# How does conservation of energy work?

I understand that the law of conservation of energy states that energy cannot be created or destroyed, it can only change from one form to another. This means that the total energy before a certain event that involves getting work done, will be the same after the event is over. So lets say i have an object of mass $M$ and i put it on the palm of my hand and i move it up by a distance of $H$ with a constant velocity $V_1$ That means that when i stop lifting the object after reaching the desired height, all the kinetic energy $\frac {MV_1^2}2$ would have been transformed to gravitational potential energy $MgH$ but since the law of conservation of energy does not take time under consideration, that would mean that if I try lifting the object again for the same distance but with greater velocity $V_2$ (not great enough for the object to leave my hand after i stop) then the kinetic energy of the object would be $\frac {MV_2^2}2$ which is greater than what it was the first time. however, the gravitational potential energy would still remain the same in both cases. So where did the extra energy go?

• The extra energy is spent by your body on accelerating your hand quicker Apr 11, 2016 at 6:35

The answer is to do with what happens when you stop moving the object upwards. You say that the upwards velocity is 'low enough that it does not leave your hand': the only value of velocity for which that is true is $0$. What this means is that, if at time $t_0$ the object is being moved upwards with velocity $v_0$ and is at $h_0$, and you suddenly stop pushing on it, then it will, of course, continue upwards with a velocity $v(t) = v_0 - g(t-t_0)$ and height $h(t) = h_0 + v_0 (t- t_0) - g(t-t_0)^2/2$.

Well, you can solve these equations for the maximum height reached, which is $h_0 + v_0^2/(2g)$. And not surprisingly there is just enough extra potential energy there to account for the kinetic energy it had at $t_0$.

• This does not make sense. If i grab an object, lets say a cup, and i take five minutes lifting it to a distance of 50 cm then it would not leave my hand when i stop. if i lift the same cup for the same distance and i only spent 3 minutes this time, the velocity wold be higher than the first time, but it won't be high enough for it to leave my hand right? Apr 12, 2016 at 9:22
• @user3769877: the reason it doesn't leave your hand is because, at the top of the lift, you accelerated your hand downwards with an acceleration whose absolute value is less than $g$ (assuming the object is just resting on your hand). At the very top of the lift the upward velocity was just $0$. During the period of the acceleration, the force exerted on your hand by the object is $m(g-a(t))$, where $a(t)$ is acceleration (+ve down), and if you integrate this suitably you'll find the same difference in work as above. If the acceleration is greater than $g$ then you have thrown the object.
– user107153
Apr 12, 2016 at 9:51

If you lift a book, then you are doing work on it. You are actually spending work on raising it's kinetic energy and on lifting it up.

$$U=K+W$$

When you reach some height and stop, then all the work done is what determines your height. If you move your hand faster in order to speed up the book to a larger kinetic energy, then you've spent more work on that and can spent less work on the lifting. In total the final height and therefore potential energy and therefore the sum of kinetic energy reached and further work done by your hand is the same.

If you happen to do so much work on the book to speed it up, that it's kinetic energy is larger than the potential energy that fits the desired height, then the book will lift off from your hand and fly a bit higher when you stop, to indeed reach this height.