Net work and energy

[justhanging]

I can't seem to be able to explain the following discrepancy.

If I raise a book a height h above the ground, I do mgh work on the book while gravity does -mgh cause its in the opposite of the displacement. The net work on this book is zero if you add up the work done by me and gravity yet the book obviously gain some potential energy. How can the book have zero net work yet have a larger potential energy?

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[Gauss44]

I can't seem to be able to explain the following discrepancy.

If I raise a book a height h above the ground, I do mgh work on the book while gravity does -mgh cause its in the opposite of the displacement. The net work on this book is zero if you add up the work done by me and gravity yet the book obviously gain some potential energy. How can the book have zero net work yet have a larger potential energy?

chemical energy to raise the book to height h
if it remains at height h, gravity has not done work (because gravity is not the force that moved the book)

chemical to potential

 

[justhanging]

chemical energy to raise the book to height h
if it remains at height h, gravity has not done work (because gravity is not the force that moved the book)

chemical to potential

Gravity does negative work on the book.

If you apply your idea to a book being pushed against friction, will you consider friction even though it doesn't push the book?

 

[DylanE]

if you were only exerting mgh force, the book wouldn't be moving. the force you have to apply has to match that of gravity, plus what it takes to raise it. so the difference in mgh1 and mgh2 is what you applied.

 

[PhysMatMan]

You convert the chemical energy in your body into gravitational potential energy. You lose the gravitational energy the book gains. When something does negative work, it puts the energy into the potential (or dissipates for friction). When you do positive work, you lose energy.

 

[PhysMatMan]

if you were only exerting mgh force, the book wouldn't be moving. the force you have to apply has to match that of gravity, plus what it takes to raise it. so the difference in mgh1 and mgh2 is what you applied.

The assumption is that this is a low acceleration/speed limit. So you can apply mg+epsilon to accelerate the book, where epsilon is very small. Then epsilon*h is very small. You can take the limit as epsilon->0.

 

[DylanE]

this could be wrong, but my take is that gravity is constantly applying -9.8 acceleration. so holding a book in place takes a force equal to 9.8*mass, or the weight of the book. 9.8 being a positive acceleration, even if there is no net acceleration. so that force (W) times the height that you then displace it, gives you the work.

 

[Gauss44]

Gravity does negative work on the book.

Gravity does NO work on the book in the example in your topic. In order for gravity to DO work on the book, gravity would have to move the book. It did not.

 

[justhanging]

For everyone who says gravity does no work on the book, take the following example:

I push a book with 10 N force one way and you push it with 5 N force the other way. After a certain amount of time the book has gained kinetic energy (no height gained, no friction). The change of kinetic energy in this situation is equal to the net work, which is equal to the work I did on the book minus the work you did on the book. How is gravity different then the work you did on the book in this situation?

 

[Gauss44]

For everyone who says gravity does no work on the book, take the following example:

I push a book with 10 N force one way and you push it with 5 N force the other way. After a certain amount of time the book has gained kinetic energy (no height gained, no friction). The change of kinetic energy in this situation is equal to the net work, which is equal to the work I did on the book minus the work you did on the book. How is gravity different then the work you did on the book in this situation?

In that case (quoted above):

Your work is the force created by you times the distance you pushed the book. W=FD

The greater the opposing force, the more force you will need to push the book the same speed. The equation for work remains the same. Your work = force from you * distance you pushed the book

If you wanted to take this a step further, you can say the work you did equals the kinetic energy of the book. In equation form, this would the force from you on the book times the distance you pushed the book equals 1/2 times mass of the book times the final velocity of the book squared. FD=1/2 mv^2

 

[justhanging]

In that case (quoted above):

Your work is the force created by you times the distance you pushed the book. W=FD

The greater the opposing force, the more force you will need to push the book the same speed. The equation for work remains the same. Your work = force from you * distance you pushed the book

If you wanted to take this a step further, you can say the work you did equals the kinetic energy of the book. In equation form, this would the force from you on the book times the distance you pushed the book equals 1/2 times mass of the book times the final velocity of the book squared. FD=1/2 mv^2

I don't think you disproved my point.

If there is an opposing force, the kinetic energy of the book = Force(right)D - Force(left)D = work done by me - work done by opposing force.

If you don't include the opposing force your overestimating the kinetic energy of the book.

 

[Gauss44]

I don't think you disproved my point.

If there is an opposing force, the kinetic energy of the book = Force(right)D - Force(left)D = work done by me - work done by opposing force.

If you don't include the opposing force your overestimating the kinetic energy of the book.

Save yourself some time and just write:

"Force(right)D = work done by me"

The rest should cancel out.

 

[justhanging]

Save yourself some time and just write:

"Force(right)D = work done by me"

The rest should cancel out.

Where are you trying to get at? Your just saying the same thing I put above.

Change in kinetic energy = work done by me - work done by the opposing force.

The question is why is the negative work done by gravity not included for a book being raised. The opposing force work in the example above was included.

 

[slz1900]

I think you could make the case that gravity is included implicitly. Work is the integral of the force function. If you were to apply a force less than the gravitational force to an object, the function would look a certain way. If you apply force greater than the force due to gravity, the curve will look different. So gravity does affect your integral, since force and its integral are different if it is present.

In other words, if gravity wasn't there and you applied the same force, you are right, force itself would not change, but it would move a lot farther. Gravity is indirectly changing part of your F*d equation.

 

[PhysMatMan]

No, I agree with op that gravity performs work on book. Since gravity does negative work and the book is at rest at the end, the gravitation PE is mgh. The net work on the book is 0, the work by gravity is -mgh (gets stored in PE), and the work by the hand is mgh (gets removed from body). Overall, book gains PE, body loses PE.

Gravity is special in that it has a PE. You don't want to use gravitational work and PE at the same time since PE is derived from g-work. Whichever way you look at it, be consistent. If a book falls a height mgh, then it loses mgh PE, and gravity did mgh work, but the book only gains mgh KE, not 2mgh. Your hand pushing an object doesn't have a PE associated and so you must only use work.