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Thread: Spring-Mass help

  1. #1
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    Spring-Mass help

    Need help solving this spring mass system. Any help is appreciated.
    Attached Thumbnails Attached Thumbnails Spring-Mass help-bungee-jump.jpg  
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  2. #2
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    Re: Spring-Mass help

    Looks like an interesting problem. But to "help" we really need to know what kind of help you need. Do you know what a "k value of 8 lbs/ft" means? Do you know what an "equilibrium position" is? The basic physics law here is "force equals mass times acceleration".
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  3. #3
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    Re: Spring-Mass help

    I'm not sure how to go about the problem at all. What formula to use, what to plug into where...it's all confusing to me. A step-by-step lay out of what to do would be very helpful.
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  4. #4
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    Re: Spring-Mass help

    First fix units (unless you choose to carry units along with all your variables): Distances are feet. Forces are pounds (lbs). Time is seconds. Remember that a person's "weight" is not the same as a person's *mass*. Weight = mass times g, where g is the acceleration due to gravity at Earth's surface.

    Second, understand the coordinate system (which isn't the one I would've chosen, but the problem specifies it, so you should go with it). The bridge is at x = -100. The non-bridge end of the free-hanging (no person attached) chord is at x = 0. The positive x direction is down. The water is at x = 120.

    The force of gravity is in the positive x-direction. The spring force from the chord is in the negative x-direction.

    The forces acting on a jumper are: gravity alone when -100 <= x <= 0, and both gravity and the chord when x >= 0.

    That "k value" means the spring constant for use in Hook's Law.

    For Question #2, I'm not entirely sure of the meaning, but I my best guess is that it means this: By equilibrium position it means the value of x where, eventually, the bungee jumper would hang after the oscillation died down... it would be some positive x-value. It's the point where there's no net force on the jumper. By "velocity at the equilibrium position" it means, I think, the instantaneous velocity at the moment when the jumper first reaches that equilibrium x value on their downward fall (and since it will be positive, the jumper will have a lot of downward velocity at that moment, and so is continuing down post it towards the water).
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  5. #5
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    Re: Spring-Mass help

    equilibrium position is where $F_{net} = 0 \implies kx = W$
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    Re: Spring-Mass help

    Appreciate the help but still lost on what equation I should plug all of this information in to.
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  7. #7
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    Re: Spring-Mass help

    solving this problem involves more than one equation ... in addition to the equation I provided for finding the position of equilibrium you will need, at a minimum, one formula from kinematics with uniform acceleration; formulas for mechanical energy (kinetic energy, elastic potential energy, and gravitational potential energy) in order to use conservation of energy principles; and equations for motion of a spring oscillator.
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  8. #8
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    Re: Spring-Mass help

    What equation to find how long she will be in free fall?
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    Re: Spring-Mass help

    Quote Originally Posted by afeldler View Post
    What equation to find how long she will be in free fall?
    one of these ...

    $\displaystyle v_f = v_0 - gt$

    $\displaystyle \Delta y = v_0 t - \frac{1}{2}gt^2$

    $\displaystyle \Delta y = \frac{1}{2}(v_0 + v_f)t$
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