Skip to main content
Engineering LibreTexts

2.1.3: Energy Units

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    It was said above that the mechanical energy is of "special importance" in physics. Why? Because the most important energy unit used in physics, the Joule, is defined as - let's again quote Wikipedia - the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of its motion through a distance of one meter.

    But why is it "the most important"? Well, the thing is that in modern physics there are special regulations, known as the "SI system", stating that the units of all quantities used in physics should be defined in terms of seven base quantities. Three of these seven are the units of length, the meter \((\mathrm{m})\), of mass, the kilogram \((\mathrm{kg})\), and of time, the seconds (s). For more details, please look at a NIST Web page (NIST = National Institute of Standards and Technology).

    So, the SI unit of force is a Newton:
    1 \mathrm{~N}=1 \frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^{2}}
    Why such a combination? In short, it all comes from something known as "the Second Newton's Law of Dynamics". And how it comes? Well, I think that rather than explaining it step-by-step, I should give you a link to a good Internet source. And if you don't yet have an idea "how big" the force of \(1 \mathrm{~N}\) is, I can tell you that the weight of something with the mass of one pound is approximately \(4.5 \mathrm{~N}\); and of 1 kilogram mass, about \(9.81\) \(\mathrm{N}\).

    Now, once we know the SI unit of \(1 \mathrm{~N}\), we can readily find the SI unit of a Joule:
    1 \mathrm{~J}=1 \mathrm{~N} \cdot 1 \mathrm{~m}=1 \frac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{2}}

    2.1.3: Energy Units is shared under a CC BY 1.3 license and was authored, remixed, and/or curated by Tom Giebultowicz.

    • Was this article helpful?