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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 $10 \,\text{ms}^{−2}$?

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    $\begingroup$ 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

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The potential energy in a uniform gravitational field is:

$$ U = mgh $$

"g" is Earth's average gravitational acceleration :

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

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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:
    $10\frac {m}{s^{2}}$

and
$10\frac {m}{s^{2}} \approx 9.80665\frac {m}{s^{2}}$, 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).

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