Elastic Forces 2-2-1 Hooke’s Law.

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Hooke’s Law.
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Presentation transcript:

Elastic Forces 2-2-1 Hooke’s Law

Hooke’s Law The restoring force exerted by an object (FS) is proportional to the amount it is stretched or compressed (x) and its ‘spring constant’ (k) FS = kx

Hooke’s Law applies to THIS region ONLY!

Example #1 A spring with a spring constant of 10 newtons per meter is stretched 0.5 meter. What force must be applied to the spring? FS = kx FS = (10 N/m)(0.5 m) FS = 5.0 N

Example #2 What is the spring constant of an elastic object if it stretches 2.0 centimeters when pulled with a force of 30 newtons? FS = kx k = FS / x k = (30 N)/(0.02 m) k = 1500 N/m

End of 2.2.1 - PRACTICE

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