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Want to understand ke formula without having to accept any assumptions

  1. May 14, 2014 #1
    To prepare for a future physics course, I googled Newton’s Laws and then found about the Work Energy Theorem. I can see how Newton’s Laws makes sense but I don’t get why the kinetic energy formula is correct. The only experiment for the Work Energy Theorem I can find does not meet the requirements of the scientific method unless you make an assumption. Can someone show me why the work energy theorem is valid without making any assumptions.
     
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  3. May 14, 2014 #2

    Dale

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    No. You have to make assumptions. That is part of science. You make assumptions and then you test those assumptions against experimental evidence.
     
  4. May 14, 2014 #3

    Doc Al

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    Typically, the work energy theorem is derived from Newton's laws. Do you have questions about the derivation?
     
  5. May 14, 2014 #4
    What assumption is that?
     
  6. May 14, 2014 #5
    The experiment I'm referring to is the one that uses a table with a pulley and an air track. Basically, I see that this experiment demands that you must accept that work (force x distance) is already a proven fact. If you assume work is scientifically valid, the experiment works (no pun intended) as advertised. However, if you do the exact same experiment and make a different assumption (measuring the time force acts instead of distance force acts through) it proves something else entirely. It shows that ft =m|v|. (I know that momentum cannot be used to represent energy but the scalar version of impulse and momentum could.)

    To me, this means the experiment I see discussed on the internet requires that I take the validity of work on faith; and that does not sound very scientific. Is there an experiment that does not require that particular assumption? Maybe, there is an experiment that converts, say, electrical energy into kinetic.

    By the way, my thanks to everyone willing to help me out.
     
  7. May 14, 2014 #6

    UltrafastPED

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    This is a definition of what we mean by work!

    See http://en.wikipedia.org/wiki/Work_(physics [Broken])
     
    Last edited by a moderator: May 6, 2017
  8. May 14, 2014 #7
    Work: [tex]W = \int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt[/tex] that is a definition.


    [tex]\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = \frac{1}{2}mv^2(t_1)-\frac{1}{2}mv^2(t_0)[/tex]

    that is a mathematical theorem, applicable when [tex]\vec{F}(t) = m\vec{a}(t)[/tex], i.e. relative to an inertial frame.


    Also, Impulse: [tex]\vec{I} = \int_{t_0}^{t_1}\vec{F}(t)dt[/tex] that is a definition.

    [tex]\int_{t_0}^{t_1}\vec{F}(t)dt = \vec{P}(t_1) - \vec{P}(t_0)[/tex]

    that is a mathematical theorem, applicable when [tex]\vec{F}(t) = \frac{d}{dt}\vec{P}(t)[/tex], i.e. relative to an inertial frame.


    These are just mathematical definitions and mathematical theorems, nothing weird here.
     
  9. May 14, 2014 #8

    A.T.

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    That would be based on the definition of electrical energy, instead on the definition of work. How would that be any different?
     
  10. May 14, 2014 #9
    I might be confused but, isn't one of the ideas about energy that it changes form and it's supposed to be conserved. Last year when I was fooling around with solenoids, I found that I could change electrical energy into what I now know as mechanical energy. After remembering I had done that, I figured that somebody must have confirmed the work energy theorem in a laboratory using precise measuring equipment doing something like that.


    I don't want to be a pain about this but if I'm going to study a science, I figure I should be allowed to see the experimental proof for something as simple as the work energy theorem. I don't want to take work or the kinetic energy formula on faith. I don't think I should be forced to accept something just because every website that I found on google tells me it is true.
     
  11. May 14, 2014 #10

    UltrafastPED

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    This is usually covered in your lab courses; for example:
    "Conservation of mechanical and electric energy: simple experimental verification"

    http://iopscience.iop.org/0143-0807/30/1/005/pdf/ejp9_1_005.pdf

    If you continue with physics into graduate school you will have the opportunity to re-derive most of the common equations during your coursework - often as part of your homework.
     
  12. May 15, 2014 #11

    A.T.

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    To confirm this quantitatively you had to use some formulas to calculate electrical energy and mechanical energy, didn't you? If you had no problem assuming these formulas, then why do you have a problem assuming the work formula?
     
  13. May 15, 2014 #12

    CWatters

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    One experiment using an air track..

    http://www.umsl.edu/~physics/files/pdfs/Mechanics Lab/Exp6.WorkEnergy.pdf

    "The objective of this experiment is to examine the conversion of work into kinetic
    energy, specifically work done by the force of gravity. The work-kinetic energy theorem equates
    the net force (gravity, friction, air resistance, etc.) acting on a particle with the kinetic energy
    gained or lost by that particle."

    "I don't want to take work or the kinetic energy formula on faith."

    I'm not sure what you mean by "I don't want to take work on faith"?

    We define work as the ability to do something useful. What we mean by useful depends on the system. For example in a mechanical system we might define it as the ability to move an object a distance against a force. In an electrical system it might be defined as an amount of charge raised a voltage.
     
  14. May 15, 2014 #13
    There is a big difference in assuming work is scientifically valid compared to how we look at electrical energy.

    It is easy to prove electrical energy is correct. I watched an electrician hook up a motor and he had the option of wiring it using 120 volts or 240 volts. At 120 volts the motor used about 10 amperes and he showed me that by rewiring the same motor to run at 240 volts, the motor used about 5 amperes. The motor worked just the same in both situations running at 1200 watts.

    My concern with work (force x distance) is that Newton's 3rd law seems to indicate there is a natural link between force and time, not force and distance. If an astronaut floating in space throws a baseball, he and the ball can only accelerate for the same amount of time. The distance each accelerates through is different.

    The more I think about it, the more and more it looks like somebody goofed a long time ago in physics. Everyone since has just gone with the flow but I am a bit of a rebel and I want to see real proof that work is the right way to measure mechanical energy. Maybe it is a good thing I started studying Newton's Laws and the work energy theorem on my own. It allows me to ask questions and spend time getting them answered. In school, we have to keep up or risk failing.
     
  15. May 15, 2014 #14

    A.T.

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    That's why both gain equal but opposite momentum.

    That's why both gain different amounts of kinetic energy.

    So yes, momentum and energy are different. That's why they have different names. But what is your point?
     
  16. May 15, 2014 #15
    I don't get what you mean. You can test the "work-energy" theorem, experimentally, in thousands of ways. You'll see it is always satisfied.

    And?

    They both are accelerating during the period of time the astronaut is pushing the ball and they both are in physical contact, call it

    [tex][t_0, t_1][/tex]

    As soon as the ball leaves the astronaut hand, ball and astronaut will be moving with uniform rectilinear motion.

    They get different accelerations during the time interval [t_0, t_1] because they have different masses. So obvioulsy the astronaut will have moved a shorter distance during that time interval [t_0, t_1] than the ball.

    You see something strange here?


    EDIT:

    For example, suppose F is the force the astronaut exerts upon the ball (and also the force the ball exerts upon the astronaut, but with different sign) during the time interval [t_0, t_1] and suppose F is constant (during that interval of time). Suppose M is the astronaut mass, m is the ball mass, a_{a} is the astronaut constant acceleration and a_{b} is the ball constant acceleration (during the time interval). You'll have:

    Astronaut:

    [tex]F = M a_{a}[/tex]

    so

    [tex]a_{a} = \frac{F}{M}[/tex]

    is the constant acceleration of the astronaut during the time interval [t_0, t_1].

    The distance the astronaut moves during that time interval will be:

    [tex]x(t_1)-x(t_0) = \frac{1}{2}\frac{F}{M}(t_1-t_0)^2[/tex]


    Ball:

    [tex]F = m a_{b}[/tex]

    so

    [tex]a_{b} = \frac{F}{m}[/tex]

    is the constant acceleration of the ball during the time interval [t_0, t_1].

    The distance the ball moves during that time interval will be:

    [tex]x(t_1)-x(t_0) = \frac{1}{2}\frac{F}{m}(t_1-t_0)^2[/tex]

    which is longer than the astronaut distance just because m<M.


    The "work" done on the astronaut by the force (the force the ball exerts on the astronaut) along the trajectory the astronaut follows during that time interval, is:


    [tex]\int_{t_0}^{t_1}F(t)v(t)dt = \int_{t_0}^{t_1}F a_{a}(t-t_0)dt= \int_{t_0}^{t_1}F\frac{F}{M}(t-t_0)dt=\frac{F^2}{M}\frac{(t_1-t_0)^2}{2}[/tex]


    Which is exactly equal to:

    [tex]\frac{1}{2}Mv(t_1)^2-\frac{1}{2}Mv(t_0)^2[/tex]

    given that

    [tex]v(t_0)=0[/tex]

    and

    [tex]v(t_1) = a_{a}(t_1-t_0) = \frac{F}{M}(t_1-t_0)[/tex]


    The "work" done on the ball by the force (the force the astronaut exerts on the ball) along the trajectory the ball follows during that time interval, is:


    [tex]\int_{t_0}^{t_1}F(t)v(t)dt = \int_{t_0}^{t_1}F a_{b}(t-t_0)dt= \int_{t_0}^{t_1}F\frac{F}{m}(t-t_0)dt=\frac{F^2}{m}\frac{(t_1-t_0)^2}{2}[/tex]


    Which is exactly equal to:

    [tex]\frac{1}{2}mv(t_1)^2-\frac{1}{2}mv(t_0)^2[/tex]

    given that

    [tex]v(t_0)=0[/tex]

    and

    [tex]v(t_1) = a_{b}(t_1-t_0) = \frac{F}{m}(t_1-t_0)[/tex]


    As you can see, the work done by F on the ball along its trajectory is greater than the work done by F on the astronaut along his trajectory, which is the same as saying that the ball gets more kinetic energy than the astronaut.


    Do you see something weird in all this?
     
    Last edited: May 15, 2014
  17. May 15, 2014 #16

    UltrafastPED

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    The same experiment can be done with mechanics; can you construct an example using levers, pulleys, or gears? I can lift the same load with half the force applied over twice the distance ...


    Newton's Third Law of Motion is certainly symmetric in time ... so applying Newton's Second Law of Motion, F = dP/dt, to our opposite but equal pair, F12 = -F21, and integrating over equal time periods we have:

    ∫dP12/dt * dt = ∫dP12= ΔP12 which is equal to -∫dP212/dt * dt = -∫dP21= -ΔP21.

    So if the bodies started at rest we have P12 = -P21: equal but opposite momentum; but in every case the changes in momentum are equal but opposite.

    You are correct that the distances covered by the two bodies are different. The Third Law of Motion implies the conservation of momentum. The Second Law of Motion is the source of the Work-Energy theorem: it's about changes in kinetic energy.


    You are correct - Isaac Newton goofed by writing his great work in Latin, and by using geometry to prove the theorems instead of the calculus. The work went through three editions during his lifetime, so you can be certain that he also felt the need for more than a few updates.

    If it was a bit easier to read beginning students might understand it more easily! :-)
     
  18. May 15, 2014 #17
    I do see something very strange and I'm very confused as to why no one else saw this before.

    The force acting on the astronaut is exactly equal to the force acting on the ball and yet, one of those two things winds up with more kinetic energy. That makes zero sense since both receive the same ACTION and to show why I say that, imagine that same astronaut throwing a different ball that is twice as massive.

    If he throws the more massive ball exactly like the other one, he will accelerate exactly the same as before but the ball's speed will be less. Compare the work done in those situations and you get the following:

    work done on astronaut with small ball EQUALS work done on astronaut with bigger ball

    work done on small ball is GREATER than work done on bigger ball


    The thing I see is that when force acts, it acts both over time and distance. The astronaut/ball situation seems to indicate that the time force acts is FAR more important than the distance force acts through. I got the idea that just because it is possible to measure the distance force acts through, it does not necessarily mean it is really that important. That should explain why I want to see the experiment that decided physics should use force x distance instead of force x time to describe energy.
     
  19. May 15, 2014 #18

    A.T.

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    Define "ACTION".

    What is exactly the same? Force? Time? Distance?

    Pull something up against gravity at constant speed and hence constant force. The energy gained depends on the height gained, not the time you need to lift it.
     
  20. May 15, 2014 #19
    What?

    A mathematical theorem is just a mathematical theorem. You could also say that 2 + 2 = 4 (in ordinary arithmetic) does not make sense to you, but that would only imply that there is something you are not understanding and/or something you are not interpreting the correct way.

    You can easily get misleaded the moment you start using vague terms. Define "action" with mathematical precision.


    Yes, in my previous example you just have to change m to m'=2m and leave F as F. Then the new ball speed will be exactly half the previous ball speed and the same with the new ball distance during the time interval [t_0, t_1] (will be half the previous distance).

    All this not only does make complete mathematical sense, but also is what actually happens in nature, as you can experimentally test in multiple ways.


    It does make complete sense if you understand what is the mathematical definition of "work". If the force F is the same in both cases (which it is in our example) and the mass M of the astronaut is the same in both cases (which it is) and the time interval this force is being exerted is the same (which it is in our example), then obviously the work will be the same in both cases (if you don't see it, then you don't understand the mathematical definition of work).


    Again, this does make complete sense if you understand the mathematical definition of work. Work done (by the same force F, acting during the same time interval [t_0, t_1] ) on the smaller ball will be obviously greater than work done (by that same F which acts during that same time interval [t_0, t_1] ) on the bigger ball just because the smaller ball (under these circumstances) travels a longer distance during that time interval, than the bigger ball.


    In the astronaut case, F is the same, M is the same, [t_0, t_1] is the same, so obviously the work done on the astronaut by those two different balls will be the same.

    In the two balls case, F is the same, [t_0, t_1] is the same, but the mass is different, m' = 2 m, so obviously the work done will be different (greater on the smaller ball, than on the bigger ball).

    Again, if you don't see all this, if it does not make complete sense to you, then you are not understanding the definition of work.


    Both are important concepts, but different concepts. Work and Impulse are different concepts, but very important and useful.

    Work is equal to increment of kinetic energy.

    Impulse is equal to increment of linear momentum vector.

    "Energy" is what we decide to define as energy (just as it is true with any other concept). We call "energy" a given entity that is conserved under concrete circumstances and that has dimension M.L^2/t^2.

    F.t=M.L/t is different, and we call it "linear momentum".

    If you start changing names, this can get messy very fast :-).
     
  21. May 15, 2014 #20
    I appreciate the effort everyone is giving but the only thing I need is the experimental proof that scientists used to verify the idea that work (force x distance) should represent energy.

    I understand that energy is thing that changes forms and so on. There is no argument with that.

    The scientific method requires that a hypothesis be independently tested. Someone must have come up with the idea of energy and other scientists confirmed it. I know the air track experiment well but it does not prove the work energy theorem is valid unless you assume work is valid. If you substitute force x time (a scalar expression), the air track experiment "proves" force time should represent energy. Obviously, or it should be, the air track experiment does not really test whether force x distance should represent mechanical energy.

    Please see that all I am asking for is the sort of experimental proof physicists would demand of me if I came up with something. For example, if I said that gravity is related not to mass but to the physical size of an object but offered no proof, no one would take me seriously. The work energy theorem came from somewhere and somebody showed it was true. By the way, if my last statement is proven true, I want credit. LOL

    Seriously, what is the experimental proof that work should be used instead of force x time? As far as I know, work has been around for at least 100 years. Is it too much to ask to see it?
     
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