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Proof of F=ma from second law of newton

  1. Feb 16, 2010 #1
    from the second law of newton we know... applied force is proportional to the change of momentum...that means F= k( mv-mu) where k is constant .....from here now proof F=ma
  2. jcsd
  3. Feb 16, 2010 #2
    The second law doesn't say that. Look at it again.
    You need to start with the correct equation before you can prove F=ma
  4. Feb 16, 2010 #3
    I think i can help you, but with my own labelling ;)

    [tex]F \sim m \cdot \frac{v^\prime - v}{\delta t}[/tex]​

    [tex]v^\prime[/tex] is the speed after a short time interval [tex]\delta t[/tex] related to the instant of time [tex]t[/tex]. So the equation above can be written in a more descriptive way

    [tex]\delta t \, F \sim m \cdot \Bigl( v(t+\delta t) - v(t) \Bigr)[/tex]​

    if the time interval [tex]\delta t[/tex] is very small, we can approximate [tex]v(t+\delta t)[/tex] according to Taylor (keyword Taylor expansion).

    [tex]v^\prime = v(t + \delta t) \approx v(t) + \dot v(t) \, \cdot \, \delta t + \mathcal{O}\bigl((\delta t)^2 \bigr)[/tex]​

    putting it in the initial equation that yields

    [tex]\delta t \, F \sim m \cdot \Bigl( ~ \bigl( v(t) + \dot v(t) \, \cdot \, \delta t \bigr) ~ - v(t) \bigr) = m \, \cdot \, \dot v(t) \, \delta t [/tex]​

    Here you see, that the time interval [tex]\delta t[/tex] must be very small, otherwise the first term of the Taylor series isn't a good approximation any longer.

    Hope that i could help you...
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