Why do we take k=1 in the derivation of F=k*ma?

  • #76
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And those experiments would quickly show that F≠maF≠maF \neq ma. The reason is because real springs only approximately obey Hooke's Law
That isn’t a problem in principle. You can simply limit the definition to small x. For small x and small v you could experimentally show f=ma. Such limitations are common.
 
  • #77
Mister T
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That isn’t a problem in principle. You can simply limit the definition to small x. For small x and small v you could experimentally show f=ma. Such limitations are common.
That's a good point. So I suppose it comes down to an issue of precision. The definition that minimizes these limitations is the better definition.

I don't think, though, that adopting this definition in any way introduces a testability of ##F=ma## that's lacking in a definition that makes use of ##F=ma##. Rather, it simply constitutes a less precise definition.
 
  • #78
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The definition that minimizes these limitations is the better definition.
Yes, which is why Newton’s law is used instead of Hooke’s law.

I don't think, though, that adopting this definition in any way introduces a testability of F=maF=maF=ma that's lacking in a definition that makes use of F=maF=maF=ma. Rather, it simply constitutes a less precise definition
You could be right. I haven’t finished thinking through this yet. But my gut feeling is still that it does introduce testability that was previously lacking, but I am mentally stuck on mass and stiffness now.
 
  • #79
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So is F=ma valid? What if a=0, as when a body is moving at constant velocity? Then is F=0? I don't think so. If I am struck by a car moving at constant velocity, I guarantee you that I will feel a force. Perhaps there's something I'm missing here.
 
  • #80
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If I am struck by a car moving at constant velocity,
If the car strikes you then neither you nor the car will move at constant velocity
 
  • #81
A.T.
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So is F=ma valid? What if a=0, as when a body is moving at constant velocity? Then is F=0? I don't think so. If I am struck by a car moving at constant velocity, I guarantee you that I will feel a force. Perhaps there's something I'm missing here.
The F in F=ma stands for the vector sum of all forces acting on the object, not for some individual force.
 
  • #82
CWatters
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So is F=ma valid? What if a=0, as when a body is moving at constant velocity? Then is F=0? I don't think so. If I am struck by a car moving at constant velocity, I guarantee you that I will feel a force. Perhaps there's something I'm missing here.
When the car hits you it accelerates you and you go flying down the road. The force you feel is due to your own acceleration.
 
  • #83
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Experiments show that a is proportional to F/M. We can write:

a = kF/M

If we had agreed units for a and M we can use the equation to define a unit for F. I might suggest that we use a unit of force called a turnip (T) defined as:

One Turnip is the resultant force that gives 12Kg an acceleration of 4.5m/s2

It's unlikely that the Turnip would be widely adopted as a unit because it would make k equal to 54 and amongst other things that can be annoying. The neatest definition would make k equal to one and hence the Newton is adopted because it does just that:

One Newton is the resultant force that gives 1Kg an acceleration of 1 m/s2.
 
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  • #84
bobob
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I think the point of the original question might be related to the equivalence principle. If you take the weight of an object to be be proportional to it's mass and that proportion is W=F=mg = m(GM/r^2) then it might be natural to ask if that is equivalent to a mass x acceleration in general. The underlying question is then, given a gravitational mass m and an inertial mass km, is k=1? In that case, the question is obviously quite reasonable given the number of experiments done look for departures from that equivalence.
 
  • #85
jbriggs444
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The underlying question is then, given a gravitational mass m and an inertial mass km, is k=1?
That question might better be posed as "is k a fixed constant".

The formula for [passive] gravitational mass is ##F=\frac{G\ m_{passive}\ m_{active}}{r^2}##.

That constant G in there means that any fixed constant k in ##m_{passive}=\frac{m_{inertial}}{k}## is indistinguishable from an adjustment to G.
 

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