When you lift your $2.5 \,\text{kg}$ laptop (a 15-inch Apple MacBook Pro, for example) by a foot, you do work of approximately $2.5 \,\text{kg} \times 10 \,\text{ms}^{−2} \times 0.3 \,\text{m} = 7.5 \,\text{J}$ which is about $4.7\times10^{19} \,\text{eV}$. This is what I saw, my reputation is to low to ask them. Where does this Value come from $10 \,\text{ms}^{−2}$?

  • 1
    $\begingroup$ See Gravity of Earth. $\endgroup$ Oct 19, 2020 at 16:40
  • $\begingroup$ BTW, under each question and each answer is the word "share". If you click that, it gives a URL to the question/answer. $\endgroup$ Oct 19, 2020 at 21:07

2 Answers 2


The potential energy in a uniform gravitational field is:

$$ U = mgh $$

"g" is Earth's average gravitational acceleration :

$$ g = 9.80665\,{\rm m/s^2}\approx 10\,{\rm m/s^2}$$


You asked: "Where does this Value come from 10ms^−2 ?"

  • 10ms^-2
  • is the same as writing:
    $10m * s^{-2}$
  • or the same as:
    $10m * \frac {1}{s^2}$
  • or the same as:
    $\frac {10m}{s^{2}}$
  • or the same as:
    $10\frac {m}{s^{2}}$

$10\frac {m}{s^{2}} \approx 9.80665\frac {m}{s^{2}}$, which is the Earth's g constant (Earth's acceleration) that is used to convert mass (in kilograms) to weight or force (in Newtons). e.g.: 1 kg on Earth weighs 9.80665 Newtons.

Newtons * Meters = Joules (the units of energy or work).

...where the height of your lift is expressed in Meters.

Once you have the Joules calculated, you can multiply them by $6.24150974×10^{18}$ to get the energy (work) expressed in Electron Volts (eV).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.