1. Energy Stored in springs E p = 1 / 2 kx 2 Where: E p – potential energy stored in spring (J) k – spring constant (N/m) x – amount of stretch/compression (m)

2. .27 J The stretch distance is.46 -.31 =.15 m E elas = 1 / 2 kx 2 = 1 / 2 (24 N/m)(.15 m) 2 =.27 J Mary H. Little-Lamb has a 24 N/m spring that is 31 cm long un-stretched. What energy does she store in it if she stretches it until it is 46 cm long?

3. 53 N/m E elas = 1 / 2 kx 2 k = 2E elas /x 2 = 2(56 J)/(1.45 m) 2 k = 53.3 N/m = 53 N/m A spring stores 56 J of energy being distorted 1.45 m. What is its spring constant?

4. 3.7 m E elas = 1 / 2 kx 2 x =  (2 E elas /m) =  (2(98 J)/(14.5 m)) = 3.7 m What amount must you distort a 14.5 N/m spring to store 98 J of energy?

5. 13.3 J E elas = 1 / 2 kx 2 (What is the difference in the stored energy of the spring?) initial energy = 1 / 2 kx 2 = 1 / 2 (23.5 N/m)(1.14 m) 2 = 15.2703 J final energy = 1 / 2 kx 2 = 1 / 2 (23.5 N/m)(1.56 m) 2 = 28.5948 J change in energy = work = 28.5948 J - 15.2703 J = 13.3245 J Or, average force = kx = (23.5 N/m)(1.35 m) = 31.725 N (1.35 is average distance) Work = Fs = (31.725 N)(1.56-1.14) = 13.3245 J How much work is it to stretch a 23.5 N/m spring from 1.14 m to 1.56 m of distortion?