CLASSICAL-MECHANICS

Built from Constants: How the Universe Defines a Kilogram

IKey Formulas

Quantity Definition Base-Unit Form
Velocity v=Ξ”xΞ”tv = \dfrac{\Delta x}{\Delta t} mβ‹…sβˆ’1\text{m}\cdot\text{s}^{-1}
Acceleration a=Ξ”vΞ”ta = \dfrac{\Delta v}{\Delta t} mβ‹…sβˆ’2\text{m}\cdot\text{s}^{-2}
Force (newton, N) F=m aF = m \, a kgβ‹…mβ‹…sβˆ’2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}
Energy (joule, J) E=F dE = F \, d kgβ‹…m2β‹…sβˆ’2\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}
Power (watt, W) P=EtP = \dfrac{E}{t} kgβ‹…m2β‹…sβˆ’3\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-3}

The three mechanics base units are the meter (m), kilogram (kg), and second (s) β€” the MKS system.

IIInteractive Visualizations

Visualization 1 β€” The Derived-Unit Cascade

Every mechanical unit is just kg, m, and s raised to integer powers, stacked one operation at a time.

Visualization 2 β€” Newton's Second Law, F=m aF = m\,a

The newton is exactly kgβ‹…mβ‹…sβˆ’2\text{kg}\cdot\text{m}\cdot\text{s}^{-2} β€” the area of a rectangle whose sides are mass and acceleration.

πŸ’‘ Challenge: find a mass and an acceleration that produce exactly 66 N.

Visualization 3 β€” Fix a Constant, Define a Unit

Fix cc at exactly 299,792,458Β mβ‹…sβˆ’1299{,}792{,}458\ \text{m}\cdot\text{s}^{-1} and the meter becomes whatever distance light covers in 1/c1/c seconds.

πŸ’‘ Same trick, the kilogram: it is fixed by declaring the Planck constant h=6.62607015Γ—10βˆ’34Β kgβ‹…m2β‹…sβˆ’1h = 6.62607015 \times 10^{-34}\ \text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-1}.

IIIQuiz Questions

Question 1

Force is defined as F=m aF = m\,a. Expressing the newton (N) purely in SI base units, which combination is correct?

βœ… Correct! A newton is kgΒ·mΒ·s⁻².

❌ Not quite. Multiply kg by acceleration (m·s⁻²), not velocity.

Show solution

Solution:

Force = mass Γ— acceleration. Mass is in kg, and acceleration is in mβ‹…sβˆ’2\text{m}\cdot\text{s}^{-2}:

F=kgΓ—mβ‹…sβˆ’2=kgβ‹…mβ‹…sβˆ’2F = \text{kg} \times \text{m}\cdot\text{s}^{-2} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

So one newton is exactly kgβ‹…mβ‹…sβˆ’2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}.

Question 2

Energy is force times distance: E=F dE = F\,d. Starting from the newton (kgβ‹…mβ‹…sβˆ’2\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) and multiplying by one more meter, what are the base units of the joule (J)?

βœ… Correct! The extra meter raises m from 1 to 2.

❌ Not quite. That is the watt β€” you divided by an extra second instead of multiplying by a meter.

❌ Not quite. Multiply the newton by one meter, so the power of m goes up by one.

Show solution

Solution:

Take the newton and multiply by one meter (the distance):

E=(kgβ‹…mβ‹…sβˆ’2)Γ—m=kgβ‹…m2β‹…sβˆ’2E = (\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \times \text{m} = \text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}

The power of m increases from 1 to 2; the seconds are unchanged. A joule is kgβ‹…m2β‹…sβˆ’2\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}.

Question 3

True or False: The 2019 redefinition of the SI units, which anchored the kilogram to the Planck constant, changed the actual size of a kilogram so that masses had to be re-measured worldwide.

βœ… Correct! Only the definition changed, not the magnitude of the unit.

❌ Not quite. The redefinition kept the unit sizes identical β€” only the definition was changed.

Show solution

Solution:

The sizes did not change. A meter is still a meter and a kilogram is still a kilogram. Only the definition changed β€” from a physical platinum-iridium cylinder to a fixed numerical value of the Planck constant. The benefit is greater precision and permanence, with no physical artifact that can drift, tarnish, or lose atoms.

The statement is therefore False.

Question 4

Power is energy divided by time: P=E/tP = E / t. Given that the joule is kgβ‹…m2β‹…sβˆ’2\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}, what are the base units of the watt (W)?

βœ… Correct! Dividing by a second adds one more s⁻¹.

❌ Not quite. That is the joule β€” remember to divide by one more second.

❌ Not quite. Start from the joule and divide by time, lowering the s exponent by one.

Show solution

Solution:

Divide the joule by one more second:

P=kgβ‹…m2β‹…sβˆ’2s=kgβ‹…m2β‹…sβˆ’3P = \frac{\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}}{\text{s}} = \text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-3}

Dividing by time adds another second to the denominator, so the exponent of s goes from βˆ’2-2 to βˆ’3-3. A watt is kgβ‹…m2β‹…sβˆ’3\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-3}.

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