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.
Best Answer
Think of the following problem: An ice-skater starts at rest in the middle of a frozen lake. They speed up to a maximum velocity $v_f$ and then they drift over the ice without friction. After some frictionless sliding the skater goes over a region of ice with sand on it. The ice-with-sand surface has certain friction with the skate blades. The skater slides for a few meters and then stops due to this friction.
What is the minimum work done by the skater of mass $m$ to reach the velocity $v_f$?
What is the work done by the friction force in the region of ice covered with sand?
If you understand that, you will see that a crane is doing the work of the skater, to accelerate the body it is carrying until a given velocity (and it has to compensate for friction on top of that) and then it has to do extra work to stop it. If the crane has no internal friction and there is no friction with the environment, then everything is symmetric and the answer is quite easy, let say $W$. In the case of friction, nothing can be said in general but that the work $W_{fr} \geq W$.