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myininaya

  • 5 years ago

Someone explain springs and work to me?

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  1. myininaya
    • 5 years ago
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    How is the following set up: Find the work required to compress a spring from its natural length of 1 ft to a length of .75ft if the force constant is k=16lb/ft

  2. myininaya
    • 5 years ago
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    I know how to do the integration

  3. myininaya
    • 5 years ago
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    Just the setup is my problem

  4. anonymous
    • 5 years ago
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    Hooke's law says that the force exerted by a spring of force constant k is \[F=-kx\]where x is the displacement from equilibrium. The work is defined as\[W=\int\limits_{x_i}^{x_f}F.dx={-k}\int\limits_{x_i}^{x_f}x.dx=-k \left[ \frac{x^2_f}{2}-\frac{x^2_i}{2} \right]\]Now, x_i=0 (there's no initial displacement from equilibrium) and x_f is 0.25 (you've displaced it from 1ft to 0.75ft). Your force constant should be negative too.

  5. myininaya
    • 5 years ago
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    So are you an engineer?

  6. myininaya
    • 5 years ago
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    If you are, maybe you can give me a rough explanation of what a spring being compressed is -kx and not kx? So I know work is force times distance. Does k or x have anything to do with the force or the distance?

  7. myininaya
    • 5 years ago
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    oh wait

  8. myininaya
    • 5 years ago
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    the force is k

  9. myininaya
    • 5 years ago
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    so x is the distance over the interval xi to xf

  10. anonymous
    • 5 years ago
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    x is the distance you displace the spring FROM its equilibrium position. k is a 'fudge' constant that's used...if you take a spring and compress over several distances, and plot against force used, you'll get (roughly) a straight line. So the force is approximately linear in displacement.

  11. anonymous
    • 5 years ago
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    x_i = 0 because it's not initially displaced from equilibrium. Some questions start with the spring already displaced and ask you to calculate the work done when you INCREASE displacement, say...in that case, x_i won't be 0.

  12. anonymous
    • 5 years ago
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    Whether k is positive or negative depends on whether you're measuring the force as the force exerted by the spring when it's displaced, or the force *you* exert on the spring when it is displaced. If it's displaced and not moving anywhere, the forces are balanced, and since the magnitude of the force is kx, if the force exerted by you is F(you)=kx, the force exerted by the spring must be F(spring)=-kx so that net force is F(you)+F(spring) = 0.

  13. myininaya
    • 5 years ago
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    Thanks for your help

  14. anonymous
    • 5 years ago
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    No probs.

  15. anonymous
    • 5 years ago
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    K is the spring constant, which is relative to each spring. In real life, that constant would be affected by things such as how thick the spring is and what the spring is made out of.

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