Clarification on work and Gravitational potential energy

Question

I’m an 8th grader, and I was studying work and gravitational potential energy (GPE) out of curiosity. I thought about the idea that when I lift an object, the work I do increases its GPE, but gravity does negative work on the object, so the net GPE should be zero. I’m wondering where I might be going wrong with this logic.

Answer 1

Gravitational potential energy and the negative work done by gravity are two ways of describing one and the same thing.

If you raise an object of mass $m$ through a distance $h$ and it is at rest both at the beginning and end of the motion then you either say:

"I have done work $mgh$ on the object and this work is stored as gravitational potential energy"

or

"I have done work $mgh$ on the object and gravity has done work $-mgh$ on the object"

If you include both gravitational potential energy and the work done by/against gravity in your calculations then you are double-counting.

Answer 2

Reference frame: Ground
Scenario: Subsystem "Hand" quasi-statically lifts subsystem "Block" within the system "Block + Hand + Earth" which has a gravitational field of constant g

Your question is: Why should the potential energy of the block remain mgh after lifting it till height h, if the block does another similar work on hand?

Understand the work done by choosing the system as Block, then Hand, then Earth, then the entire system. You are headed correctly in studying physics, but the work done BY block ON hand does not add / change the value of work done ON the block. They are 2 separate systems.

1. BLOCK: Two forces are acting: Applied normal force, gravity.
   Initially the block had zero energy. When the block reaches h height, total work done by gravity = (mg) (h) cos 180 = -mgh, total work done by hand = (mg) (h) cos0 = mgh. Thus net work done = work done by all force = zero (= change in KE, hence proved work - energy theorem)

2. HAND: Two forces acting are: Weight of block, gravity. (Assume you = hand, or there would be another force from muscles on the hand... complex) (even then the position of the hand is changed to your COM, so imagine your COM quasi-statically ascending. But the system is too complex to justify. It is complex just like studying a question like "What form of energy is lost by an engine to give KE to the vehicles?") Just know that this is a Separate system, work done on hand does not add to work done On block. Moreover, as you see one of the forces acting on the hand is the weight of the block, downwards, which thus does a work on the hand that is equal to $(mg)(h)cos180$ = $-mgh$. But $-mgh$ is NOT THE ONLY FORCE ON THE HAND. The problem is even physicists have to work hard on this to solve it.

3. ENTIRE SYSTEM: For any system, we check the forces acting on the system, as in 2 cases above. But have you noticed, we check only on the external forces? As you see, on this entire system, NO EXTERNAL FORCE ACTS, only INTERNAL FORCES ACT. And we know, internal forces have the following characteristics:

• Internal forces are always balanced. That is why, change in momentum in an isolated explosion is always zero.
• Change in momentum is always zero for internal forces acting, implying, not net force acting, so if the system is isolated, it is in mechanical equilibrium.
• That means that energy change only takes places within the system between different forms. It can also get dissipated, so mechanical energy may not be conserved (e.g., explosion), but energy in general (all forms) is conserved. That dissipated energy is also included in the total energy of the big system.
• This logic directly arises from the Principle of Conservation of Energy and the Definition of Internal Force.
• Internal forces can also become non-conservative (hence disspation), in that case mechanical energy is not conserved, though total energy is conserved (e.g., friction dissipates energy as heat, sound, etc). However, even that will not happen in your scenario, where there (in ideal case) appears no non-conservative force that actually dissipates some energy. Most dissipation occurs from within the human body. As you see, by calculating energy of the block, it is so clearly conserved. The dissipation of energy from human body means the person is alive, which needs oxygen flow giving rise to air drag and pressure difference... leave that.
  In one sentence, since we are omitting any external force on the earth, there is no energy loss from the system into any other system.

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Why the block is not losing energy by doing work on the hand?
This is certainly true that a subsystem loses energy by doing work on another subsystem, but that is for doing POSITIVE work only. By doing –ve work, it MUST GAIN energy. The thinking process for that is: Block does –ve work on hand $⇒$ Hand does positive work on the block (Now this "doing +ve W means being done –ve work upon" is clear from the formula of work, at least in this case). Now the hand must be losing some energy to give energy to block. The situation becomes more complex because this delves into a deeper mechanism.

Answer 3

I concur with the answer by contributor Gandalf61.

Thought demonstration:
Let's say a mass $m$ is put on a platform that can be lifted pneumatically. There is a reservoir of compressed gas ready to be used for operating a pneumatic cilinder. You open a valve and and the expanding gas lifts the mass over a height difference $h$.

For simplicity: treat as idealized case: no friction.

The amount of work done by the expanding gas is $mgh$; mass times gravitiational acceleration times height difference.

The total energy of the system of both pneumatic machinery and mass $m$ is a conserved quantity.

The energy of the mass is raised by an amount of $mgh$

As pointed out by Gandalf61: in a system with multiple components: for each component the work done should be counted once.

The recommended practice:
In any situation where you can enter the work done as positive work you should enter it as positive work.

Discussion

While in general the concept of negative work should not be used in evaluation, there is a reason for having the concept of negative work available.
Here is an example of a case where entering work done in the form of negative work is a necessity: the mechanics of orbital motion.

As vivid example: Halley's comet

Perihelion: point of closest approach to the Sun
Aphelion: point of largest distance to the Sun

From aphelion to perihelion the Sun's gravity is doing work, increasing the velocity of the comet.

The celestial body in orbit is subject to a single force: the Sun's gravity. So: during the motion from perihelion to aphelion the decrease of the velocity of the comet is accounted for in terms of the Sun's gravity doing negative work.

Answer 4

while studying gravitational potential energy we only consider work done by gravity.you don't focus on how you displaced the body and by how much force.

Answer 5

What you have to understand is you did not increase the GPE of the object, but of the Earth-object system. GPE (and all forms of PE) is a system property. Although it is common to describe GPE as $mgh$ the actual general equation for GPE is

$$U=-\frac{GmM}{r}$$

where $G$ is the universal gravitational constant and $r$ is the distance between the centers of the 2 masses.

If you substitute for $G$, the mass of your object as $m$ and the mass of the Earth for M, and $r$ for the radius of the Earth (assumes the radius of the object is << less than the radius of the Earth), then the general equation for GPE becomes $U=mgh$. See John Rennie’s derivation in the following link. Near Earth vs Newtonian gravitational potential

The bottom line is you did positive work transferring energy to the object while gravity did an equal amount of negative work taking the energy you gave the object and storing it as GPE in the object-Earth system. Per the work energy theorem, the net work is zero meaning there is no change in kinetic energy of the object (assumes the object began and ended at rest before and after lifting).

Hope this helps.