SpringsContents:

•Force on a spring•Whiteboards

•Energy stored in a spring•Whiteboards

Force on springs

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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 = kx15 N = k(.42 m), k = (15 N)/(.42 m) = 35.7 N/m

Whiteboards: Force on springs

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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?

.59 m W

F = ma, weight = mg F = kxF = (2.1 kg)(9.8 N/kg) = 20.58 NF = kx, x = F/kx = (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?

11000 N/m W

F = ma, weight = mgF = kxF = (75 kg)(9.8 N/kg) = 735 Nk = 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?

Energy Stored in springs

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Force vs Stretch

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0 0.2 0.4 0.6 0.8 1 1.2

Stretch distance in m

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rin

g F

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Work = Fs, but which F to use?

F = 0

F = kx

Average force:F = 1/2kx

Let’s use the average force: F = 1/2kxW = Fs = (1/2kx)(x) = 1/2kx2

Eelas = 1/2kx2

Whiteboards: Elastic Potential Energy

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.27 J W

Eelas = 1/2kx2 = 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?

53 N/m

Eelas = 1/2kx2

k = 2Eelas/x2 = 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?

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3.7 m

Eelas = 1/2kx2

x = (2 Eelas/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?

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13.3 J

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 Jfinal energy = 1/2kx2 = 1/2(23.5 N/m)(1.56 m)2 = 28.5948 Jchange 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