An ideal spring of negligible mass is 12.00 cm long when nothing is attached to it. When you hang a 3.15-kg weight from it, you measure its length to be 13.40 cm. If you wanted to store 10.0 J of potential energy in this spring, what would be its ?total? length? Assume that it continues to obey Hooke’s law.

Solution 14E Hooke’s law is given by F = kx, where k is the force constant of a spring and x is its elongation. When the 3.15 kg weight is hung from the spring, its length changes by x = 13.40 12.00 cm = 1.40 cm = 0.014 m Now, the force needed to extend the spring will be balanced by the weight of the object, So, mg = kx k = mg/x k = 3.15 × 9.8 N/0.014 m = 2205 N/m Now, the potential energy of a spring with force constant k and elongation x is given by, 1 2 P = k2 …..(1) Given that, P = 10.0 J From equation (1), 1 2 10.0 J = 2 × 2205 N/m × x x = 2205m 2 x = 0.009 m 2 x = 0.095 m x = 9.5 cm Initial length of the spring is = 12.00 cm Total length of the spring afterward = 12.00 + 9.5 cm = 21.5 cm Thus, the total length of the spring after the elongation is 21.5 cm.