1. Springs Contents: Force on a spring Whiteboards Energy stored in a spring Whiteboards

2. Force on springs TOC F = kx F = restoring force (in N) k = spring constant (in N/m) (spring stiffness) x - Amount the spring has been distorted (in m) (stretched,/compressed) (show stretch amount, and force) A spring requires 15 N to stretch 42 cm. k = ? F = kx 15 N = k(.42 m), k = (15 N)/(.42 m) = 35.7 N/m

3. Whiteboards: Force on springs 11 | 2 | 323 TOC

4. 6.9 N W F = kx = (53 N/m)(.13 m) = 6.89 N = 6.9 N Ali Zabov stretches a 53 N/m spring 13 cm with what force?

5. .59 m W F = ma, weight = mg F = kx F = (2.1 kg)(9.8 N/kg) = 20.58 N F = kx, x = F/k x = (20.58 N)/(35 N/m) =.588 m =.59 m Nona Zabov allows the weight of a 2.1 kg mass to stretch a 35 N/m spring. What distance does it stretch?

6. 11000 N/m W F = ma, weight = mg F = kx F = (75 kg)(9.8 N/kg) = 735 N k = F/x = (735 N)/(.068 m) = 10808 N/m = 11000 N/m Fyreza Goodfellow has a mass of 75 kg. When he gets into his car the springs settle about 6.8 cm. What is the aggregate spring constant of his suspension?

7. Energy Stored in springs TOC Work = Fs, but which F to use? F = 0 F = kx Average force: F = 1 / 2 kx Let’s use the average force: F = 1 / 2 kx W = Fs = ( 1 / 2 kx)(x) = 1 / 2 kx 2 E elas = 1 / 2 kx 2

8. Whiteboards: Elastic Potential Energy 11 | 2 | 323 TOC

9. .27 J W E elas = 1 / 2 kx 2 = 1 / 2 (24 N/m)(.15 m) 2 =.27 J Mary H. Little-Lamb stretches a 24 N/m spring 15 cm. What energy does she store in it?

10. 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? W

11. 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? W

12. 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? (2) W