Gravitational Potential Energy and weight

[Bashyboy]

Is what I am currently reading about. The book I am reading presents it by giving an example; it proposes lifting a mass without any acceleration through a height, h, with an upward force equal and opposite to its weight. Does that mean the net work done on this mass is zero? My hand still does work on the mass as well as gravity, but the amount of work does is zero, right?
Also, where does the mass get its potential energy from? I know its position contributes to its potential energy, but what about the energy I use to to lift it up? Does the energy I use transfer and convert into potential energy as I lift the mass up?

 

[CWatters]

it proposes lifting a mass without any acceleration through a height, h

If h is not zero then the object has acquired PE. So some energy has been expended by the system doing the lifting.

...with an upward force equal and opposite to its weight. Does that mean the net work done on this mass is zero

No. It means the object is rising at a constant velocity. Remember..

F=m*a

so if the forces sum to zero the acceleration is zero but the velocity need not be zero.

If the velocity is non zero then the lifting device expends energy..

Work = F*h

The power is the rate of doing work so..

Power = F*h/t

where t is the time taken to move h and h/t is the vertical velocity

 

[CWatters]

My hand still does work on the mass as well as gravity, but the amount of work does is zero, right?

That's not correct. Your hand does work = mgh

It might help to understand this point...

In theory if you hold a weight at a constant height you do no work on it. Consider a book shelf. It needs no power source to hold the books at a constant height. In practice the human body isn't very efficient. Your body burns energy just creating a force even if it's doing no work. So it feels like you are doing work holding up a weight at a constant height but you don't actually do work on the weight until you raise it higher.

Also, where does the mass get its potential energy from? I know its position contributes to its potential energy, but what about the energy I use to to lift it up? Does the energy I use transfer and convert into potential energy as I lift the mass up?

Yes. Thats correct.

 

[Bashyboy]

How can there be a net work if there is no change in velocity and, presumably, the mass remains the same?

 

[Drakkith]

In theory if you hold a weight at a constant height you do no work on it. Consider a book shelf. It needs no power source to hold the books at a constant height. In practice the human body isn't very efficient. Your body burns energy just creating a force even if it's doing no work. So it feels like you are doing work holding up a weight at a constant height but you don't actually do work on the weight until you raise it higher.

Just to elaborate and clarify the explanation, the human body uses muscles to hold things up. These muscles are made up of many muscle fibers that each contract and then relax. To hold up an object you have fibers contracting then relaxing then contracting again, burning energy in the process of contraction. This contrasts with something like a table which does not burn energy to hold an object up.

How can there be a net work if there is no change in velocity and, presumably, the mass remains the same?

Work = force x distance

Applying a force that raises the object does work, even if no acceleration is done. (After the initial acceleration of course)

 

[Doc Al]

Is what I am currently reading about. The book I am reading presents it by giving an example; it proposes lifting a mass without any acceleration through a height, h, with an upward force equal and opposite to its weight. Does that mean the net work done on this mass is zero?

Yes, the net work on the mass is zero.

My hand still does work on the mass as well as gravity, but the amount of work does is zero, right?

You do an amount of work equal to +mgh. Gravity does negative work equal to -mgh. The net work--due to all forces--is thus zero. (If that's what you mean.) That's why there's no change in kinetic energy.

Also, where does the mass get its potential energy from? I know its position contributes to its potential energy, but what about the energy I use to to lift it up? Does the energy I use transfer and convert into potential energy as I lift the mass up?

Yes. The work you do in lifting the mass becomes the increased gravitational potential energy of the mass/earth system.

 

[Drakkith]

Yes, the net work on the mass is zero.

You do an amount of work equal to +mgh. Gravity does negative work equal to -mgh. The net work--due to all forces--is thus zero. (If that's what you mean.) That's why there's no change in kinetic energy.

Yes. The work you do in lifting the mass becomes the increased gravitational potential energy of the mass/earth system.

Hmmm. Was my statement about work in my post above incorrect? I thought you did in fact do work on the object by lifting it.

 

[Doc Al]

Hmmm. Was my statement about work in my post above incorrect? I thought you did in fact do work on the object by lifting it.

Your post was fine, but I think you missed that the OP was asking about net work.

 

[Bashyboy]

Yes. The work you do in lifting the mass becomes the increased gravitational potential energy of the mass/earth system.

But doesn't gravity do equal and opposite work on the lifted box, somehow counter-acting the energy I put into lifting it?

 

[Doc Al]

But doesn't gravity equal and opposite work on the lifted box, somehow counter-acting the energy I put into lifting it?

Depends on what you mean by 'counter'. Obviously there's no change in kinetic energy, since the net work is zero. So in that sense, gravity does 'counter' the work you do. But you did do work and that energy did go somewhere.

But if you think of gravity as being represented by a potential energy term, then you can think of the work you do as becoming gravitational potential energy.

Work done by all forces (including gravity) = ΔKE

Work done by all forces (except gravity) = ΔKE + ΔPE

These two formulations of the Work-Energy theorem are equivalent. In the second version, work done by gravity is already included in the PE term.

 

[Bashyboy]

So, gravitational potential energy is more a less an accounting system to measure the potential work that could be done due to the objects position. And so this gravitational potential energy doesn't actually become "real" unto the object starts falling? which we have an account of the energy it possesses the the potential work it could do

 

[Doc Al]

So, gravitational potential energy is more a less an accounting system to measure the potential work that could be done due to the objects position. And so this gravitational potential energy doesn't actually become "real" unto the object starts falling? which we have an account of the energy it possesses the the potential work it could do

I think that's a reasonable way to think of it.

 

[Bashyboy]

Thank you all for you help.