Hi everyone! How do these two types of energy work for things that are being lowered down slowly? I've been told that whenever one drops something (let's say it's 1kg) from 5 meters, it will be traveling at 10 meters per second just before it stops at the ground. And I've also learned that this is true even when the object is sliding down a slide, it's just that the velocity may be going in another direction. And this is all expressed with the conservation of energy: PE at the top will be mgh, 1 * 5 * 10 (I'm using 10 for g just to make it easy for this question), and it will be the same as the KE at the bottom 1/2mv^2, 1/2 * 1 * v^2. So 50 = 1/2 * v^2 and v = 10, right? So this is how it relates to my question: How does the conservation of energy relate to this object at this height if the object is lowered down slowly? If I exerted an upward force with my hand (let's say 5 newtons) on the same object from the same height, the KE at the bottom wouldn't be 50 joules would it? So where did the original potential energy of the object go? This question popped into my mind when I was thinking about a helicopter landing so if you want to use that as in example that would be great! Thanks so much for taking the time t look over this, I really appreciate it!