I finally got the values!
vi = 0
d = 5.1m
m = 1.00 kg (nice and round!)
t = 0.92
k=(1/m)*(F/v)^2
Assuming the "v" in the formula is the end speed.
a = (vf-vi)/0.92
2a + vi = vf
vf = 0.92(9.8)
vf = 9.016
k = (1/(1.00)) * (F/(9.016)^2
Is F the net force experienced by the box...
Not at all. I know:
V1 = 0m/s
a = 9.8 m/s
\Deltad = 5.1m (reformed, not 5.2)
\Deltat = ?
Thus, using the equation
\Deltad = V1\Deltat +(1/2)a\Deltat2
I can solve for time:
5.1 = (1/2)(9.8)\Deltat2
\sqrt{}(5.1/4.9) = \Deltat
THerefore \Deltat = approx 1.02 seconds.
That would be the time if...
Alright. Can I safely subtract from the kinetic energy if I take into account that the force is reduced due to crumple zone and parachute, without messing anything up?
For instance if the kinetic force without reducers upon hitting the ground is 6m/s, but with reducers is 5 m/s, therefore...
Correct. I figured out that an egg cannot sustain more than 4.9 Newtons of force.
Unfortunately I will not be able to discover the values today, it will have to wait for tomorrow.
Well, that's good news!:smile:
Not too worry, I will get all the values I can tomorrow, try it out, and then return to this forum to either confirm the results (unlikely), or get more help (very likely).
Thanks to you three guys for helping me!
I will be using a parachute and crumple zone which will reduce the force dramatically, I just wanted to use a simple example to help introduce myself to this new concept.
So, just to clarify, will the force I find be less than if I just dropped the box?
Alrighty, so:
h = 5.2m
m = 1 kg
g = 9.8
Eg = mgh
Eg = 50.96
50.96 = (1/2)kx2
101.92 = 50x2
x = sqrt(5096).
x = 71.39
Now, what do I do? Do I work my way back to force, and find the difference?
oh, using the equation (1/2)kx^2, I can figure out the elastic potential per spring, if I discovered x using a 1kg weight on 1 spring. And from elastic potential to...
Assuming that:
Mass = 1 kg
Acceleration = 9.8 m/s^2
I know that F = 1 N.
Thus, if F = kx, and x is an assumed 0.02m, then:
k = 50.
so... now what? (and what unit is k measured in?)
Yes.
You are given two values, and you need to solve for one of them using an equation with four variables, three of which are known (the other being g).
x is the displacement of the end of the spring from its equilibrium position. So, if I were to press down on the springs with the same force it experiences when hitting the ground, would I measure from where the top of the spring was to where it is now, or from the middle to where it is now...
Unfortunately we will not be covering Hooke's law nor other momentum questions.
So (based on your link), are you suggesting that I use the F = dp/dt equation in order to find the reduced force on the box due to the springs?
BTW, is the force itself reduced, or just spread out over a...
I'm in grade 11... I've tried multiple other forums (such as XKCD's) and sites (such as Answers.Yahoo.com), but either there was insufficient knowledge or bashing. I figure I'd give this forum a try.
As for the actual math, I am aware of its complexity and the fact that calculus may be...