The surprise of some engineers about comparing kinetic energy to lift?!?.
Yes, let's look at the formulas, and how we apply them. In steady state flight, we talk about the 4 FORCES of flight. F = kg m/s$^2$ = ma: Lift, Gravity, Thrust, Drag
So why not Kinetic Energy KE = kg m$^2/s^2$ = F × d, or even Power P = kg m$^2/s^3$ = F × d/t?
Because, in describing forces on an aircraft in steady state flight, everything happens in the same distance at the same time, so d/t cancels out, leaving a balance of forces.
However, we can easily determine total drag from the glide ratio.
The energy state of an aircraft is its potential energy mgh + its kinetic energy 1/2mv$^2$.
Assuming it maintains constant speed in a glide and flys in a straight line at constant AOA to its landing point, energy consumed is all mgh: kg m$^2$/s$^2$.
There for total drag force × distance = mgh
So if the glide ratio is 10 to 1, total kinetic energy converted from potential energy is around kg m/s$^2$ × distance = kg × gravity (m/s$^2$) × height. Canceling units we have,
for a 1000 lb airplane (thrust = drag): 100 lbs of thrust force is required for level flight.
To determine lift to drag ratios, we move to the wind tunnel, and develop data points for
the formula Ctotal drag = Cdrag form + Cdrag induced. As seen in Peter Kampf's graph, this will vary depending on airspeed and angle of attack.
A better wing will have a higher lift to total drag ratio, and considering that these are aircraft, a better question may be how much "forward energy" is lost NOT creating lift.
Drag is a product of motion, and apportioned to lifting AND moving forward, nothing is wasted or lost. "Forward energy" is thrust × distance. What one can do next is to begin to work on efficiency, so one may need less to go as high and as far.