Here is the payoff. Gravitational $U = mgh$, kinetic $E_k = \tfrac{1}{2}mv^{2}$, elastic $U = \tfrac{1}{2}kx^{2}$ — three accounts, one currency, and when friction is absent the total never changes. It only moves. Height becomes speed; speed becomes squeeze; squeeze becomes height again. That is conservation of mechanical energy, and it lets you skip every intermediate step: you never need the force, the acceleration or the time, only the two ends of the trade.

The famous consequence: set $mgh = \tfrac{1}{2}mv^{2}$ and the mass appears on both sides, so it cancels — $v = \sqrt{2gh}$. Everything sliding down the same smooth slope arrives at the same speed, whatever it weighs. A question that hands you a mass and then does not need it is not being careless; it is checking whether you know that.

Set $\tfrac{1}{2}mv^{2} = \tfrac{1}{2}kx^{2}$ instead and the halves cancel, leaving $x$ proportional to $v$ — a straight line, and a rare mercy in a chapter this full of squares. And when friction IS present, nothing about the method breaks: friction's negative work simply appears as a withdrawal on one side of the ledger.