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Starting off the movement

  1. Dec 7, 2008 #1
    The power is the amount of energy being transferred, the energy change P = Fv at specific moment of time. The kinetic energy is proportional to speed: E = mv2/2. I wonder how a finite force starts moving an object at rest. The zero velocity implies zero power P=0, which implies zero kinetic energy inflow dE = 0 * dt = 0. Ultimately, as the energy does not change, the speed persists 0. In other words, the speed does not change because it is zero and it remains zero because it does not change. We cannot start off. Where is the mistake?
     
  2. jcsd
  3. Dec 7, 2008 #2

    russ_watters

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    Nowhere in any of that was a mention of acceleration and velocity is a dependent variable. Those are output calculations, not inputs. What makes an object accelerate is f=ma (a=f/m).
     
  4. Dec 7, 2008 #3
    To expand upon what Russ wrote:

    At the moment the object is at rest,
    Power P=0
    Kinetic Energy E=0
    But dE is not zero.
    dE = d(mv^2)/2 = mdv
    Just because v is zero does not mean that dv is zero. By definition dv/dt is acceleration. If you are about to start pushing on the object then acceleration = F/m. So after a finite period of pushing, v is not zero either and E and P now have finite magnitudes too.
     
  5. Dec 7, 2008 #4
    Russ, introducing a new term should not break the solid mechanics theory, which, as I have explained above, does not need you acceleration entity to prove that no acceleration exists :) Use Accam razor. Do you claim that Newton mechanics in inconsistent? :)
     
    Last edited: Dec 7, 2008
  6. Dec 7, 2008 #5
    I may completely mistake but dE/dv = m/2 d(v^2)/dv = mv rather than m, so dE = mv dv. That might me a clue to the answer. I mean that dv = dE / mv = 0/0 indefinite. In other words, the speed may change despite dE is zero. Right?


    dE is equal to power: dE = P dt = Fv dt. It cannot be that the power is 0 while energy is changing. The power is the 'speed' of energy, energy is not moving if its speed is 0. From the latter formula it is clearly seen that dE is zero when v = 0.
     
    Last edited: Dec 7, 2008
  7. Dec 7, 2008 #6
    dE is not zero when v = zero. that's my whole point. This last equation for power is wrong. Check your units. Power can't be equal to Fv and Fvdt.
     
    Last edited: Dec 7, 2008
  8. Dec 7, 2008 #7
    Please check the formula now. Tell me if it is correct and how can dE be different from 0 if v = 0? Don't try to dodge anyone by suggesting that dE is not proportional to v. ;)
     
    Last edited: Dec 7, 2008
  9. Dec 7, 2008 #8
    You also made a mistake in declaring dE = mv dv. Again, check your units. dE = mdv. dE is NOT proportional to v. dE is proportional to dv.

    Really smart people think this stuff is easy. Myself, I had a hard time learning it. Looks like you are having a hard time too. Let me assure you there is no contradiction. I have used this stuff to design machinery for many years. Sometime I screw up, but the physics always works.
     
  10. Dec 7, 2008 #9

    russ_watters

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    Huh? Solid mechanics theory? Occam's razor? Newton mechanics inconsistent? That entire post is meaningless gibberish!
     
  11. Dec 7, 2008 #10
    I believe that both x and dx measure in the same units. Hence, both E and dE are measured in [joules] or [kg meters^2] rather than your [kg meters], which is a unit for momentum. In my galaxy, mdv = dp rather than dE. May be it is good that I do not design machinery.
     
    Last edited: Dec 7, 2008
  12. Dec 7, 2008 #11
    I give up, voljok. Maybe someone else can help you. I can't.
     
  13. Dec 7, 2008 #12

    russ_watters

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    Probably not, but one more try before we lock the thread:

    Mike is correct: the units of x and dx are not the same. Taking the derivative (the "d") turns units of meters (for example) to meters per second. The derivative of distance is velocity. Further, you said above that the derivative of energy is power - but now you say the units are the same. So do you think power and energy are the same thing?

    Valjok, your entire line of reasoning needs to be tossed in the trash. Your idea doesn't hold water.
     
  14. Dec 8, 2008 #13
    What? "d" does not stand for 'deriviative'!! it stands for 'differential'!!! I can explain to the children that deriviative is a ratio of differentials: dx/dt. This is how we get [meters/sec]. Neither dx alone nor dt are measured in meters per second. Where do I say that power and energy measure in the same units?

    If you are a moderator on a phys-math forum, it is the forum which must be tossed into the trash along with the whole education system, which produces such awesomely illiterate "experts".
     
  15. Dec 8, 2008 #14

    russ_watters

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    It's been a while since I took calc and I forgot the definition, so I'll give you "derivative" vs "differential", but in this case, that it was dx/dt was assumed. However, finding a nitpicky mistake of mine does not change anything about your entire point being wrong. Thread locked.
     
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