Gravitational Potential Energy And Work Done

[Miraj Kayastha]

This question may sound weird but when we lift an object with a force equal to the weight of the object up to a certain height the Earth is doing negative work on the object as well. So shouldn't the net work be zero?

 

[Simon Bridge]

"net work" done by what?

When you lift an object, you are doing work on the object, you lose energy, and that energy is stored as gravitational potential energy. We say that energy is "conserved".

 

[MrMatt2532]

The potential energy between two objects is -G*m*M/r. If you put work into the system you can increase the separation vector r, or you can take work out of the system and decrease the separation vector r. No contradictions here.

You seem to be thinking that you are doing work on one object and that object is in turn doing work on another. That is the wrong line of thinking. You are changing the potential of a two object system. Usually, we can approximate the potential as m*g*h, where g=G*M/r^2, here where r is the radius of the Earth and M is the mass of the earth.

 

[Miraj Kayastha]

What I meant was:
Consider a particle on the ground. This particle is raised by a force of magnitude mg to a height h above the ground. At this point, the work done on the particle by the force is mgh, which is equal to the potential energy of the particle. But, during this period, the force of gravity also acts on the particle and is displaced by h, and so does a work of −mgh on the particle. Shouldn't the two cancel and no net work should be done on the particle?

If they don't cancel, then where did the energy that came from the work done by the force of gravity go?

 

[Dale]

The magnitude of the net force is 0 so, according to the work energy theorem, the change in KE is 0. The positive work done by the external force is balanced by the negative work done by gravity with a transfer of energy from the external force's source of energy to the gravitational potential energy.

 

[agenthulk]

Hey, okay I like to break it down start with momentum defined as the product of the mass of an object and it's velocity

The total momentum in any isolated system before any event is equal to the total momentum after the event
There is an important limit to the application of the conservation of momentum- the event must be isolated from other forces Now look at work defined as the product of the net force and the displacement through which that force is exerted. I'm not going into the unit of work called joule.

What is net force? What is displacement? Answer is net force- equal magnitude in size but pulling in an opposite direction so this is equal to zero if not equal in size then add the one that is great displacement we all know

Now power is defined as work done divided by the time used to do work

Energy is defined as the ability to do work

Example of energy: gravitational potential energy and kinetic energy and there it is

Potential energy: ability to do work because of position: PE = weight times height

The total energy at the end of any event is equal to the total energy before the event or can't create or destroy energy

With all that summed up go look at a pendulum and I believe you will have your answer (conserved)

 

[sophiecentaur]

What I meant was:
Consider a particle on the ground. This particle is raised by a force of magnitude mg to a height h above the ground. At this point, the work done on the particle by the force is mgh, which is equal to the potential energy of the particle. But, during this period, the force of gravity also acts on the particle and is displaced by h, and so does a work of −mgh on the particle. Shouldn't the two cancel and no net work should be done on the particle?

If they don't cancel, then where did the energy that came from the work done by the force of gravity go?

Your simple question just involves work done ON and not BY. Your notion of "cancelling out" has no meaning because Work Done ON does not take into account where the work originated from.

This question gets asked (in one form or another) many many times on PF. If you want to resolve your difficulty, you need to look - strictly - at the definition of work done. The work is done ON the particle and the agency that caused the Work Done is of no consequence. Any real machine, doing work ON something will end up using more Energy or work done BY (Efficiency is always less than 100%) but the work done ON, will be the same.
In the frame of reference of the Earth, if you stand on the Earth and lift up a heavy mass, you are actually doing work on the mass/Earth system by pushing them apart.
What work is done ON the mass? Its weight (force) times the distance moved by the mass.
What is the work done ON the Earth? It is the same force times the distance that the Earth moves relative to you. How much is that? Realistically, you could say it's zero so the work done is zero.
If you stand on soft ground and stretch your body by the same distance, you have still done some work ON the mass but you will also have done measurable work ON the soggy ground. The work done ON the mass is no longer the same as the work done BY you.

 

[Miraj Kayastha]

But what about the negative work done by the Earth on the mass when the mass is lifted? Shouldn't that decrease the energy of the earth-mass system?

 

[agenthulk]

Not really the mass of Earth is so tremendous, it's change would be very very small undetectable but not nonexistent I would not even worry

 

[Miraj Kayastha]

I meant the work done by Earth not work done on earth.

 

[agenthulk]

Sorry misread

 

[Dale]

But what about the negative work done by the Earth on the mass when the mass is lifted? Shouldn't that decrease the energy of the earth-mass system?

This is backwards. Positive work done by some system S decreases the energy of S. So negative work by S would increase the energy of S.

 

[Miraj Kayastha]

So that means the gain in potential energy of the earth-mass system should when we lift an object should be 2mgh. I know I am missing something. Could you clarify me?

 

[Dale]

Why would you think it should be 2 mgh? It is just mgh.

The force does work mgh on the Earth mass system. The source of the force loses energy mgh. The Earth mass system gains potential energy mgh.

 

[Miraj Kayastha]

This is backwards. Positive work done by some system S decreases the energy of S. So negative work by S would increase the energy of S.

Here you said the system gains energy because Earth also does negative work on the mass.

 

[Dale]

A system can't do work on itself. The Earth cannot do any work on the Earth mass system, by definition. The only work done on the Earth mass system is from the external force. That work increases the energy of the Earth mass system by mgh.

Now, you are free to define your system differently, as long as you are consistent. You could define your system as the mass. Then the external force does positive work on the mass and the gravitational field does negative work. In this formulation the energy of the gravitational field increases due to the negative work it does.

 

[Simon Bridge]

You are noticing the symmetry in the conservation of energy.
Parts of a system gain and lose energy but the entire system conserves energy.
The concept of "work" only applies to parts of a system, not the system as a whole.

 

[sophiecentaur]

Here you said the system gains energy because Earth also does negative work on the mass.

Why would you say that the Earth does work any more than the mass does work? If you had two large equal masses, which one would be doing negative work and which would be doing positive work?

You have got yourself enmeshed in a classification worry, rather than getting down to the real problem - which involves Energy Transfer. Use the strict definitions of Work, Energy Force etc. and you can describe all classical systems perfectly well and make correct predictions about what will happen. The Classification thing can be avoided totally.

Classification is not the same as Definition.