Spring-Mass help

Apr 2015
7
0
New Jersey
Need help solving this spring mass system. Any help is appreciated.
 

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Apr 2005
20,249
7,909
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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Apr 2015
7
0
New Jersey
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.
 
Sep 2012
1,061
434
Washington DC USA
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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Jun 2008
16,216
6,764
North Texas
equilibrium position is where $F_{net} = 0 \implies kx = W$
 
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Apr 2015
7
0
New Jersey
Appreciate the help but still lost on what equation I should plug all of this information in to.
 
Jun 2008
16,216
6,764
North Texas
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.
 
Apr 2015
7
0
New Jersey
What equation to find how long she will be in free fall?
 
Jun 2008
16,216
6,764
North Texas
What equation to find how long she will be in free fall?
one of these ...

\(\displaystyle v_f = v_0 - gt\)

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

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