Problem — True or false (dimensional analysis)

Problem

True or false (dimensional analysis). (a) $[F]=[m\,a]=\text{kg}\cdot\text{m}\cdot\text{s}^{-2}$. (b) A dimensionally correct formula is necessarily physically correct. (c) An equation $E = mc^2 + mv$ is dimensionally correct. (d) Energy and work have the same dimensions.

Solution

(a) TRUE. By the second law $F = m\,a$, so

$[F] = [m][a] = \text{kg}\cdot\text{m}\,\text{s}^{-2} = \text{N}$

This is exactly the definition of the newton.

(b) FALSE. Dimensional analysis is a necessary but not a sufficient condition. A formula can have the right dimensions and still be wrong: it may be off by a pure numerical factor (e.g. a missing $2$ or $\pi$), or combine the right quantities in a physically incorrect way. E.g. $E = \tfrac12 mv^2$ and $E = 5\,mv^2$ are both dimensionally correct, but only the first is physical.

(c) FALSE. Let us compare the dimensions of the two terms on the right-hand side:

$[mc^2] = \text{kg}\cdot(\text{m}\,\text{s}^{-1})^2 = \text{kg}\,\text{m}^2\,\text{s}^{-2} \ (\text{energy}), \qquad [mv] = \text{kg}\cdot\text{m}\,\text{s}^{-1} \ (\text{momentum})$

The two terms have different dimensions: they cannot be added, so the equation is dimensionally incorrect.

(d) TRUE. Work is force times displacement, $W = F\,s$, and energy is measured in the same unit:

$[W] = [F][s] = \text{kg}\,\text{m}\,\text{s}^{-2}\cdot\text{m} = \text{kg}\,\text{m}^2\,\text{s}^{-2} = \text{J} = [E]$

$\ev{\text{(a) T}\quad\text{(b) F}\quad\text{(c) F}\quad\text{(d) T}}$

Linked notes

Topics: Grandezze fisiche e misura Skills: Analisi dimensionale