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myininaya
 5 years ago
Someone explain springs and work to me?
myininaya
 5 years ago
Someone explain springs and work to me?

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0How 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

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0I know how to do the integration

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0Just the setup is my problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hooke'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.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0So are you an engineer?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0If 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?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so x is the distance over the interval xi to xf

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x_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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whether 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0K 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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