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3. A spring has a spring constant k=8.75 N/m. If the spring is displaced 0.150 m from its equilibrium position, what i...

3. A spring has a spring constant k=8.75 N/m. If the spring is displaced 0.150 m from its equilibrium
position, what is the force that the spring exerts? Show your work.
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Answer #1

Force Exerts on Srping is givne by

 

F = -kx    = - 8.75 * 0.150   =  -1.3125 N

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Answer #2

You use the equation F=1/2(kx^2). When you plug your numbers in you should get


F=(1/2)(8.25N/m)(0.150m)^2= 0.0928

answered by: Madi
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