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According to Newton's 2nd law of motion, F=k(m)(a), where k= constant. Now when F=1,m=1,a=1, then k=1. But when F,m,a each≠1 then the formula is not valid. So how can we say that F=(m)(a) is true everywhere in every case.We also solve questions by taking this formula only.

Qmechanic
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DEB SANKAR ROY
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jun 14 '22 at 04:47
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    This is very unclear... What is the definition of $k$? In which unit system do you take $F=1$, $a=1$ and $k=1$, and why do you take such a special case? What do those extra parenthesis mean in $F=(m)(a)$? – Miyase Jun 14 '22 at 04:49
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    K is a constant. According to Newton's 2nd law of motion, F=(m)(a), where m=mass, a=acceleration. Also F is directly proportional to the product of mass and acceleration. So, F=k(m)(a). Also in the book it is written that when F=1,m=1,a=1, k=1. Also F=(m)(a), is used to solve the problem. But why will I take F,m,a each equal to 1 always. Different values of F,m,a will give different values of k. – DEB SANKAR ROY Jun 14 '22 at 05:04
  • We also say that H = 12 J where H is your height in inches and J is your height in feet. This works when your heights in inches and feet are 60 and 5. But what if your height in inches is 70 and your height in feet is 4? Then the formula is not valid, so how can we say it's true in every case? – WillO Jun 14 '22 at 05:34
  • Learn mathjax so F=(m)(a) can become $\vec F = m \vec a$. – JEB Jun 14 '22 at 19:00
  • Possible duplicates: https://physics.stackexchange.com/q/104101/2451 and links therein. – Qmechanic May 27 '23 at 13:17

2 Answers2

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I'm not sure where your confusion lies. According to Newton's second law of motion, the rate of change of momentum is directly proportional to the force applied.

$$\vec{F} \propto \frac{d \vec{p}}{dt}$$

If the mass of the system remains constant

$$\vec{F} \propto m \frac{d \vec{v}}{dt}$$

Or we can write it as

$$\vec{F} = k m \vec{a}$$

where $k$ is a constant of proportionality.

We define a unit force as a force which can produce a unit acceleration in a body of unit mass. Using $k = 1$, we get the familiar equation

$$\vec{F} = m \vec{a}$$

Mechanic
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    But why we will always take k=1. k is a constant. k can be 2 ,3,4,... also. Why we will always take k=1. – DEB SANKAR ROY Jun 14 '22 at 05:34
  • We define the unit of force this way. For example, in the SI system, we define one newton of force as the force which produces an acceleration of 1 m/s^2 in a body of mass 1 kg. Now you can invent your own system of units where you define a unit of force which produces, let's say, half a unit of acceleration in a body of unit mass or a unit acceleration in body of half a unit mass. In that case k =2 and so on. – Mechanic Jun 14 '22 at 05:43
  • @DEBSANKARROY Do you know what a law is in physics theories? A law is an extra axiom imposed on the mathematical theory so that the units on the numbers in the solutions correspond to observables and measurements in the lab. These extra axioms have not been falsified in fitting with classical mechanics the data and observations, so they are considered true by construction of the theory of classical mechanics. – anna v Jun 14 '22 at 05:45
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k the constant of proportionality is ALWAYS taken to be 1 because force is directly proportional to acceleration.

If we use your equation F = kma. This can be seen by graphing force you dependent variable against acceleration you independent variable. The gradient k, passes though the origin with a gradient of 1 due to it's direct proportionality.

(The gradient is k because mass is a constant)

Therefore, 1 (the gradient of your line) = k when this occurs this is called a linear relationship.

Hope this helps.

Callum
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