A spring with a force constant of 250 N/m is compressed 4.00 cm from its natural length. How much elastic potential energy is stored in it?

1. Elastic potential energy stored in a Hooke's law system is PE = (1/2) k x squared. 2. Convert the compression: 4.00 cm = 0.0400 m. 3. Square it: (0.0400 m) squared = 1.60 x 10^-3 m squared. 4. PE = (1/2)(250 N/m)(1.60 x 10^-3 m squared) = 0.200 J. 5. Forgetting to square the displacement gives 5.00 J, and dropping the factor of one half gives 0.400 J, so both shortcuts mislead. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section Energy in Hooke's Law of Deformation_