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Conservation of energy derived from Newton's Laws?

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data

    "Show how conservation of energy (with only gravitational potential energy and kinetic energy) can be gotten from Newton's Laws.

    This was the question that was asked of me, and sorry for the wording not being terribly precise.

    2. Relevant equations

    I am not exactly sure which equations are allowed and which aren't.
    I assume F=ma, but i am not sure about Fg = GmM/r^2, W=Fd, or kinematic equations.

    By Newton's Laws do they mean more than the "three laws"?

    The student had not studied calculus yet.

    3. The attempt at a solution

    For instance, if using (i) F=ma, (ii) vf^2=vo^2+2ad, and (iii) W=Fd, we can do the following:
    i&ii) Fd = m(vf^2 - vo^2)/2 ==> Fd = (1/2)*m*vf^2 - (1/2)*m*vi^2
    iii&previous) W = (1/2)*m*vf^2 - (1/2)*m*vi^2

    But this required more than simply F=ma, nammely (ii) and (iii).

    I read online that conservation of momentum can be derived from newton's third law. How?
    Newton's third law is not really an equation, but more of a 'concept'.
    I am failing to understand something here.

    Thanks for your help to discuss/answer this vague question,
  2. jcsd
  3. Dec 9, 2008 #2
    The full definition of Newton's third law (translated from the Latin but not by me!) states:

    If a body impinges upon another, and by its force changes the motion of the other, that body also (because of the equality of the mutual pressure) will undergo an equal change, in its own motion, toward the contrary part. The changes made by these actions are equal, not in the velocities but in the motions of the bodies; that is to say, if the bodies are not hindered by any other impediments. For, as the motions are equally changed, the changes of the velocities made toward contrary parts are reciprocally proportional to the bodies.

    So what is equal is "changes in the motions". This sounds to me like "overall momentum is conserved".
  4. Oct 6, 2009 #3
    using F=ma we can easily show that [tex]\int[/tex]F.ds=(1/2)mv2
    Now d/dt{[tex]\int[/tex]F.ds} = d/dt{(1/2)mv2}
    [tex]\Rightarrow[/tex] d/dt{[tex]\int[/tex]F.ds-(1/2)mv2}=0
    [tex]\Rightarrow[/tex] (1/2)mv2-[tex]\int[/tex]F.ds=constant

    Use this to prove conservation of energy. In the case of a body moving under gravity the constant will be initial kinetic energy and the integral will become mgh. for a collision problem u can use newtons third law to prove conservation of energy.
    Last edited: Oct 6, 2009
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