Lifting things: turning work into stored energy

Hold a bag of shopping at arm's length for a minute. Nothing moves, yet your arm aches — annoying, but physics has a neat explanation.

Now lift that bag from the floor to a shelf. You push up, the bag moves up, so you do work on it.

Where does that work go? Not into speed (the bag ends up still). It goes into a store: the bag is now higher, and if you let go, that height turns back into motion.

This lesson shows that the energy stored is exactly the work you did, and that it comes out as a beautifully simple formula.

To lift a mass $m$ at a steady speed, your upward force must exactly balance the downward pull of gravity.

The gravitational force on a mass $m$ is its weight, $W_\text{weight} = mg$, where $g = 9.81\ \text{m s}^{-2}$ near the Earth's surface.

So the lifting force you must apply has size

$F = mg$

Push any harder and the bag accelerates; push any less and it stays on the floor. For a steady lift, $F = mg$ exactly.

The bag moves vertically through a distance equal to its change in height, which we write $\Delta h$.

The Greek letter $\Delta$ (delta) just means "change in". So $\Delta h = h_\text{final} - h_\text{initial}$.

Only the vertical change matters here, because the lifting force and the weight both point vertically. Carrying the bag sideways at constant height does no work against gravity at all — more on that in a moment.

Work done by the lifting force:

$W = Fs$

Substitute the force $F = mg$ and the displacement $s = \Delta h$:

$W = mg\Delta h$

All of that work is transferred to the bag's gravitational potential energy store, so the change in that store is

$\Delta E_P = mg\Delta h$

That is the whole derivation. Three lines, one idea: the work you do lifting becomes stored energy.