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

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

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

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

  2. Force on springs • 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 TOC

  3. Whiteboards: Force on springs 1 | 2 | 3 TOC

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

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

  6. 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? 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 W 11000 N/m

  7. Energy Stored in springs F = 0 F = kx Average force: F = 1/2kx Work = Fs, but which F to use? Let’s use the average force: F = 1/2kx W = Fs = (1/2kx)(x) = 1/2kx2 Eelas = 1/2kx2 TOC

  8. Whiteboards: Elastic Potential Energy 1 | 2 | 3 TOC

  9. Mary H. Little-Lamb stretches a 24 N/m spring 15 cm. What energy does she store in it? Eelas = 1/2kx2 = 1/2(24 N/m)(.15 m)2 = .27 J W .27 J

  10. A spring stores 56 J of energy being distorted 1.45 m. What is its spring constant? Eelas = 1/2kx2 k = 2Eelas/x2 = 2(56 J)/(1.45 m)2 k = 53.3 N/m = 53 N/m W 53 N/m

  11. What amount must you distort a 14.5 N/m spring to store 98 J of energy? Eelas = 1/2kx2 x = (2 Eelas/m) = (2(98 J)/(14.5 m)) = 3.7 m W 3.7 m

  12. How much work is it to stretch a 23.5 N/m spring from 1.14 m to 1.56 m of distortion? (2) Eelas = 1/2kx2 (What is the difference in the stored energy of the spring?) initial energy = 1/2kx2 = 1/2(23.5 N/m)(1.14 m)2 = 15.2703 J final energy = 1/2kx2 = 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 W 13.3 J

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