What I understand is that work is not the same as a car using gas or a crane lifting a car high up into the air. Let's use the crane as an example. And let me write out a few lines from the book.
$$W_{\Sigma F} ~=~ \Delta E_k ~=~ \frac{1}{2}mv^2 - \frac{1}{2} mv_0^2. $$
So basically this means that there really isn't any work done by the crane. The start speed for the car being lifted is 0, and so is the end speed. It makes me cringe, because I want to write out the energy the crane used to lift the car. Same as the potential energy for that given car at that height. And the same goes for a car driving up a mountain at a constant speed.
$$E ~=~ E_0 + W_A.$$
This means that the mechanical energy ($E_p + E_P$) is the same after a change if you add whatever work is done by other forces. If I apply this to the crane lifting $E_0$ would be zero because there is no potential at height=0. Which implies the mechanical energy at the top is equal to the work done by the crane. But this thinking is wrong. I can't see the difference between a guy pushing a box something with a specific force a specific distance at a constant speed and the crane lifting this car under the same parameters. I think work is only applied when something is being accelerated, but then again the book shows example of how much work is done when picking up a backpack.
Ironically I can get around my lack of understanding and do the tasks needed doing, but I would very much like to understand this properly before my practical exam next week.
