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Potential Energy

Imagine yourself on skis, poised motionless at the top of a snow-covered hill: one way or another, you are deeply aware of the potential of the hill to increase your speed. In Physics we like to think of this obvious capacity as the potential for gravity to increase your kinetic energy. We can be quantitative about it by going back to the bottom of the hill and recalling the long trudge uphill that it took to get to the top: this took a lot of work, and we know the formula for how much: in raising your elevation by a height  h  you did an amount of work  W = m g h  ``against gravity'' [where  m  is your mass, of course]. That work is now somehow ``stored up'' because if you slip over the edge it will all come back to you in the form of kinetic energy! What could be more natural than to think of that ``stored up work'' as gravitational potential energy

$$V_g = m \, g \, h$$ (11.13)

which will all turn into kinetic energy if we allow  h  to go back down to zero?^11.13

We can then picture a skier in a bowl-shaped valley zipping down the slope to the bottom   $[V_g \to T]$  and then coasting back up to stop at the original height   $[T \to V_g]$  and (after a skillful flip-turn) heading back downhill again   $[V_g \to T]$. In the absence of friction, this could go on forever:   $V_g \to T \to V_g \to T \to V_g \to T \to$  . . . .

The case of the spring is even more compelling, in its way: if you push in the spring a distance  x, you have done some work   $W = {1\over2} k \, x^2$  ``against the spring.'' If you let go, this work ``comes back at you'' and will accelerate a mass until all the stored energy has turned into kinetic energy. Again, it is irresistible to call that ``stored spring energy'' the potential energy of the spring,

$$V_s = {1\over2} k \, x^2$$ (11.14)

and again the scenario after the spring is released can be described as a perpetual cycle of   $V_s \to T \to V_s \to T \to V_s \to T \to$  . . . .