Newtonian mechanics - where does the energy come from?

I am imagining an object on the floor. If I pick up the object moving it upwards with a constant velocity then a net force must be applied to that object in order to have it go from not being in a state of rest to being in motion. It has accelerated and F(net)=ma. It now has kinetic energy equal to 1/2mv^2. As the object moves upwards (at constant velocity) it gains gravitational potential energy (GPE). I am under the impression that the work done on that object is equal to the GPE that it gains as it moves upwards.

So if the work done has gone into storing the GPE and they are equal, where did the energy come from which made it accelerate and gain kinetic energy in the first place? It seems as though the energy put into lifting it must be equal to the acceleration caused and thus the kinetic energy gained AND the GPE too. I am also under the impression that the kinetic energy and GPE are equal. So it seems as though I have done work on the object and the work done is equal to the sum of K.E and GPE. How is that possible? Surely that way I have gotten out twice the energy that I put in which is clearly nonsense.

Also, if the K.E and GPE are equal, how is it that it can keep moving upwards with a constant velocity and hence a constant K.E and the GPE is increasing?

The more I write and the more I go over this in my head, the more confused I get and the more frustrated I get which just makes it more difficult. Not to mention the headaches!

Again, sorry if this is really stupid and i'm fairly certain that what I have asked here will make little sense but it's the best that I can do with the mess in my head. Even just writing this I have thought of many other apparent problems and contradictions which I can't even begin to formulate into a coherent question.

Any help that anybody can offer will be greatly appreciated.

The main equation for you is that of energy conservation:

$$K_1+U_1+W=K_2+U_2$$

The numbers represent before and after, or situation 1 and situation 2.

• Kinetic energy $K$ is "motion" energy. Objects having a speed (not acceration, that doesn't matter) have kinetic energy: $$K=\frac12 mv^2$$
• Potential energy $U$ is "stored" energy. This appears when objects "want to" move somewhere else - in this case the object wants to fall down again because of gravity, so by lifting it up there is "stored" energy which can come in use by letting it go: $$U=mgh$$
• Work $W$ is energy added. And work is done by forces $F$: $$W=F x$$ where $x$ is the distance moved. So, if a force - like the one applied by your hand when lifting the object - lifts the object the distance $x$, then it has done the work $W=Fx$ on the object; in other words, it has added this mount of energy to the system.

To begin with there is no motion and the object is at the bottom, so $K_1=0$ and $U_1=0$. You then do work $W$, which turns into kinetic energy while some of it is also stored as potential energy. The equation becomes:

$$K_1+U_1+W=K_2+U_2 \quad\Leftrightarrow\quad 0+0+W=K_2+U_2 \quad\Leftrightarrow\quad Fx=\frac12mv^2+mgh$$

So the work is equal to the sum of the final energies in situation two. All that energy comes from the energy input in the form of work.

The acceleration does not have an influence on all this. The only thing acceleration has an influence on is how big the force will be - from Newton's 2nd law, a larger acceleration requires larger forces. So if the object is accelerated largely, then that would have caused a larger force and thus more work $W$ done.

Even just writing this I have thought of many other apparent problems and contradictions which I can't even begin to formulate into a coherent question.

Don't give up!

You've got some things mixed up, but you already know that. Here's what they are in particular:

I am under the impression that the work done on that object is equal to the GPE that it gains as it moves upwards.

This is not true in general, but only at times when the object is stationary — when it has zero kinetic energy. So before you start lifting it, and after you stop, this is true.

where did the energy come from which made it accelerate and gain kinetic energy in the first place?

You can look at your action as having three parts, overlapping in time:

1. You accelerated it upward, increasing the K.E. above zero. You did work (over a short distance) to do this.
2. You are doing more work to keep it moving upward at a constant velocity — as opposed to it slowing down and falling under the influence of gravity. This work is where most of the final gravitational potential energy comes from.
3. At the end, you stop it (put it on a shelf, just hold it in midair, whatever). Now the K.E. is zero again, and this is the reverse of step 1.

Now, there are actually two different ways step 3 could happen.

• Suppose that you move stiffly and ensure that the object comes to an abrupt stop. In this case, the energy that was the K.E. of the object returns to your arm (and turns into heat as you oppose it, because sadly muscles aren't reversible systems).

• Or, you could (and probably will) stop pushing upward before the object reaches its final height. Then the K.E. turns into G.P.E. over time, just like in a free-flying projectile, except that you will stop it before it starts falling down again. This is more efficient, and means that all of the work you did on the object (neglecting air resistance and such) becomes G.P.E. eventually.

The extreme version of this is throwing an object to a higher place: you're lifting it with your arm a little bit, but mostly you're adding K.E. which gradually turns into G.P.E., and when it hits its landing spot it has only G.P.E. (However much more energy it had than it needed — reflected in how far it falls beyond its peak height — is dissipated in the impact when it stops.)

When you lift the object from a stationary position, you cant move it at a constant speed, not at the beginning at least. You start from V=0 and depending on the force you apply which is required to be larger than object's weight (mg) object accelerates up to your desired final speed, Vf at the height of H.

The work you have done is two parts: lifting the object to elevation of H which is preserved as potential energy and the extra work that has gone into giving the object its speed, Vf.
Now if you let go of the object at elevation H it will keep going up decelerating by this equation $$1/2m.Vf^2=m.g.dH$$ and dH is additional height object climbs before it burns the kinetic energy and stops at height of $$H_final =H+dH$$ and then it will fall down.

There is no mystery as you see! Any and all work you have done on the object has been added together and converted to its potential energy!

OK, you're mixing things together in your mind.

To start with, forget about velocity and acceleration. Just think about force and distance. Work = Energy = Force times distance.

Now, make a distinction between being in a gravity field like on the surface of the earth, versus being way out in space. What makes them similar is general relativity, but for heaven's sake, don't worry about that.

When some weight is lifted through a distance, what is the force? The weight, right? So if you climb a ladder 1 meter, and you weigh 1kg, you have transferred 1 kg-meter of potential energy from your muscles to your gravitational potential energy. (Force should really be measured in Newtons, but I don't want to add confusion.)

Now if you're in space, and you push on some weight with some force for some distance, then you accelerate it up to a certain velocity, then the work you did on it (which is energy) went from your muscles into its kinetic energy.

OK, back to earth.

When you lift an object, you're exerting a force because of gravity, and that makes potential energy.

When you let go and it drops, the force of gravity accelerates it downward, and as it gets lower and speeds up, its potential energy converts into kinetic energy. That is, until it smashes into the ground, and its kinetic energy converts into heat and into breaking all the chemical bonds holding the object together.

• As the question has the sentence "If I pick up the object moving it upwards with a constant velocity then a net force must be applied to that object in order to have it go from not being in a state of rest to being in motion.", I would mention it explicitly that when you lift an object on Earth, both things happen: In the brief moment between the object being in rest and the object being in constant motion, you exert a force that is larger than the weight of the object, thus also contributing to its kinetic energy. – JiK Sep 3 '16 at 15:29
• "Force should really be measured in dynes," Newtons are better they are standard. – Caridorc Sep 3 '16 at 18:34

To start use Newton's second law $\vec F~=~m\vec a$ with the force given by Newton's gravity principle, $$\vec F~=~-\frac{GMm\vec r}{|\vec r|^3}.$$ Here the vector $\vec r$ is between the two masses $m$ and $m$. The motion of the small mass is of particular interest, and for $m~<<~M$ the motion of $M$ can be ignored. Now integrate the force along a path $$\int\vec F\cdot d\vec r~=~m\int\vec a\cdot d\vec r.$$ This defines energy as the work done by a force displaced through a distance.

With this integration we first work on the right hand side of the equation. The acceleration is the time rate of change in momentum $\vec a~=~d\vec v/dt$ and the right hand integral is $$m\int\vec a\cdot d\vec r~=~m\int\frac{d\vec v}{dt}\cdot d\vec r.$$ We now use $d\vec r~=~\vec v dt$ to perform a change of variables $$m\int\frac{d\vec v}{dt}\cdot \vec v dt~=~\frac{m}{2}\int\frac{d(\vec v\cdot v)}{dt}\cdot dt~=~\frac{m}{2}\vec v\cdot v~+~C,$$ for $C$ a constant of integration. We recognize the integration result as the kinetic energy $K~=~\frac{1}{2}mv^2$.

Now consider the left hand side with the force. This integration is $$-GMm\int\frac{\vec r\cdot d\vec r}{|\vec r|^3}~=~\frac{GMm}{|\vec r|}~+~C'.$$ Now I combine the two constants of integration and call that $E$ and express the integration of $\vec F~=~m\vec a$ with position as $$\frac{GMm}{|\vec r|}~+~E~=~\frac{m}{2}\vec v\cdot v,$$ or $$E~=~\frac{m}{2}\vec v\cdot v~-~\frac{GMm}{|\vec r|}.$$ This term $E$ is a constant of integration. There are fundamental ways of looking at this, particularly with Hamiltonian mechanics and Noether's thoerem.

if the work done has gone into storing the GPE and they are equal, where did the energy come from which made it accelerate and gain kinetic energy in the first place?

When the object is in acceleration, part of work done by you goes to increase KE and part goes to increasing GPE.

I am also under the impression that the kinetic energy and GPE are equal. So it seems as though I have done work on the object and the work done is equal to the sum of K.E and GPE. How is that possible? Surely that way I have gotten out twice the energy that I put in which is clearly nonsense.

if the K.E and GPE are equal, how is it that it can keep moving upwards with a constant velocity and hence a constant K.E and the GPE is increasing?

This is true only for isolated system. If total mechanical energy (KE+GPE) of the system is increasing, the system isn't isolated (or, force field is non-conservative). In this case (gravity is conservative), clearly you are adding work to the system.

protected by Qmechanic♦Sep 3 '16 at 13:42

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