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Hooke's Law

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morgan ohland

on 26 April 2010

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Transcript of Hooke's Law

Hooke's Law Sample Problem A F=-kx x= the displacement of the end of the spring to equilibrium position

F= the restoring force exerted by the material

k= the force constant States that the extension of a spring is in direct proportion with the load added to it as long as the load does not excess the elastic limit. m= .55kg
g=9.81 m/s^2
x=-2.0cm or .020 m
k=? Choose an equation
F net=0=F elastic +f g
F elastic = -kx
F g =-mg
-kx-mg=0 Rearrange the equation to isolate the unknown
kx=-mg
k=-mg/x Substutiee the values and solve
k=-(.55kg)(9.81m/s^2)/-.020 m
k=270 N/m
Spring C g=9.81m/s^2
m=500g
x=1.20 cm.012 m Spring A g=9.81m/s^2
m=50g
x=1.50cm
k=? Spring F x=10.1
m=500g
g=9.81m/s^2
k=? 1.6cm .016 m
294.5 g .2945 kg 2.6 cm .026 m
499.1 g .4491 kg 3.0 cm .030 m
69.1 g .0691kg 22.6 N 180.6. N 180.6 N
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