Potentional energy is work done by gravity?

[primarygun]

1.Potential energy is defined as the work done by gravity, right?
2.Energy gain refers to potential energy gains,right? but no kinetic energy gain.

 

[chroot]

Potential energy is defined as the ability to do work.

"Energy gain" is an ambiguous term, since the energy could be in any form.

- Warren

 

[primarygun]

So mgh is the change in potential energy, when it is positive, it is done by gravity but not human , right?
How about a man lift up a box(1kg) vertically upward 10m , what's the work done by the man and what's the work done by the gravity?

 

[chroot]

When you lift a box, you're doing work. When a box falls, gravity is doing work.

- Warren

 

[primarygun]

Thanks for quick response.
When I lift that box, the force supplied by me should be what?
I believe that this question evolutes to mgh.
If the supplied force by me is mg, the object should not be moved.
I think one better explanation for mgh is that the height the object gains, a corresponding amount of work done (energy) gains. But not say that I gave work done(energy) which is equivalent to mgh to it, right?

 

[chroot]

Force and work/energy are different concepts. If you lift the box 10 meters, you will do a specific amount of work -- you'll expend a specific amount of energy. The total amount of work done doesn't depend in any way on how you lift it 10 meters.

If you have a specific problem you're having trouble with, you should just post the whole thing so we can help you.

- Warren

 

[primarygun]

My teacher said that a box 1kg leave up by a man 50m, then he just said the potential energy= work done , so PE=500 J. The explanation may be a bit vague that cause me think of the work done is by human efforts against the gravitational force. Contradiction arises as F by human not = mg, otherwise, no movement or continuous movement.

Here is my illustration. I think that after a certain distance rose, if it is released, the Earth pull F=mg , this is the source of the work done.
And where have the human efforts gone? Some stored in object as PE, where are the others, KE?

 

[chroot]

Once again, force and work are different concepts. Work is force integrated over distance.

If all you're interested in is the energy required to lift a box a specific distance, force is totally irrelevant.

- Warren

 

[futb0l]

The Work = Fx ... so when there's a force going down F = mg... so then you have to multiply the distance so it becomes Work = mgh.

When the object is held on your hand, then there is potential energy mgh. When you drop it.. in the middle of the flight, there is both kinetic energy (because it's moving) and potential energy (mg(1/2)h).. and then when it reaches the ground, there is no kinetic energy and potential energy, all the energy is turned into the work done by gravity.

This may help you: http://www.physicsclassroom.com/mmedia/energy/se.html

 

[cepheid]

Contradiction arises as F by human not = mg, otherwise, no movement or continuous movement.

primarygun...you seem to be stuck here. Keep in mind that mgh represents the minimum amount of work a man/woman could do in lifting a box of mass m. I know what you are thinking...if he/she exerts a force of mg exactly equal to the box's weight, then the net force on the box is zero, and it is currently at rest on the ground, so how could the box possibly move upward?

What is not mentioned is that he/she must supply an initial force greater than the box's weight for a very short time just to give it a "kick" to heave it upwards, accelerating it to whatever final velocity, v, (as fast or as slow as the person wants) at which he/she will lift it uniformly after that. So the force can immediately be reduced to mg after this initial "hoisting" so that the net force on the box is zero and it rises uniformly with velocity, v. With the exception of the initial kinetic energy you impart to the box, the rest of the work you do goes to lifting the box against gravity, resulting in an increase in potential energy. (You could supply a force greater than mg all the way through, thereby doing more work, but the gain in potential energy up to height h would be the same, the rest of your work would be "wasted" accelerating the box, ie. increasing its kinetic energy)

Why is this initial force never mentioned? Because it gets "cancelled out" at the end of the trip. Think about it...if you're going to lift it to height 'h', and it's currently rising with velocity v, you've got to stop it at the end
This involves applying another barely discernable force to rapidly decelerate it. At the beginning, the change in velocity is v - 0 = v. At the end, the change in velocity is 0 - v = -v. Bottom line: the total work done on the box:

W = 1/2m(v² - 0²) + mgh + 1/2m(0² - v²) = mgh

= the change in potential energy of the box

 

[primarygun]

Thank you.
Before your help, I can only figure out that I should understand the formula mgh=PE is the energy stored to be released. I could not think that the work done by human.(A slightly push first).
PE=mgh where the object is in uniform motion. But the total energy=mgh + 1/2mv^2
, right?

 

[cepheid]

Thank you.
Before your help, I can only figure out that I should understand the formula mgh=PE is the energy stored to be released. I could not think that the work done by human.(A slightly push first).

Potential energy is, if you like, energy stored to be released (i.e. in this case energy of a system by virtue of the position of masses in a graviational potential). All I was pointing out is that the minimum amount of work required by a human or whatever is lifting the object to a height, h, is mgh. In other words, the potential energy does not come from nowhere...work has to be done to lift it to that height at which it has that potential energy. Think about this carefully...the work done lifting the object is at least equal to the negative of the work done by gravity when the object rises by that much. From the definition of gravitational potential energy, this is equal to the object's change in potential energy.

PE=mgh where the object is in uniform motion. But the total energy=mgh + 1/2mv^2
, right?

I don't understand what you are trying to say here. PE = mgh...period. It depends on the height of the object...whether it is moving or stationary.

As for total energy...if the object is moving, then it has kinetic energy as well (kinetic energy...energy of motion). It's total mechanical energy is given by mgh + 1/2mv^2.
But I thought that you already knew all of this when I launched into my previous explanation. It seems that we are taking steps backwards now?

Finally...note that what I was saying to you before is consistent with what chroot and everybody else has been telling you...I was just trying to give a specific example to illustrate the point.

 

[primarygun]

I got all these. Thank you.