# Part of physics answer I didn't understand, and don't have reputation to ask them. $10\,\text{ms}^{−2}$

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 $$10 \,\text{ms}^{−2}$$?

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

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

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).