Worked examples — SI units — seven base units and all derived units

Link back to the parent: SI units — seven base units and all derived units. This work leans on Dimensional Analysis (checking equations by units), Significant Figures and Errors (rounding answers), Kinematics — velocity and acceleration, Newton's Laws of Motion and Work, Energy and Power (the equations that define the units).

---

The scenario matrix

Before we work anything, here is the full map of case-classes a "unit derivation" question can throw at you. Every worked example below is tagged with the cell(s) it covers.

Cell	Case class	What makes it tricky
A	Straight forward-derivation ($F=ma$ style)	just substitute base units
B	Rearrange-first (unknown is buried)	isolate the target, THEN substitute
C	Negative / fractional powers appear	dividing units lowers exponents; watch signs
D	Dimensionless factor present ($\frac{1}{2}$, $2\pi$, $\sin \theta$)	numbers & pure ratios carry NO unit
E	Degenerate / zero input	what happens when a quantity is $0$ or a unit "cancels out"?
F	Prefix trap (km, cm, g, hour)	convert to base units first
G	Real-world word problem	pull the equation out of a story
H	Exam twist (verify a claimed formula)	is this even dimensionally legal?
I	Limiting / consistency check	do two routes give the same base form?

The rule that survives every cell:

---

Example 1 — Cell A (straight derivation)

Forecast: guess the exponent of the second before reading on — is it $-2$ or $-3$?

1. Write the recipe: $[P]=\frac{[W]}{[t]}=\frac{{\text{kg⋅m}}^2{\text{⋅s}}^{-2}}{\text{s}}$. Why this step? Power's defining equation is energy over time; the unit is that same division.
2. Divide by a second → subtract $1$ from the exponent of $\text{s}$: $-2-1=-3$. Why this step? Unit algebra: dividing by ${\text{s}}^1$ lowers the $\text{s}$ power by one.
3. Result: $[P]={\text{kg⋅m}}^2{\text{⋅s}}^{-3}\equiv 1\text{W}$.

Verify: cross-check via $P=Fv$: $[F][v]=({\text{kg⋅m⋅s}}^{-2})({\text{m⋅s}}^{-1})={\text{kg⋅m}}^2{\text{⋅s}}^{-3}$. ✔ Same answer, different equation — that's a consistency win (Cell I).

---

Example 2 — Cell B + C (rearrange, negative powers)

Forecast: capacitance is charge divided by voltage — will $\text{kg}$ end up positive or negative?

1. Substitute known base forms: $[Q]=\text{A⋅s}$ and $[V]={\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1}$. Why this step? We already derived these in the parent note; reuse, don't rederive.
2. Divide: $[C]=\frac{\text{A⋅s}}{{\text{kg⋅m}}^2{\text{⋅s}}^{-3}{\text{⋅A}}^{-1}}$. Why this step? The defining equation is a division, so the unit is a division.
3. Subtract exponents base by base: =\text{A}^{2}\,\text{s}^{4}\,\text{kg}^{-1}\,\text{m}^{-2}.$$ **Why this step?** Dividing by a quantity flips the sign of *its* exponents, then we add. The $\text{kg}$ was upstairs in the denominator, so it lands as $\text{kg}^{-1}$ — a **negative power**, exactly Cell C.
4. Result: $[C]={\text{A}}^2{\text{⋅s}}^4{\text{⋅kg}}^{-1}{\text{⋅m}}^{-2}\equiv 1\text{F}$.

Verify: matches the reference table in the parent note. ✔

---

Example 3 — Cell D (dimensionless factors)

Forecast: all three look like energy. Any impostors?

1. (a) $[\frac{1}{2}m{v}^2]$: drop the $\frac{1}{2}$ (pure number, Cell D), then $\text{kg}\cdot ({\text{m⋅s}}^{-1}{)}^2={\text{kg⋅m}}^2{\text{⋅s}}^{-2}=\text{J}$. ✔ Why this step? A dimensionless coefficient can never change a unit; only the variables count.
2. (b) $[2\pi mgh]$: drop $2\pi$ (also dimensionless), then $\text{kg}\cdot ({\text{m⋅s}}^{-2})\cdot \text{m}={\text{kg⋅m}}^2{\text{⋅s}}^{-2}=\text{J}$. ✔ Why this step? $g$ is an acceleration ${\text{m⋅s}}^{-2}$; $2\pi$ is just geometry, no unit.
3. (c) $[p^2/(2m)]$: $\frac{({\text{kg⋅m⋅s}}^{-1}{)}^2}{\text{kg}}=\frac{{\text{kg}}^2{\text{⋅m}}^2{\text{⋅s}}^{-2}}{\text{kg}}={\text{kg⋅m}}^2{\text{⋅s}}^{-2}=\text{J}$. ✔ Why this step? Square the momentum unit (double every exponent), then divide by mass (subtract).

Verify: all three collapse to ${\text{kg⋅m}}^2{\text{⋅s}}^{-2}$. All are genuine energies. ✔

---

Example 4 — Cell F + G (prefix trap in a word problem)

Forecast: will using grams and cm make the number 1000× or 100× off — or both?

1. Convert to base units first. Mass is already $1200\text{kg}$ (do NOT feed grams into an SI equation). Height $45\text{m}$ (do NOT use $4500\text{cm}$). Time $30\text{s}$. Why this step? The equation $mgh/t$ produces watts only if every input is in base SI. Grams and centimetres would silently multiply your answer by wrong powers of ten (Cell F).
2. Substitute: $P=\frac{(1200)(9.8)(45)}{30}\text{W}$. Why this step? With clean base units the number comes out directly in watts — no extra conversion needed at the end.
3. Compute numerator: $1200\times 9.8=11760$; $\times 45=529200$. Divide by $30$: $P=17640\text{W}=17.6\text{kW}.$ Why this step? Rounded to 3 significant figures (Significant Figures and Errors).

Verify: unit check $\frac{\text{kg}\cdot {\text{m⋅s}}^{-2}\cdot \text{m}}{\text{s}}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}=\text{W}$. ✔ And $17640\text{W}$ ≈ 24 horsepower — believable for a car on a hill. ✔

---

Example 5 — Cell E (degenerate / zero input)

Forecast: if there's still a force, is any work done? Does the unit even exist when $d=0$?

1. Work is $W=Fd$, so $[W]=\text{N⋅m}={\text{kg⋅m}}^2{\text{⋅s}}^{-2}=\text{J}$ — the unit is unchanged. Why this step? A unit is a type; it does not disappear just because a value is zero. Even $0\text{J}$ is measured in joules.
2. The value: $W=F\times 0=0\text{J}$. Why this step? Cell E — a degenerate (zero) input collapses the number but not the dimension. Zero displacement ⇒ zero work, regardless of how hard you push.
3. Contrast with $W=Fd\cos \theta$ at $\theta =90^{\circ }$: $\cos 90^{\circ }=0$, so again $W=0\text{J}$ even with motion — because the force is perpendicular to the path. Why this step? $\cos \theta$ is dimensionless (Cell D), so it changes the value to zero but never the unit.

Verify: $F\cdot 0=0$ and $\cos 90^{\circ }=0$ ⇒ $W=0\text{J}$ both ways. ✔

---

Example 6 — Cell H (exam twist: is the formula even legal?)

Forecast: one of these gives seconds; one gives something absurd. Which?

1. Handle the original. Inside the root: $\frac{L}{g}=\frac{\text{m}}{{\text{m⋅s}}^{-2}}={\text{s}}^2$. Why this step? Dividing metre by (metre per second²): the metres cancel and ${\text{s}}^{-2}$ in the denominator flips to ${\text{s}}^{+2}$.
2. Take the square root: $\sqrt{{\text{s}}^2}=\text{s}$. Multiply by dimensionless $2\pi$: still $\text{s}$. Why this step? A square root halves every exponent ($2\to 1$); $2\pi$ carries no unit (Cell D). Result: seconds — a valid period. ✔
3. Handle the rival. Inside: $\frac{L}{g^2}=\frac{\text{m}}{({\text{m⋅s}}^{-2}{)}^2}=\frac{\text{m}}{{\text{m}}^2{\text{⋅s}}^{-4}}={\text{m}}^{-1}{\text{⋅s}}^4$. Why this step? Squaring $g$ doubles its exponents; then divide into $\text{m}$.
4. Square root: $\sqrt{{\text{m}}^{-1}{\text{⋅s}}^4}={\text{m}}^{-1/2}{\text{⋅s}}^2$. Why this step? Halving the exponents gives $-\frac{1}{2}$ on metre and $2$ on second — this is not a time. The rival formula is dimensionally impossible. ✗

Verify: original → $\text{s}$; rival → ${\text{m}}^{-1/2}{\text{⋅s}}^2$. Only the original is a valid period. ✔ (See the figure — the two "unit towers" end at different heights.)

---

Example 7 — Cell C + I (fractional powers, two routes agree)

Forecast: a square root of (force ÷ linear density) — does it really come out as a speed?

1. Units inside: $[T_{\text{en}}]={\text{kg⋅m⋅s}}^{-2}$, $[\mu ]={\text{kg⋅m}}^{-1}$. Why this step? Tension is a force; $\mu$ is kilograms per metre.
2. Divide: $\frac{{\text{kg⋅m⋅s}}^{-2}}{{\text{kg⋅m}}^{-1}}={\text{m}}^{1-(-1)}{\text{⋅s}}^{-2}={\text{m}}^2{\text{⋅s}}^{-2}$. Why this step? $\text{kg}$ cancels ($1-1=0$); metre exponent $1-(-1)=2$; the ${\text{s}}^{-2}$ is untouched.
3. Square root halves exponents: $\sqrt{{\text{m}}^2{\text{⋅s}}^{-2}}={\text{m⋅s}}^{-1}$. Why this step? $\frac{2}{2}=1$ and $\frac{-2}{2}=-1$ — a genuine speed (Cell C fractional step landing on whole powers).

Verify: ${\text{m⋅s}}^{-1}$ is exactly the unit of velocity from Kinematics — velocity and acceleration. ✔ Consistency with Example 1's cross-checks confirms our unit-algebra rules are being applied the same way everywhere (Cell I).

---

Example 8 — Cell B + G (word problem, hidden rearrange)

Forecast: 2 minutes hides a prefix trap — spot it before computing.

1. Convert time to base units: $2\text{min}=120\text{s}$ (Cell F — never leave minutes in an SI power calculation). Why this step? The watt is joules per second; feeding "2" would over-count by $60\times$.
2. Apply $P=\frac{W}{t}=\frac{360000}{120}\text{W}$. Why this step? Power is energy per unit time — the defining equation.
3. Compute: $P=3000\text{W}=3\text{kW}$. Why this step? A 3 kW kettle is a realistic value — a sanity anchor (Cell G).

Verify: unit chain $\frac{\text{J}}{\text{s}}=\frac{{\text{kg⋅m}}^2{\text{⋅s}}^{-2}}{\text{s}}={\text{kg⋅m}}^2{\text{⋅s}}^{-3}=\text{W}$. ✔ Number: $360000/120=3000$. ✔

---

Recap

Recall Which cell is which?

Straight substitution = Cell ::: A Rearrange to isolate the target first = Cell ::: B Negative or fractional exponents appear = Cell ::: C Ignore $\frac{1}{2}$, $2\pi$, $\cos \theta$ in a unit check = Cell ::: D Zero displacement gives $0\text{J}$ but the unit stays joules = Cell ::: E Convert km/cm/g/minutes to base units before substituting = Cell ::: F Base-unit form of the farad? ::: ${\text{A}}^2{\text{⋅s}}^4{\text{⋅kg}}^{-1}{\text{⋅m}}^{-2}$ A pendulum period $2\pi \sqrt{L/g}$ has units of? ::: seconds ($\text{s}$) Wave speed $\sqrt{T_{\text{en}}/\mu }$ has units of? ::: ${\text{m⋅s}}^{-1}$ Kettle: $360000\text{J}$ in $2\text{min}$ → power? ::: $3000\text{W}$ ($3\text{kW}$)