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Energy conservation

  1. Jul 2, 2007 #1
    Energy can be neither created nor destroyed.
    What is energy?. ability to do work.
    what is work ? work = force x displacement.
    Consider a body of mass 'm' moving at a uniform velocity 'v'.work done here is zero(no acceleration).So energy is zero.
    DONT SAY KINETIC ENERGY is there.please check the derivation of kinetic energy.

    Since work = F*d
    and
    F = m*a
    then
    work = m*a*d.

    From kinematics we know that
    d = 1/2 * a*t2
    and
    t = v/a.

    so d = (1/2* a *v2)/ a2 = (1/2* v2)/ a

    so work = m* a * (1/2* v2)/ a = 1/2 * m * v2.

    in this derivation please note that t = v/a, is only applicable when initial velocity is zero. here initial and final velocity are v ,and kinetic energy will be zero.

    now come back to the point.what will happen ,if this body hits another body at rest.?(mass is irrelevant-body at rest have no energy stored,isn't it?).

    After collision, both of the bodies have energy(since there is an acceleration).

    from where does this energy comes from.?
     
    Last edited: Jul 2, 2007
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  3. Jul 2, 2007 #2

    russ_watters

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    How did the body come to be moving at a velocity 'v'?
     
  4. Jul 2, 2007 #3

    Doc Al

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    You'd better check the definition of kinetic energy. If the body is moving, it has kinetic energy.
     
  5. Jul 2, 2007 #4
    If the collide then the law of conservation of momentum comes into play, no energy is gained or lost?

    [tex]m_{a}v_{oa}+m_{b}v_{ob} = m_{a}v_{fa}+m_{b}v_{fb}[/tex]

    If one car has a velocity of 0, then the other car would have to have some sort of velocity to hit it. Since one of your cars is not moving:

    [tex]m_{a}v_{oa} = m_{a}v_{fa}+m_{b}v_{fb}[/tex]

    This means that since the masses cannot change there wil be a change in velocities, the change is going to vary depending on a few factors such as is the collision elastic or inelastic? Technically speaking if there was no friction or gravity the cars would never stop (I think), basically what I'm getting at is that there is no acceleration as a result of the collision, just from friction of the street (I'm assuming your cars are on a road) which is why the cars are going to accelerate negatively until their velocities are 0 and both are at rest.
     
    Last edited: Jul 2, 2007
  6. Jul 2, 2007 #5
    whatever way you look,you admit there is a change in momentum,which means a force and displacement is there.this finally means a work is done.So role of energy is relevent.
     
  7. Jul 2, 2007 #6
    This is not the way ,a scientist should look at the things.If you say there is kinetic energy in a moving body, you have to substantiate this with some logic or calculations

    As i mentioned in my previous post, acceleration,a = (v2 - v1)/t , or t = a/(v2 - v1). In this case since v2 = v1 ; t = 0 ; so if you substitute this ,we can find out the kinetic energy is zero.

    Otherwise Could you please derivate the kinetic energy formula ,for a body moving with a uniform velocity?or please give me a link.
     
    Last edited: Jul 2, 2007
  8. Jul 2, 2007 #7
    Change in momentum yes, but no energy is "gained" or "lost" it's simply (forgive me but it's the best word I could think of) transfered.

    [tex]a = \frac {\Delta{v}}{\Delta{t}}[/tex]
    Since it's a change in time I'd venture to guess that's an average acceleration not an instantaneous one. But last time I checked the forumla was

    [tex]KE = \frac{1}{2}mv^{2}[/tex]

    If a body is moving it has a velocity that is NOT 0, therefore it has kinetic energy. You proved it yourself, acceleration doesn't matter you derived the equation in a why which acceleration "cancels out".
     
    Last edited: Jul 3, 2007
  9. Jul 3, 2007 #8

    rcgldr

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    This formula is independent of a (non-accelerating) frame of reference:

    [tex]work = 1/2\ m\ (\ V_1^2\ -\ V_0^2\ )[/tex]

    The simpler form

    [tex]KE = 1/2\ m\ (\ V_1^2\ )[/tex]

    Is really just the first equation with [tex]V_0 = 0[/tex], so by definition, KE is the equivalent of the work it takes to accelerate or decelerate an object from [tex]V_0[/tex] to [tex]V_1[/tex]. This allows KE to be defined (not derived) as a form of potential energy relative to a (non accelerating) frame of reference moving at [tex]V_0[/tex].
     
    Last edited: Jul 3, 2007
  10. Jul 3, 2007 #9
    Thank you Jeff,that's it.
    I was expecting the same answer from many.
    I will come to my point based on this answer.
    What will happen if we use the same concept for force?
    say the force acting on the mentioned body,moving with a uniform velocity v1 is F= mass x acceleration (from zero to final velocity).
    So now force on the body is F = m * v1/t.
    now consider this body accelerated to a velocity of v2 in the next interval of 't1' seconds.
    Now at this instance , force on body is F2 = m*v2/t1
    force on body if it was still travelling at v1 for t1 seconds will be F1 = m*v1/t1.
    so what is the net force /residual force or whatever,which caused this acceleration?

    net F = (m*v2/t1) - (m*v1/t1) = m * a .

    Why? and what is the use of changing the perspective.

    Isn't it a clear answer for Inertia of body moving at uniform speed?Is it because of this force(already acting),a body which is at rest or in uniform motion tries to resist any changes in the motion?

    Well ,i hope it is clear.please don't see it with a prejudice and correct me ,if i am wrong anywhere
     
    Last edited: Jul 3, 2007
  11. Jul 3, 2007 #10

    Doc Al

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    You're kidding me, right? :rolleyes:

    You are confusing change in kinetic energy with kinetic energy. From one second to the next, a body moving at constant speed has zero change in KE. The kinetic energy is certainly not zero.

    Look up the work-energy theorem, which relates work done to the change in KE. (Work-Energy Principle) As far as KE itself, it is defined to be [itex]1/2 m v^2[/itex] (at least in Newtonian physics).
     
  12. Jul 3, 2007 #11

    Doc Al

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    This is gibberish. Is the body moving with constant velocity or not? Acceleration is change in velocity over time, not a particular velocity over a particular time.

    If the body is moving at a constant velocity, the net force on it is zero.


    In order for a body to change its velocity, a net force must act on it. That force is given by [itex]F_{net} = m a[/itex]. Basic stuff! No need for vague concepts such as "residual force".

    Nope. Newton figured this out centuries ago. No force is needed to maintain a constant velocity.
     
    Last edited: Jul 3, 2007
  13. Jul 3, 2007 #12
    So we're saying F=m*a, which I'm sure you know is equivalent to F=dp/dt. Im just putting it out there for later use...
    I'm sure you mean F = m*dv/dt = m*a. Why that v1? Don't answer yet, I'll get to it in a second.
    Remember F= m*dv/dt, mass times the Change in velocity over the Change in time. So with your method you'd say : F = m*(V2-V1)/(t2-t1)

    The force doesn't change. Its stays constant so long as acceleration (change in velocity) is constant.
    Simplify that. F=m*(v2-v1)/t1=m*(dv)/(dt)= m*a
    This is correct. Given a force F the mass "m" will accelerate from "v1" to "v2" over some period of time "t1" is correct.

    You can't think of it as F(v), Force isn't a changing function of "v".
    F(v1)=F(v2)=ma >> its constant in this example.



    I don't see what you're asking. Any mass needs a force to change its velocity. Thats Newtons 1st Law.
     
  14. Jul 3, 2007 #13

    rcgldr

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    Ok, so v0 = 0 in your example. The way you've written it, the component (v1/t) doesn't look like an acceleration. It should be
    F = m * (v1 - v0) / (t1 - t0)
    so that it's clear that the acceleration is represented as a change in velocity versus a change in time, even if v0=0 and t0=0.
    This is almost correct, write it as
    F = m *(v2 - v1)/(t2 - t1).

    These examples are assuming a constant force.
     
    Last edited: Jul 3, 2007
  15. Jul 3, 2007 #14
    What is the cause of inertia?Does anybody figured this before?
    Inertia,for a body moving at constant velocity ,is the force acting on it ,which is the cause of displacement of the body.(whatever may be the cause.somebody pulled it or push it).
    Our force in regular equation,F = m*a;is the additional force required to change its velocity in a given interval.
    Inertia, for a body at rest (in space,where there is no gravitational field- A body at rest is a concept ony,it can be relatively at rest)is zero.
     
    Last edited: Jul 3, 2007
  16. Jul 3, 2007 #15
    please... t , t1 ,t2 etc. mentioned here are time intervals.(may be i am using wrong notations)
     
  17. Jul 4, 2007 #16

    russ_watters

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    That's still gibberish. Inertia is a stand-alone physical property of matter. "Cause" is irrelevant. Newton's first law states explicitly that no force is required to keep a body in motion, only to accelerate it. Don't think that because you are asking a question that physics doesn't care about that that automatically makes it new and profound. It isn't - it is simply irrelevant. I'm not sure how we can make you accept how inertia works, but you can see it in everyday life. It just does.
    I don't see "additional force" in that equation, I only see "force". "Additional force" is something you have made up and it doesn't exist. It doesn't matter what you call it - "additional force" or "residual force": it doesn't exist.
    A body is always at rest with respect to itself. Newton's first law is not picky on this point: whether at rest or just in a uniform/constant state of motion, it works the same.

    Heck, this is a pretty clear flaw in your idea that uniform motion requires a constant force. Since you understand that uniform motion is frame of reference dependent, so too would the force be frame of reference dependent. And you can't have force be frame dependent. A scale reads what a scale reads and only that value.
    No. They are not time intervals, they are just times - what you see when you stare at your watch. t0=0 seconds, t1=1 second, t2= 2 seconds, etc. The interval from t1 to t2 is (t2-t1)=2-1=1 second. So the acceleration over that interval is the change in velocity divided by the time interval in which the change occurred.

    Jeff wrote it exactly correctly.
     
    Last edited: Jul 4, 2007
  18. Jul 4, 2007 #17
    Why weight?-because of gravity(everybody knows)
    Why conductivity?-because of valence electrons(everybody knows)
    Why Inertia?-Nobody knows-so don't ask that question-it is irrelevant.

    Is that what you mean?

    And Let us consider one of everyday situation.You are travelling in a car.Suddenly you applied break.(You forget to wear your seat belt).You are throwned forward.
    why? because of inertia.-finished.
    If i say,that force which was acting on you,which cause you to move at a velocity equal to that of the car is still acting on you and remains until another force acts upon you,is there anything wrong?-you can call it inertia.
    if there was no gravity ,or any other forces,the force will continuously act on you ,keeping you to move at same velocity.
    And regarding rest, i stated that relatively a mass can be at rest.
    you stated it in another way that a body is always at rest with respect to itself(relatively at rest).
    Finally,regarding the time intervals ,it was something regarding my derivation and i have all the right to tell somebody that by that notations i mean time intervals (i remember i appologise for using wrong notations).
     
  19. Jul 4, 2007 #18

    Doc Al

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    You are welcome to ponder the "ultimate" source and meaning of inertia, but first learn a bit of physics.


    Somewhere you've picked up the misconception that a force is required to keep you moving forward. Some ancient Greeks thought this, but we've since discovered otherwise.

    You seem nominally interested in physics, but not enough to pick up a textbook and learn what we already know. Constant velocity does not require a force!
    Nope. Again: Learn (or at least read about) Newton's laws. Learn what we mean by a "force". A force requires an agent: something doing the pushing or pulling. If you are moving at a constant velocity either all the forces acting on you have canceled out or there are no forces acting on you. (For the latter, imagine a spaceship coasting in outer space, away from all masses.)

    What's relevant here is whether the body is accelerating or not.
     
  20. Jul 4, 2007 #19
    What would have been the scenario if human beings stopped thinking out of box(from what they learned till then).?

    Tell me why you require a force to stop an object moving at uniform velocity?
    With your force you have to cancel some other force.what is the other force.it is inertia.if you try to find an equation to find out that force,you will end up with my conclusion.

    Ok.A force require an agent.yes did i told you no?
    To acheive a constant velocity,you require a force .isn't it?Agents role is finished there.
    shall i explain?
    consider a golf ball (at rest-relatively).
    you hit it with the club(sorry if i am right-club is the stick i mean)
    Now the ball is accelerating positively ,then negative acceleration and ultimately comes to a hault.
    If you see any intervals,you can see there is a force acting on the ball.
    But you cannot say,the club was hitting on the ball throughout the way.
    Learning is something and understanding is something else.
    Please don't undermine peoples.There is something to learn from every layman.
     
  21. Jul 4, 2007 #20

    Doc Al

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    You should open the box and see what's in it. But first you must find the box. :wink:

    If you are asking why the world is the way it is, I can't answer that.
    There is no force (at least no net force) acting on an object moving at uniform velocity. So there is no "other force" that you have to cancel.
    Your conclusion is based on misunderstanding.


    No!

    I have no idea what this example is supposed to tell us. If you wish to accelerate the golf ball, you must exert a force on it. True. So?
    Might I dare suggest that there is something to learn from studying basic physics?
     
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