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Silly question, derivation of energy

  1. Jul 21, 2008 #1
    Here's a silly question I ended up thinking about. Maybe I learned it in college, but it's long forgotten by now ^~^; It might be quite a silly question, something I shouldn't have forgotten so easily, but alas, here I am, I've gotten so accustomed to simply using the algorithms without much thought. What a bad thing to happen T_T

    Where did the "divided by 2" go?

    [itex]{\color{red} a := \frac{m}{s^{2}} := \frac{m^{2}}{s^{2}m} := \frac{ \left( \frac{m^{2}}{s^{2}} \right) }{m} := \frac{ \left( \frac{m}{s} \right) \left( \frac{m}{s} \right) }{m}} \;\; \Rightarrow \;\; a = \frac{v^{2}}{s}[/itex]

    [itex]E = \frac{mv^{2}}{2}[/itex]

    [itex]E = Fs = mas = m \frac{v^{2}}{s} s = \frac{mv^{2}s}{s} = mv^{2}[/itex]


    a := acceleration
    E := energy
    F := force
    m := mass
    m := metre
    s := space
    s := second
    v := velocity

    Is my mind deteriorating, or why am I unable to see what is wrong? o_O
     
  2. jcsd
  3. Jul 21, 2008 #2
    You're mixing up ( red ) s as seconds with ( black) s as displacement. See definition of Joule ( energy ) and Newton ( force ). In fact, some of your definitions are quite reckless. Get rid of m = metre. Prefer s = displacement. Get rid of s = second. Prefer t = time.
     
    Last edited: Jul 21, 2008
  4. Jul 21, 2008 #3

    rock.freak667

    User Avatar
    Homework Helper

    In your finding an equation for accleration, you used dimensional analysis. Using that doesn't give you the value of constants, so it not safe to assume that the constant is always 1.

    Your equation for acceleration should have been [itex]a=\frac{kv^2}{s}[/itex]

    But I see that you were deriving the formula for kinetic energy.
    There are two ways to do it, with a constant force or variable force.

    For constant force.

    E=Fs=mas

    now use [itex]v^2=u^2+2as[/itex]

    If you wanted to use calculus (for a variable force)

    [tex]E= \int F ds= \int (ma) ds[/tex]

    and [itex]a=\frac{dv}{dt}[/itex] (So use the chain rule to get 'a' in terms of d(something)/ds)
     
  5. Jul 21, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    All this demonstrates is that v^2/s has the same units as acceleration. For an object uniformly accelerated from rest, [itex] a = v^2/2s[/itex] where v is the final speed. (Derive this from the definition of acceleration and average velocity.)
     
  6. Jul 21, 2008 #5
    I haven't really done much derivation in my life, so yes, never expected to get it right the first time. Demonstrating that the units are the same was probably not the best idea either. However, thank you for the help ^_^

    I'm also using a algorithms/equations book from college, so I never realised to look elsewhere. A few things I haven't seen before. I took all the physics classes in college, either I have forgotten quite a bit, or there were some very important parts left out.

    I am working on ballistic algorithms, so that is why I've suddenly began with derivation. It makes it a lot easier for users to understand the algorithms if they see how it's derived, rather than just in-your-face.
     
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