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Springs and Hooke’s Law Physics 11

Springs and Hooke’s Law

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Springs and Hooke’s Law. Physics 11. Springs. A mass-spring system is given below. As mass is added to the end of the spring, how would you expect the spring to stretch?. Springs. Springs. 2 times the mass results in a 2 times of the displacement from the equilibrium point… - PowerPoint PPT Presentation

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Page 1: Springs and Hooke’s Law

Springs and Hooke’s Law

Physics 11

Page 2: Springs and Hooke’s Law

Springs

A mass-spring system is given below. As mass is added to the end of the spring, how would you expect the spring to stretch?

Page 3: Springs and Hooke’s Law

Springs

mgFg

gmFg 1

gmFg 2

gmFg 3

x x

x

x

springg FF

springF

Page 4: Springs and Hooke’s Law

Springs 2 times the mass results in a 2 times of the

displacement from the equilibrium point… 3 time the mass… 3 times the

displacement…

springg FF kxFspring

xkgmkxmg

22

Page 5: Springs and Hooke’s Law

What kind of energy is this? Potential Energy

Elastic Potential Energy to be exact!

Page 6: Springs and Hooke’s Law

What else besides springs has elastic potential energy? Diving boards Bows (bow and arrows) Bungee cord

Page 7: Springs and Hooke’s Law

Hooke’s Law

Fspring: Applied forceX : displacement of the spring from the

equilibrium position (units: m)K: the spring constant (units: N/m)

kxFspring

Page 8: Springs and Hooke’s Law

Hooke’s Law the restoring force is

opposite the applied force. (negative sign)

Gravity applied in the negative direction, the restoring force is in the positive direction

kxFspring

Page 9: Springs and Hooke’s Law

Example An archery bow requires a force of

133N to hold an arrow at “full draw” (pulled back 71cm). Assuming that the bow obeys Hooke’s Law, what is its spring constant?

Page 10: Springs and Hooke’s Law

F = kx 133 = k(0.71) k = 133/0.71 k = 187.32 N/m 190 N/m

Page 11: Springs and Hooke’s Law

Restoring Force The restoring force is the force that is

needed to put the spring back to equilibrium.

Example: If you stretch a spring by 0.5m and you had to use 150N of force, the restoring force is -150N.

Page 12: Springs and Hooke’s Law

Practice Problems Textbook

Page 258 35-37

Page 13: Springs and Hooke’s Law

Elastic Potential Energy of a Spring Formula: Ee = ½ kx2

Units: Joules (J)

Page 14: Springs and Hooke’s Law

Example: A spring with spring constant 75 N/m

is resting on a table. A) If the spring is compressed a

distance of 28cm, what is the increase in its potential energy?

B) What force must be applied to hold the spring in this position?

Page 15: Springs and Hooke’s Law

Answer: A) Ee = ½ kx2

Ee = ½ (75)(0.28)2

Ee = 2.9 J B) F = kx F= 75(0.28) F = 21 N

Page 16: Springs and Hooke’s Law

Practice Problems Page 261, questions 38, 39, 40 Page 261 (Section Review)

1, 2, 3, 4, 7

Page 17: Springs and Hooke’s Law

Conservation of Energy with a Spring Ex. 1: A 4.0 kg block slides across a

frictionless table with a velocity of 5.0m/s into a spring with a stiffness of 2500 N/m. How far does the spring compress?

Page 18: Springs and Hooke’s Law

Answer X = 0.20m

Page 19: Springs and Hooke’s Law

Example 2: A 70. kg person bungee jumps off a

50.m bridge with his ankles attached to a 15m long bungee cord. Assume the person stops at the edge of the water and he is 2.0m tall, what is the force constant of the bungee cord?

Page 20: Springs and Hooke’s Law

Answer: 64 N/m

Conservation of Energy Worksheet

Page 21: Springs and Hooke’s Law

Practice Problems Textbook

Page 261 38-40

Section review (p 261) 1-10