elongation = 15 cm-12 cm = 3 cm Spring force, F = kx, where k is spring constant and x is elongation, thus k = F/x = 30 N/ 3 cm = 10 N/cm Work done = 1/2 kx^2 In this case, x = 20 cm - 12 cm = 8 cm thus work done, W = 1/2 . 10 N/cm . 8 cm . 8 cm = 320 N.cm = 3.2 N m = 3.2 J.