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I understand fully where the constant comes from and why it is defined as $k=1\ N\ kg^{-1}\ m^{-1}\ s^2$

But does it have an official name?

If I were to give it a name I suppose I would maybe call it one of the following:

  • Constant of inertial mass

  • The inertia constant

  • Newton's constant of inertia

  • Newton's 2nd law constant

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    $\begingroup$ "The definition of the newton" $\endgroup$
    – The Photon
    Jul 5 '20 at 16:02
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It's built in to the definition of the newton.

If the newton is defined as "the force that produces 1 m/s2 acceleration when acting on 1 kg" then clearly no further constant of proportionality is required in the formula $F=ma$ .

If you're working in some other unit system, the unit of force will be defined similarly to avoid the need for any further scaling factor in this formula.

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It is sometimes called a definitional constant, but this term may not be universally understood. See for instance [1].

[1] T. Quinn, Physical quantities, in Metrology and Fundamental Constants, Proceedings of the International School of Physics "Enrico Fermi", Course CLXVI, 2007.

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The proportionality constant to Newtons second law is defined as the mass. There is no 'k'. It's just $f\propto a \implies f = M a$ where M is some constant of proportionality.

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  • $\begingroup$ Wow , I never thought about mass like that before. Really interesting. $\endgroup$
    – Kantura
    Jul 5 '20 at 16:09
  • $\begingroup$ I don't think this correct at all. $M$ is not a constant and really $f\propto a$ AND $f\propto M$, so $f\propto Ma$. $\endgroup$
    – Gert
    Jul 5 '20 at 17:16
  • $\begingroup$ You’re right, mass doesn’t have to be constant. Just like for any proportionality it doesn’t have to be constant. E.g. a spring has a spring constant k which is the constant of proportionality. But you can change the spring constant if you use another sprint, or deform your original one. It’s just that we define that constant based on the proportionality. So even though it can change, the relationship is always true. $\endgroup$
    – Jake Rose
    Jul 6 '20 at 13:18

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