This document discusses Hooke's law of elasticity and provides examples of its application. [1] Hooke's law states that the extension of a spring is proportional to the applied load as long as the elastic limit is not exceeded. [2] Materials that obey this law are known as linear-elastic or "Hookean" materials. [3] The document then provides the formula for Hooke's law, F=kx, and gives two examples of using it to calculate spring force and spring constant.

1. Christine Mae M. Lagumbay
Roxan M. Bahingawan
Ernalyn Sanico
Norjamela Ali
Hanna Grace Dayak

2. Hooke's law of elasticity is an approximation that states
that the extension of a spring is in direct proportion with
the load applied to it. Many materials obey this law as
long as the load does not exceed the material's elastic
limit. Materials for which Hooke's law is a useful
approximation are known as linear-elastic or "Hookean"
materials. Hooke's law in simple terms says that strain is
directly proportional to stress.

3. F= -kx
x is the displacement of the spring's end from its equilibrium
position
S.I unit is meters
F is the restoring force exerted by the spring on that end
S.I unit is N or kg.m/s2
k is a constant called the rate or spring constant
S.I unit N/m or kg/s2
The negative sign indicates that the force exerted by the spring is in
direct opposition to the direction of displacement.

4. Hooke's law is named after the
17th century British
physicist Robert Hooke. He first
stated this law in 1660 as
a Latin anagram, whose solution
he published in 1678 as Ut
tensio, sic vis, meaning, "As the
extension, so the force".

5. What is the force required to stretch a spring whose
constant value is 100 N/m by an amount of 0.50 m?
Using the formula F=kx solve the question
F=force(N)
k=force constant(N/m)
x=stretch or compression(m)
F=-(100N/m)(0.50m)
F=-50 N

6. When a 13.2-kg mass is placed on top of a vertical spring,
the spring compresses 5.93 cm. Find the force constant of
the spring.
Using Hooke's Law: F = -kx
Derivation: -k= F/x
F=force(N)
k=force constant(N/m)
x=stretch or compression(m) 5.93cmx1m/100cm
-k = F/x =0.0593m
= (13.2x9.8)(0.0593m)
k = -2181 N/m

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