I agree with the Dude, but wanted to try to quantify an example for my own interest.

Please don't read further if you aren't interested in the maths, it's just a different way to learn for us dorky engineers.

Think of how easy it is to put a ski tail on the ground hold the tip and bend it into reverse camber with your hand/arm over the time it takes to complete even a short turn. Not a lot of energy required or therefore stored.

To convince myself of the force required to hold a ski in reverse camber, I just stood on some bathroom scales and measured an increase of 40kg to bend my skiis 50mm (2 inches) at the binding with tip and tail supported under two adjacent shelves. So the force is, F=ma= 40x9.81=392N, (acceleration due to gravity approx 9.81)

Now think of trying to stop a skier with the momentum, lets say 100kg for convenience doing 30 km/hr , (8 m/s) at the end of the turn. Try to stop that with your arm! So intuitively, I suspect that the bent ski does not contribute much to "The Force" at release.

A bent, edged ski is storing an amount of energy, but it is transferring the force necessary to hold all force created by the moving skier, (force due to both gravitational acceleration and centripetal acceleration), to the snow/ground.

F(gravity) =ma = 100x9.81=981N

(acceleration due to gravity approx 9.81 m/s ^2)

F (centrifugal) = ma(centripetal)

a(centripetal) = v(tangential)^2 / r

Lets say the turn radius is 15m

and the tangential velocity is 8 m/s

a(centripetal) = 8 ^2 /15 = 4 m/s^2

so F (centrifugal) = 100 x 4= 400N

Do the vectors, gravity is straight down, centripetal acceleration is up the fall line. The slope of the hill? say 40degrees from horizontal. We have force due to gravity straight down to the centre of the earth and force due to the turn straight down the fall line.

I get total force is approx 1236 N at 76 degrees below the horizontal.

This means the ski must be tipped 14 degrees above horizontal, or if we add the slope of the hill, 54 degrees above the slope.

So: we have the ski requiring a force of 392N to keep it bent, but transferred through the ski to the snow, and with stored energy available to re-camber the ski at release.

I think that the energy in the bent ski simply helps the stance leg retract at the release/transfer. As the ski is no longer on the snow resisting the forces, the body is propelled down the hill in the direction of the forces at an acceleration of this force divided by the mass of the skier.

a=F/m= 1236/100 = 12.36m/s^2 or about 1.25 g's.

No wonder we need to re-centre during this part of the turn. Of course if we tighten the radius, increase the tangential velocity or increase the mass of the skier this number quickly increases and so does the required tipping of the ski.

If I have this right, The "Force", is mostly the combined forces on the skier mentioned above and the force used to bend the ski simply helps the retraction of the stance leg at release.