kazaa81
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Just a simple physics question....
Hi,
potential energy of a body at a certain height is measured in joules and calculated from the formula:
Ep = m*g*h
where:
m=body mass
g=gravity ac. (9,81m/s2)
h=height where the body is.
Now, if I want the result in kilograms instead of joules, what do I do?
Think, for example, which I weight 50kg; I fall from 2 meters of height. What is the force (strenght) in kilograms which I've released to the
ground (pressure doesn't interest me now, first I need to know how much kg, then kg:cm2=pressure)?
P.S. As you've read, it's an example....because if I fall/jump from 2 metres I will probably die, not like in films, where people jump like
cats!
Thanks at all for help!
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Blind Angel
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| Quote: | From Wikipedia
Definition
The joule is a derived unit defined as the work done,or energy required, to exert a force of one newton for a distance of one metre, so the same
quantity may be referred to as a newton metre or newton-metre (also with meter spelling), with the symbol N·m or N m. It can also be written as
kg·m2·s−2. However, the newton metre is usually used as a measure of torque, not energy.
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so divide the Joule (J) by your mass (M) (thus eliminating the body mass in you calcul) and you'll be left with Newton (N). And pressure is N*m²
(a Pascal (Pa)). So your final calculation would mostly be:
Pa = Mgm²
where:
Pa is the pressure you'll exerce
M is your mass
g is the gravitic constant
m² is the area of your feet
It seem that the final speed doesn't matter in that, maybe I'm not righ, physic isn't my strong point. But i'll post this anyway
so somebody will surely correct me if I'm wrong.
Oh, and fallinf from 2 meter isn't that a big deal, you'll probably break a leg if you fall on cement, but not on grass (migh be hard on the
vertebral collum though), and if you know how to hit the ground well, usually by rolling, you can get close to nothing.
[Edited on 8-10-2005 by Blind Angel]
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kazaa81
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Thank you, Angel!
So, it will be (if I weight 50kg, and have 700cm2 of feet):
Pa= 50*9,81*0,07
so Pa = 34,335 Pascal per m^2.
However, supposing I fall from 10cm and weight 50kg on a scale (person-weighter thing....), what weight would be reached in the strict time (peak of
maximum weight showed)?
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12AX7
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You can't answer to that question. You're fussing around with units in a way the units don't like to work. In particular it's
the time frame.
What you're dealing with here is an impulse. If your body were perfectly rigid, then the moments before you touch the ground, force would be
zero; the instant you hit the ground, force goes to infinity as it stops you in exactly zero time. (This is a Dirac delta function, since the total
is finite, but to get there, you have an infinity of zero -- clearly not a computable situation!)
If however you define your legs as springs (such that F = kx), then the instant your feet first touch the ground, force is zero. (As well it should
be -- force should go smoothly from zero, before there was contact, up to some maximum value. Nature doesn't allow for mathematically abrupt
events.) As your center of mass continues to fall, the spring is compressed (x rises) and consequently, F rises in proportion. This force begins to
decelerate you, and there will come a point in time when velocity towards the ground reaches exactly zero.
(Note that, at this time, the spring continues to exert its force, meaning after this point in time, you will rise -- with a perfect spring and no
wind resistance, you will in fact bounce forever. With realistic legs, the force is not proportional to compression, and the force is variable by our
muscles.)
Now to expand on this event, the time between the moment your feet touch the ground and the moment your mass stops moving is how long it took to
decelerate. As concerns energy, you had a certain kinetic energy before the event, and now you have zero. (By conservation of energy, it is now
stored in the spring as potential energy, or in reality, simply dissipated as friction and biological work by your legs.)
Let me introduce the concept of power: one joule spent in one second is one watt. One joule spent in one-tenth of a second is ten watts. Power is a
very important concept because a few pounds of TNT might have the exact same energy as a candle, but damned if one isn't horribly destructive
while the other is weak and slow. The difference is time: a detonation might release all its energy in one milisecond (or less), while the candle
burns for perhaps days continuously.
This applies to our situation by exchanging gravitational potential energy to heat (i.e., total loss, we aren't using it for anything else). The
same gravitational energy is released whether you walk down a flight of stairs or jump out of a window of the same altitude, but the speed is
different. Also, the impact takes the total of that energy, while when walking down the staircase, the energy is dissipated at the same time it is
given. Nonetheless, the only difference is time in which the total amount of energy is released - when walking down, it's over the entire trip
of say ten seconds; when falling, it's over the entire time of impact, maybe 1/100th the time.
To return to my very first sentence, you cannot answer this question exactly -- as you can with E = mgh above -- because you need to
know the duration of time you will stop in, the effective spring constant of your legs (if you even land on your legs!) and so on.
Angel: "force on the ground" is sum of the pressure, so there isn't much use of area of the feet here. This is an impulse problem and
the magnitude of the impulse depends on time.
To help you imagine this, just consider that it takes F = mg to keep you at the same altitude. This is the average force your legs (and the ground)
have to support to stand there. If you jump, your legs leave the ground and force drops to zero - until you land, in which case you now need to
decelerate, which costs additional force. In summary, if you are alternately off the ground for 0.9 second and standing for 0.1 second repeatedly,
you'll have to exert 10 times your weight's force during that 0.1 second. This is a duty cycle thing.
Tim
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kazaa81
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Thank you 12AX7,
you've cleaned up a little my ideas....so, it's not possible to obtain straight-way a result...
Excluding friction, in the fall the important varibles are mass, gravity ac. and probably final speed?
So, if we obtain speed when about to reach the ground, supposing your starting speed was 0, from 2 metres it will be
Vf=9,81*time
Now, can we apply this to kinetic energy to obtain a valid result of what I'm searching for?
Ek = ½*m*Vf^2
Please correct me if I've made mistakes.
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12AX7
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Yes that will get your kinetic energy at the bottom, just before impact.
Tim
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neutrino
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E<sub>k</sub> = 1/2mv<sup>2</sup>
I'm not sure where the f in your equation came from.
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Ramiel
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potential energy is converted to kinetic energy as the body falls towards the massive object... E<sub>TOTAL</sub> =
E<sub>k</sub> + E<sub>p</sub>
Caveat Orator
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