lucKY2b wrote:
Sorry, but you are absolutely wrong on this. Change in gravitational potential energy IS mgy_f-mgy_i. If y_f = y_i, there is no net change in gravitational potential energy and you've not done work against gravity; end of story. Any additional energy expended while the treadmill is inclined is just because you do need to lift your knees a little higher, and you do cut off your stride on the back end a little, but that's it. If you were taught differently, you were misinformed.
You're confusing the net change in gravitational energy, and the added energy cost of having to run up/down hill.
It is true that, on the street, you can approximate the energy cost by the gravitational energy lost or gained. However, that doesn't mean that there isn't an energy cost on a treadmill with an incline. There is just not a net change because the treadmill motor is busy replacing the energy as you go.
On the street, the system is simple, it's just the runner, so you can get away with E=mgh. However, on the treadmill, there is another energy input into the system, so you also have to look at the energy from the motor.
Here are a few illustrative examples:
Energy (work) of a body moving a straight line is E=FD (force times distance). Since you don't go anywhere on a treadmill, are you going to assert that it requires ZERO energy input to run on a treadmill?
You could make the example more obvious by thinking about a car driving on a (larger) treadmill. Or, even better, a dyno! By your logic, because the car doesn't go anywhere, it requires zero energy. I may just be missing out, but I've never seen a car make a dyno pull without using any fuel. This is because, like running on the treadmill, there are other bodies in the system which need to be part of the energy balance.
The car example also works to illustrate the treadmill incline/decline. If you set the decline high enough, the car (or the runner) would just roll right off the treadmill (decreasing potential energy). Turning the belt on doesn't change this, it just balances out the potential energy loss with the input from the motor to result in a net static energy (but the car still acts, feels, and behaves as though it is going downhill).
Finally, I encourage you to set up a rope over a set of pulleys and a motor, so that you can have it move down at the speed at which you can climb the rope. By your logic, it will take zero energy to climb it forever, since you never go up, and thus there is no energy cost. I suspect you will find this is not the case. ;)