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What does F=ma really means?

  1. Oct 17, 2009 #1
    F=ma
    How do we define m and F? If their definition both come from this equation, then the equation doesn't really mean anything...We can also definite a F', and say F'=ma^2, and there isn't anything wrong... However, we have F=ma as something true, then F and m must be something true, indentifiable and measureable. So what does F and m really means??
     
  2. jcsd
  3. Oct 17, 2009 #2
    Er, where to even start.

    F=ma is a mathematical fit for the force experienced by an accelerating object. It isn't arbritraty as it can be found experiementally.

    F = force m = mass a = acceleration.

    You can assign any letter you want to any mathematical value. So it's perfectly valid mathematically to say that F' = Ma^2, but it dosent show anything in reality, its purely a mathemetical construct.

    F=MA shows something.
     
  4. Oct 17, 2009 #3
    In classical physics F and m are phenomenological quantities which are specified by other relations. So, you can take some amount of matter and on the basis of that assign to m some value. You can postulate some force law and then make predictions for the motion.

    Deeper explanations for m and F can only be obtained from fundamental physics (quantum field theory).
     
  5. Oct 17, 2009 #4

    f95toli

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    Sort if true. But the question asked by the OP is by no means trivial. There were whole books written about this back when classical physics was still "state of the art"( as far as I remember the "The Science of Mechanics" by Ernst Mach discusses this at great lenght) and and far as I know they never reached a generally accepted conclusion.
    I guess one could say that back then they were having the same sort of issues with classical mechanics (and thermodynamics) as we are having now with quantum mechanics.

    Also, some of the questions that come out of this are still very much "hot" topics in physics, e.g. why "heavy" (gravitational) mass is the same as inertial (the m in F=ma) mass. Meaning it touches upon topics such as the Higg's boson etc.
     
  6. Oct 17, 2009 #5

    arildno

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    xxChrisxx:

    This is totally wrong.

    cocosisi's question is very meaningful, and goes, for example, deep into the philosophical underpinnings of Einstein's General Relativity.

    Einstein noted that we are using essentially two distinct definitions of mass, namely inertial mass and gravitational mass (a fundamental assumption within G.R is that these have the same value).


    To take the case of inertial mass:

    Cocosisi:

    1. Suppose you have a mechanism that can be observed to impart accelerations to objects in contact with it (note that ACCELERATIONS are measurable (not that easy, but it CAN be done!))

    2. We say that this mechanism, BY WAY OF A FORCE, imparts acceleration to various objects.

    3. Let us ASSUME that the mechanism gives the SAME force to different objects; for example that our mechanism is a spring that we have compressed 1cm prior to bringing it into contact with the object.
    (That assumption can be given empirical verification that by performing NUMEROUS experiments on some set of objects, each of those objects gains the same acceleration it did on a prior trial)

    4. We observe, however, that different objects gains DIFFERENT acceleration from what we can arguably say is the same force.

    We may therefore construct, within this particular experimental setting, an object-connected scale of ratios F/a, i.e, a "mass scale" We can pick out one object, O_{0}, whose acceleration equals [itex]a_{0}[/itex], and we set the value of the common force, [itex]F_{0}[/itex], equal to a_{0}.

    Thus, that object is given "unit mass", and the rest will be assigned various mass values directly related to the acceleration they experience.

    5. Now, some non-trivial results can be observed within the setting:
    If we take two objects, and makes them stick to each other so that they move as one object, then that compound object's ACCELERATION will be found as if we could add the masses of the two objects together and find the compound mass (i.e, degree of resistance to acceleration) from that.
    Thus, mass is seen to be an ADDITIVE property, that certainly is consistent with the (as yet unproven) idea that mass is some constant property belonging to the object itself, rather than some meaningless, accidental correlation!

    6. By fiddling with two objects, 1 and 2, we find that in the above experiment, the relation:
    [tex]\frac{m_{1}}{m_{2}}=\frac{a_{2}}{a_{1}}[/tex]
    holds.

    Now, SUPPOSE that masses are something intrinsic to objects, "forces" something that this mass-property is independent of, something that merely "happens to" the object.

    What would this entail?

    It would entail that if we did ENTIRELY DIFFERENT experiments (than the spring example, compression of 1cm), then the ratio of observed accelerations would REMAIN CONSTANT, also when we have absolutely no reason of thinking that the forces are generated by the same mechanism as before, and for that matter, have another value than the one in the spring example.

    7. And this is indeed what we see!
    The ratio [tex]\frac{m_{1}}{m_{2}}[/tex] (and therefore acceleration ratio) remains CONSTANT, irrespective of what type of force is applied to both objects.

    8. The accelerations are seen to change, but we are fully entitled to regard the masses as being constant properties of the objects themselves.

    9. Now, we can utilize this discovered property to gauge forces of a variety of forms, and thus to create force scales as well.
     
  7. Oct 17, 2009 #6
    You are talking to an engineer here not a physicist. I run with arms flailing from GR, Newtonian mechanics all the way.

    On reflection it does seem that the OP's question could be asking something deeper than what is 'F' and 'M' in the most basic context.
     
  8. Oct 17, 2009 #7

    arildno

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    And an engineer ought to know everything about how to construct mass and force scales, and WHY they can be regarded as meaningful constructions that most likely have some important relation to actual reality.
     
  9. Oct 17, 2009 #8
    All I need to know from F=MA is; shove something and it moves. Shove it twice as hard it accelerates twice as fast.
    F=ma can be regarded as meaningful becuase... thats what happens in reality. We know this to be true.
    For practical purposes it really doesnt need to be more complicated than that.
     
    Last edited: Oct 17, 2009
  10. Oct 17, 2009 #9

    atyy

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    This equation is the definition of F. m is just a constant of proportionality, defined to make the equation true.

    The goal of classical physics is to find how properties of objects (such as their colour, texture, size etc) cause them to produce or react to forces. For example, the the electric force is produced by a property called "electric charge", and F=(q1.q2)/(r.r).
     
  11. Oct 17, 2009 #10

    arildno

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    And on what basis do you really define what it means to "shove twice as hard"??

    cocosisi's question is highly apposite.

    IF, for example, we REALLY had a relation:

    F=m(F)*a,

    then shoving "twice as hard" wouldn't necessarily result in twice the acceleration.
     
  12. Oct 17, 2009 #11
    There is really no point in going into this deeper. As what 'shove it twice as hard' means is pretty much what you put a few posts above (post 5), albeit in a very wordy fashion.

    The fact that you can explain it better than I can makes no odds and is the reason I would make a crap teacher.
     
  13. Oct 17, 2009 #12

    Dale

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    In most consistent systems of units (including the SI system) F=ma would define force, and mass would be defined in some other way (e.g. by reference to a standard mass).
     
  14. Oct 17, 2009 #13
    Mass is the total energy content of an object. The (generalized) force is defined as minus the derivative of the energy content w.r.t. the external variable.

    If you have two atoms at some distance, then the force is given by minus the derivative of the total energy of the system w.r.t. the distance. And the energy can be obtained by solving the Schrödinger equation. A fully relativistic treatment will give you the total energy and thus the mass.
     
  15. Oct 17, 2009 #14

    arildno

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    Well, you could for example use Newton's 3.law to postulate the equality of two forces within a couple, and then use the ratio of observed accelerations to define a scale for inertial mass.
     
  16. Oct 17, 2009 #15
    SI units are not consistent. In SI units we use inconsistent units for Time, Length, Mass, Temperature etc. etc. This comes at the price of having to introduce extraneous conversion factors (c, G, hbar, k_b, etc.) in formulae.

    From a metrological point of view, however, it is advantageous to have different physical standards for physical quantities, even if they are related by fundamental physics. As technology advances, it can happen that we can rely on fundamental physics to perform accurate measurements. If that enhances the acuracy whit which you can perform measurements, the SI system will be revised. That happened in 1983 when the meter was redefined in terms of the second by spcifying a value for the speed of light.

    For historical reasons we still use different units for lengths and time intervals, but we could just as well measure time in meters.
     
  17. Oct 17, 2009 #16

    Dale

    Staff: Mentor

    The word you are looking for is "natural"; the SI system of units is consistent but not natural.

    Consistent simply means that the base units and the derived units in the system don't have any conversion factors between them. For example, in the SI system the unit of power is 1 W = 1 kg m²/s³, but in the US system the unit of power is 1 hp = 550 ft lb/s. The US system is inconsistent because it needs a conversion between a derived unit and the base units with the same dimensions.

    If a system of units is "natural" then some of the dimensionful universal constants are numerically equal to 1. You can think of these as conversion factors between units of different dimensions. A "geometrized" system of units takes this one step further and considers the universal constants to be not just numerically equal to 1, but also dimensionally equal to 1 (i.e. c is unitless). This means that all of the fundamental units have dimensions of length.
     
  18. Oct 17, 2009 #17
    I see! I prefer to use geometrized units. :approve:
     
  19. Oct 17, 2009 #18
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  20. Oct 17, 2009 #19
     
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  21. Oct 17, 2009 #20
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  22. Oct 17, 2009 #21
    I think that's very good question for which one has to learn to get away from "taking everything your high school teacher said as law" to "self-made insights"

    Force is defined as the flow of momentum
    [tex]F=\frac{\mathrm{d}p}{\mathrm{d}t}[/tex]
    and together with the law of action and reaction, it turns out your question is completely equivalent to:
    ...asking why momentum
    [tex]
    p=mv
    [/tex]
    is conserved and how to determine the m in that equation.

    Well, I assume all know forces between two objects (gravity, electromagnetism,...) experimentally obey
    [tex]
    v_1=av_2+b
    [/tex]
    at any moment. And therefore we can define mass ratios by
    [tex]
    \frac{m_2}{m_1}=a
    [/tex]
    which yields [itex]m_1v_1+m_2v_2=\text{const}[/itex] which we wanted.
    It also turns out the mass of a particle does not change over time. So now we know how to determine the mass of a particle (which we can tabulate) and also we accept conservation of [itex]mv[/itex] as a law.

    All these consideration are of course classical. I'll think about what one has to keep in mind taking into account relativity...
     
    Last edited: Oct 17, 2009
  23. Oct 18, 2009 #22

    arildno

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    It tells us that it is a consistent, i.e, meaningful construction.

    Furthermore, by working FROM that construction, we have found that it is an extraordinarily USEFUL construction.

    I think that's quite a lot!
     
  24. Oct 18, 2009 #23
    What is tricky thing that mass, velocity exist in any small time interval. But acceleration in infinityly small time interval turns to zero, when dt limt->0 then acceleration apears to disapearing and instead accleration in this small interval you see constant speed v. Acceleration is in our heads, but not in some time interval or in some certain time interval. Force good describing gravity F=mg. If we add h then we know how long force was upon object of mass m. For instance, if m=1 kg and g=10 m/s^2, then after 10 seconds object speed will be 100 m/s. And average speed will be 50 m/s after 10 seconds. S=v*t=50*10=500 meters will fall 1 kg stone withing 10 seconds on the ground from height 500 meters. So force was acting upon stone 10 seconds, so how would you call such action on 1 kg stone? ?=F*t=m*a*t. I would think it could be somthing between kinetic energy W=mvv/2 and momentum p=mv. Why ones better than overs? Maybe I should call it job A=mat. But seems we already have work formula. So if 1kg stone after 10 s reach speed 100 m/s, so this fine gives momentum m*v=1(kg)*100(m/s)=100 (kg*m/s) and m*a*t=1(kg)*10(m/s/s)*10(s)=100 (kg*m/s).

    Edit: I would describe job A=m*a*t or A=m*g*t if there no any contra force. For example if stone fall in vacum tube onto earth from 500 m height then his job is velocity, which stone will reach till fall on ground multiplied with mass, acceleration and time A=mat=1*10*100=100 kg*m/s. And now what is work? Work is when rocket with acceleration a=20 m/s^2, agains gravity reaching height 500 meters. Air friction ignore (supose rocket flying up in some vacum tube). So force which is contra force to rocket raise force is gravity force or acceleration g=10 m/s/s. So rocket mass m=1000 kg and mass disipation because of fuel decreasing is almost invisble or let just ignore it. So gravity force is G=10(m/s^2)*1000(kg)=10000 N. And rocket lifting force F=20(m/s/s)*1000(kg)=20000 N. So total force is R=F-G=20000-10000=10000 N. So rocket will be seen as falling like stone on eeart in vacum, but in oposit direction. So rocket will reach height 500 meters in 10 seconds. So rocket kinda engine make job A=mat=20(m/s/s)*1000(kg)*10(s)=200000 kg*m/s. This is real work-job which made rocket reaching such height.
    Now supose we add friction of air to rocket which will be a force counter to force which lifting rocket, this force may be about 4 m/s/s. So then rocket after first second will got speed 20-10-4=6 m/s. After next second rocket from first to second second will reach speed 2 m/s and total speed will be 8 m/s (after two seconds when rocket start to fly from ground). But here need integration since speed of rocket also deacrease and for second-next second since it flying at bigger speed friction too will be bigger. So as you gues here need real work formula W=mvv/2=1000*100*100/2=5000000 J and this units teoreticaly equal to job, which rocket made without friction so with air friction it takes 5000000/200000=25 times more energy to reach same speed with same momentum of rocket.
    Or another maybe simpler example. If I want just to accelerata 1 kg stone in cosmos to 100 m/s then for it need A=mv=1*100=100 (energy units). And if I put this stone on some very long desk and again trying it accelerate to speed 100 m/s, then here again we meet desk and air friction, which is say F=ma=1(kg)*3(m/s^2)=3 N counterforce. So if I just push stone with 3 N force then stone will not move, but if I push with 6 N force then stone will move as if he would be in vacum, in free space and I would push stone with 3 N force or a=3 m/s^2 acceleration. So this time need to use kinetic energy formula to find out how need energy to push stone 500 meters distance. And this would look like this: W=mgh=1(kg)*3(m/s/s)*500(m)=1500 J. Such distance will be reached after (let me see 3*10(s)=30 m/s, average v=30/2=15; 3*100=300 m/s, average v=150 m/s; 500/150=10/3; (3*100*10/3)=1000 m/s, average v=1000/2=500 m/s | let me try different way, 3 m/s after 1 second and average speed 1.5 m/s, from 1 s to 2 second speed will chanes from 3 to 6, so 6/2=3 meters per second second will move stone, 6 per third second, 12 per four s, so 1.5+3+6+12+24+48+96+192=382.5 meters after 8 seconds or 382.5+192=574.5 m after 8.5 s) about 8.3 seconds. And velocity will be about 8.5(s)*3(m/s/s)=25.5 m/s. So kinetic energy W=mvv/2=1*25.5*25.5/2=325.125 J. Somthing wrong 500/3=166.(6) m/s. W=mvv/2=1*(500/3)^2/2=13888.(8). mvv/2=1500, v=2*1500^(1/2)=77.46 m/s. Whatever...
    Edit2: Speed raise: 3 m/s, 6 m/s, 9 m/s, 12 m/s, 15, 18, 21... Distance raise: (0+3)/2+(3+6)/2+(6+9)/2+(9+12)/2+(12+15)/2... or 1.5+4.5+7.5+10.5+13.5+16.5+19.5+22.5+25.5... Or 1.5*(3+6+9+12+15+18+21+24+27+30+33+36+39+42+45)=1.5*315=472.5 m. And this take 15 seconds. 45*1.5=67.5 meters will move per fifteen second. 3*15=45 m/s. Still less than 77.46 m/s. What the hell it is?
    Edit3: It's ok I just wrong vv/2=1500, vv=1500*2, v=3000^(1/2)=54.77 m/s, so very good with almost 45 m/s agree. But still need more.
    Edit4: mah=1*3*472.5=1417.5 J. vv/2=1417.5; v*v=2835; v=53.2447 m/s. Even closer to go. No per first second reaching speed 3 m/s, per second second 6, per third 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45 and this all is 15 seconds per which stone reaching 472.5 m, but still cann't get that mah=mvv/2.
    Edit5: Maybe everybody wrong and acceleration a means, not how much speed increasing in one second, but how much flying distance increasing in one second? Let's check this out: 3(m)+6(m)+12(m)+15(m)+18+21+24+27+30+33+36+39+42+45+48+51+54=513 meters and this we get in 17 seconds. But then we do not know speed of stone after 17 seconds. But we can calculate it. If 3 m falling per first second then average speed is v=3 m/s. And speed in the end of first second is 6 m/s. And average speed when per second second stone fall 6 m, average speed means 6 m/s, so at ve end we get 54 m/s speed. And now it need just put into two kinetic energy formuls comparision. mah=1*3*513=1539 J. mvv/2=1539; vv=3078; v=55.4797 m/s. Still do not much with 54 m/s. What I doing wrong (supose 1kg stone falling on the ground from 500 m heigh without air resistance and need to know stone velocity when it reach/hit ground, when g=3m/s^2)?
    Edit6: Seems I wrongly calculate here: 1.5*(1+3+6+9+12+15+18+21+24+27+30+33+36+39+42)=1.5*316=474 m and it's after 15 seconds. So speed will be 3*15=45 m/s. mgh=1*3*474=1422 J. vv/2=1422, vv=2844, v=53.329. Still wrong.
    Let's try everything to calculate clearly. (0+3)/2(m)+(3+6)/2+(6+9)/2+(9+12)/2+ (12+15)/2+(15+18)/2+(18+21)/2+(21+24)/2+ (24+27)/2+(27+30)/2+(30+33)/2+(33+36)/2+ (36+39)/2+(39+42)/2+(42+45)/2+(45+48)/2+(48+51)/2+(51+54)/2= 1.5+4.5+7.5+10.5+13.5+16.5+19.5+22.5+25.5+28.5+31.5+34.5+37.5+40.5+43.5+46.5+49.5+52.5=486 m. And it's after 18 seconds. mah=1*3*486=1458 J. mvv/2=1458, vv=2*1485=2916, v=54 m/s. And now it's finaly everything much. After 18 seconds stone falling with a=3 m/s/s will reach ground with speed 54 m/s. And if there no resistance, but only force in cosmos F=ma=1*3, then job will be A=mat=1*3*18=54 kg*m/s or energy units. So with friction and moving stone on table 486 metrs will cost 1458/54=27 times more energy. Almost all energy will go to friction, only small part of it will be used for stone acceleration. As far as I can see now.

    Edit7: Back to the rocket problem. Rocket of mass 1000 kg have engine which in free space accelerating with force F=ma=1000*13=13000 N. So a=13 m/s/s. Ignore air friction and rocket fuel waist. Only gravity acceleration is g=10 m/s/s. So if I would put 1000 kg object on table and if I would push it agains force of gm=10000 N and distance 500 meters then I would waist W=mgh=1000*10*500=5*10^6 J. Since there is gravity then rocket will fly with acceleration a-g=13-10=3 m/s^2. Such distance rocket will doen in 18 seconds or let it be 486 meters distance from previous example with stone. So after 18 seconds rocket will get speed 54 m/s and would fly distance 486 . Flying such distance with no gravity force (because gravity force we calculate separatly and get 5 milions joules) rocket engine will spend same amount of energy as if it I moving on table against force 3+3+3+3+3... meters distance sine rocket flying 18 seconds then it's willl be like distance 18*3=54 meters. So energy will be spend A=m*a*S=1000*3*54=162000 J or A=m*a*t=1000*3*18=54000 J - we will later figure out which is not wrong. So rocket total waist 5*10^6 + 162000=5162000 J. And get moment p=mv=54000 kg*m/s. This moment can lift rocket distance of like if rocket would falling after some amount of seconds. This distance of falling with 10 m/s/s acceleration of gravity will be (0+10)/2+(10+20)/2+(20+30)/2=5+15+25=45 m. So after about 3 or more precisly 3.1 second rocket would fall distance 54 m. And rocket speed from moment is also 54 m/s. So rocket will fly another second 54 m height till it stop. So rocket total height will be 486+54=540 meters and it's only with energy of 5162000 J (more precisly it would be mgh+maS=1000*10*486+1000*3*54=5022000 J).
    Now assume rocket engine pushing rocket in free space with acceleration of 20 m/s^2 and with gravity it would gives us only 10 m/s^2. Rocket spend energy for overcoming gravity all the time will be mgh=1000*10*500=5*10^6 J. Till rocket reach 500 meters it till took (0+10)/2+(10+20)2+(20+30)/2+35+45+55+65+75+85+95=500 meters. So in egzactly 10 seconds rocket will reach height 500 m and speed 100 m/s. So rocket moment this time will be about two times bigger mv=1000*100=10^5 kg*m/s. And rocket when turn off engine at 500 m height will fly up about 5 seconds because (0+10)/2+(10+20)2+(20+30)/2+35+45=125 or in 4 seconds it will be 80 m heigh, so in about 4.5 seconds rocket would fall 100 meters. And it have 100 m/s speed. So rocket will lift itself 100 more meters and will stop. On the over hand perhaps I should calculate what average speed rocket gonna make that it would be 100 m/s if rocket would fall some amount of time because of gravity. But this is also is average speed or average distance flyied per roughly 4.5 second. So this time total wasted energy will be mgh+maS=5*10^6+1000*10*S, S this time will be a*t=10*10=100 so W=mgh+maS=5*10^6+1000*10*100=6*10^6 J. Or mgh+mat=5*10^6+1000*10*10=5100000 J. Which answer is true I will try to figure out by if rocket always fall from 500 m height and how much energy need for acceleration and after what time it would stop if instantly accelerated rocket (with moment mv=1000*v) would be puted on the ground and for it would be let to fly until it will stop.
    Edit8: So if would rocket fall from 500 m height without friction on earth ground and if g=10 m/s^2. Then rocket would fall with final speed 100 m/s. To reach such speed acording to A=maS=maat formula need A=m*a*a*t=1000*10*10*10=10^6 J. And rocket have 10^5 kg*m/s moment. Now if we instantly teleportate rocket witch in space flying with 100 m/s speed to the earth ground then teoreticaly it should fly to height of 500 m and to stop. So I calculate, that energy would be wasted 5 times less. How it can be? Maybe because after first second acceleration took from rocket 10 m/s speed and after first raising second rocket speed will drop to 90 m/s, but not to 95 m/s like if would be with average deaceleration gravity speed 100-5=95. So then after ten seconds rocket speed will fall to 0 m/s, because after second second it would be 80 m/s. But even with average speed of deaceleration each time substrackting should lift rocket to desired 500 meters. 100-5-15-25-35-45-55-65-75-85-95=-400, so now I see, that it's not enough such a speed of 100 m/s to fight with gravity till 500 m. So rocket should be accelerated to 500 m/s to overcome gravity and reach 500 m and only then to stop.
    Edit9: So to accelerate 1000 kg rocket with engine which accelerating rocket with 3 m/s/s acceleration to 54 m/s need 18 seconds and now how much seconds need to accelerate rocket to 486 m/s? Don't you think guys it would be easier to find out with mvv/2, but I not sure if h supose to be 486 m in mah formula, so need to do old work 3, 6, 9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,61, or just simply 486(m/s)/3(m/s^2)=162 seconds. So after 162 seconds rocket reach 486 m/s speed if rocket engine in empty free space accelerating it with 3 m/s^2. So according A=maat=maS=1000*3*3*162 and we get then energy against gravity. And here m is rocket mass 1000 kg, first a means toghter with m which force F=ma=1000*3=3000 kg*m/s^2 resist through the distance when we pushing object of certain mass m=1000 kg. Second a means that we also pushing objects 3 m/s speed, because since is friction object speed do not increasing and times t we get a*t=S - the distance which we push 1000 kg object against force 3000 N at constant speed 3 m/s. This is real job if you put on table 1000 kg rocket and will push it against friction which would be 3000 N at speed 3 m/s then you will do job A=maat=1000*3*3*162=1458000 J. And rocket with moment mv=1000*486=486000 kg*m/s is able to reach 486 m height if it would be teleportated with such moment on the ground and instantly will fly up. So after first second rocket speed will deacres from 486 m/s to 486-(0+10)/2=481 m/s, after second second deacreas to 481-(10+20)/2=466 m/s, after third second to 466-(20+30)/2=441 m/s, then to 441-35=406 m/s, then 406-45=361 m/s, then 361-55=306 m/s, 306-65=241, 241-75=166, 166-81, 81-95=-14 m/s. So after 9 seconds speed decrease to 81 m/s and after about 9.8 second speed will deacrease to zero and rocket will stop at 486 m height.
     
    Last edited: Oct 19, 2009
  25. Oct 18, 2009 #24

    Dale

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    They do have a certain appeal, and I actually don't have trouble giving time units of length, but I could never quite get my head around doing the same for mass and charge for some reason. I guess it is just a failure of imagination on my part.
     
  26. Oct 18, 2009 #25

    arildno

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    I think geometrized units are silly. Why not chargified units instead?
    They would sizzle, at least..:smile:
     
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