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My question is very similiar to this one: Work done by gravity on falling object does not seem to equal change in mechanical energy

As I've understood it, work is only done on an object if the object experiences a change in its mechanical energy. This means that if energy is added to an object or if energy has left an object, some force must have acted on the object and thus done work on it.

So now onto the question: Let's pretend that we have an object of mass 10 kg and we drop it from a height of 2 meters. Using the formula for gravitational potential energy (EP = mgh), we get that the object has a potential energy of 196,4 J before being dropped. Right before it hits the ground, all of the potential energy has been converted to kinetic energy (assuming there is no friction from the air).

If we choose the object as the system, then the force of gravity is external to the system. It does external work on the object and thus increases its energy (kinetic energy). From the thread I mentioned earlier, it is stated that the object cannot have an initial potential energy, simply because potential energy is measured relative to a reference point. If the object is the system, then there is no reference point. This makes sense to me, but what doesn't is the fact that kinetic energy is also relative. The kinetic energy of an object is proportional to the velocity of the object, and we all know that velocity is measured relative to a reference point, right? My question is therefore; How can one measure the kinetic energy gained by an object that is falling from a height, if the object is the system and nothing else is included in the system?

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    $\begingroup$ A height relative to what? The concept of height is without a metric unless it's in relation to something. $\endgroup$ Oct 30, 2022 at 12:46
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    $\begingroup$ That's what I'm saying, so what's the question if there is no fall as there is no height? $\endgroup$ Oct 30, 2022 at 12:54
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    $\begingroup$ The object you don't mention is causing the gravity and is a reference point for height. $\endgroup$
    – Jim Clark
    Oct 30, 2022 at 13:11
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    $\begingroup$ Potential energy is not associated with a single object. You don't have any potential energy if your object is the system. $\endgroup$
    – across
    Oct 30, 2022 at 13:34
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    $\begingroup$ Gravity is acting on the object and drags it down so that the object is falling downwards Got you. If there's no external reference points, how do you know that object falls downwards ? If there's no any reference points, then any direction makes no sense. $\endgroup$ Oct 30, 2022 at 20:00

3 Answers 3

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As I've understood it, work is only done on an object if the object experiences a change in its mechanical energy.

Per the work-energy theorem, net work is only done on an object if the object experiences a change in its kinetic energy. Mechanical energy consists of kinetic plus potential energy. An object does not possess potential energy because potential energy is a system property, not a property of an object.

This means that if energy is added to an object or if energy has left an object, some force must have acted on the object and thus done work on it.

Again, this only applies to the kinetic energy of an object and work done is the net work done.

So now onto the question: Let's pretend that we have an object of mass 10 kg and we drop it from a height of 2 meters. Using the formula for gravitational potential energy (EP = mgh), we get that the object has a potential energy of 196,4 J before being dropped.

It is the combination of the object and earth, i.e., the object-earth system, that has the potential energy of 196.4 J, not the object alone.

If we choose the object as the system, then the force of gravity is external to the system. It does external work on the object and thus increases its energy (kinetic energy).

That's correct. But it is also correct for the earth-object system where gravity is an internal force of the system. For the earth object system, if there is no net external force acting on the system and no internal dissipative forces (e.g., friction) acting on the system, the increase in kinetic energy of the object equals the decrease in gravitational potential energy of the earth-object system.

From the thread I mentioned earlier, it is stated that the object cannot have an initial potential energy, simply because potential energy is measured relative to a reference point.

Though I haven't read the thread, again gravitational potential energy is the property of the earth-object system. It depends on the relative position of the object and the earth. It is independent of any reference point outside the earth-object system.

If the object is the system, then there is no reference point. This makes sense to me, but what doesn't is the fact that kinetic energy is also relative.

The reference point is the location of the observer. The observer can be anywhere inside or outside the system. So to say there can be no reference point if the object is the system is incorrect.

The difference is gravitational potential energy, since it depends only on the relative position of the earth and object in the earth-object system, is independent of any reference point outside the earth-object system. On the other hand, velocity (and thus kinetic energy) depends on reference points both within and outside of the earth-object system.

The kinetic energy of an object is proportional to the velocity of the object,..

Kinetic energy is proportional to the velocity squared.

and we all know that velocity is measured relative to a reference point, right?

Correct.

My question is therefore; How can one measure the kinetic energy gained by an object that is falling from a height, if the object is the system and nothing else is included in the system?

Again, you can measure the kinetic energy with respect to any reference point. It is purely arbitrary. What is measured will depend on the relative motion between the object and the observer.

Hope this helps.

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  • $\begingroup$ So a system and a reference point are not the same? Maybe thats my confusion? When I selected the object to be the system, I thought it implied that the object was now also our only reference point. In other words, no external reference points. However, this confuses me even more because if we are allowed to have an external reference point, then we object will have a height relative to that point and thus it will have potential energy. But it is a well established fact that potential energy is not defined if the object is the system $\endgroup$
    – Gabriel
    Oct 30, 2022 at 17:00
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    $\begingroup$ The system and a reference point can be the same or not be the same. It’s completely arbitrary. In physics a reference point is simply a point in space used to describe the position or motion of other things relative to this point $\endgroup$
    – Bob D
    Oct 31, 2022 at 1:51
  • $\begingroup$ @Gabriel The gravitational system Gabriel & The Earth has a potential energy. The gravitational system Gabriel & Alpha Centauri has a different potential energy. Both change as you (and Earth or Alpha Centauri, respectively) move through time and space. Potential energy is not a property of Gabriel, it's a property of the system you're observing. When you think about a Gabriel falling towards Earth's surface from a low height, it's usually implied you're talking about the potential energy of the Gabriel & Earth system, but that doesn't mean there aren't infinite other potential energies. $\endgroup$
    – Luaan
    Oct 31, 2022 at 7:53
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If the object is the system then you have things which are internal to the system and things which are external to the system.

System: Object
So you can measure the speed and height of your object relative to something which is external to your system, eg the floor in your laboratory.
One external force on the object - the gravitational attraction of the Earth, $mg$.
If the object falls a height $h$ then the work done by the external force is $mg\times h = mgh$ and this produces the increase in the speed and hence the kinetic energy of the object.
Now what about taking the measurements?
As well as the force on the object due to gravitational attraction of the Earth there is the force on the Earth due to gravitational attraction of the object.
Why does that matter?
It is because as well as your object falling down the Earth rises up and so does a possible reference point - the floor in the laboratory.
However if the choose the centre of mass of the Earth and the object as your reference point you know that that point does not move because taking the Earth and the object as a system no external forces act on that system.

System: Object & Earth.
No external forces but there are two internal forces - force on object due to gravitational attraction of the Earth and force on Earth due to gravitational attraction of the object.
Since there are no external forces the linear momentum of the system $m_{\rm Earth}\vec v_{\rm Earth} + m_{\rm object}\vec v_{\rm object}$ cannot change so as the object falls downwards the Earth must rises upwards.
You can find the speed of the object by considering the work done by the force on object due to gravitational attraction of the Earth but must use the centre of mass of the system as the reference point.
As there is a greater than one mass system under consideration the concept of gravitational potential energy can also be used.

In practice all this does not matter because the motion of the Earth is so much smaller than the motion of the object.

However, your reservations about the reference point are well founded in that a free-fall type method is used to find a value for the acceleration due to gravity very accurately.
The problem with trying to make very accurate reads is that there are other influences which affect the floor of the laboratory as a reference point.
For example, in the paper A new absolute determination of the acceleration due to gravity at the National Physical Laboratory, England it is explained that the experimenters had to consider, what effect, if any,

storms in the Atlantic [with] a peak amplitude of about $0.5\,\rm \mu m$ at a period of about $5\,\rm s$ (with some seasonal variation) and the short period [oscillation] generated locally by machinery and traffic, and [with] a peak amplitude of about $0.1\,\rm \mu m$ at a period of about $0.2\,\rm s$,

had on their value for the acceleration due to gravity.

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There are two ways of looking at this.

If you consider gravity an "external force", then you only count kinetic energy of your object as mechanical energy. Gravity is doing work on your object and increasing its kinetic energy.

If, on the other hand, you consider your object as having gravitational potential energy, then gravity is not an external force. It's your object using gravitational potential energy to do work on itself. It's transferring potential energy to kinetic energy. It's as if your object was composed of two parts, and one part is doing work on the other part.

Now, you're also saying that there is no such thing as velocity when you think of gravity as an external force. That makes no sense. There is always a reference frame you have to use. Your object is accelerating in that reference frame due to the gravitational force. The reference frame doesn't have to be "attached" to any physical object, but if you say there is a force of gravity, it implies you're using some reference frame in which you are seeing that force.

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  • $\begingroup$ When you say "the reference frame doesn't have to be 'attached' to the object", do you mean that one is allowed to use an external reference frame even though the system is defined as being the object and nothing else? This seems to imply that we can measure the height from which the object is being dropped (with respect to the reference point) and thus we can deduce the object's potential energy with respect to the reference point. But we know that potential energy is not defined for a single object and we just defined our system to be the object and nothing else. How does this make sense? $\endgroup$
    – Gabriel
    Oct 30, 2022 at 16:12
  • $\begingroup$ One not only is allowed to use a reference frame for a single object, newtonian physics makes no sense without using a reference frame. All the equations are in some reference frame. Potential energy is perfectly well defined for a single object in a gravitational field: it's $U = mgh$. $\endgroup$ Oct 30, 2022 at 16:25
  • $\begingroup$ But if we consider the object to be the system, what is the value of h? h is height and we need a external reference point in order to find out. The problem is, we don't have an external reference point when the object is defined as the system. Thus there is no potential energy. $\endgroup$
    – Gabriel
    Oct 30, 2022 at 16:32
  • $\begingroup$ In Newtonian mechanics every object -- even if there is only one -- has a property called position and another property called velocity. $h$ is one coordinate of the position. If you "don't have a reference point" then you can't model the situation using newtonian mechanics. Pick a reference point :) In your OP you said "we drop it from a height of 2 meters" which means you must have already picked a reference point, and your object was 2 meters above it. $\endgroup$ Oct 30, 2022 at 16:36
  • $\begingroup$ Of course you can also choose a frame of reference such that your object is at the center of it. In that frame there is no gravitational field, there is 0 acceleration, your object has 0 kinetic energy and 0 potential energy. $\endgroup$ Oct 30, 2022 at 16:39

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