Why is the work done and force applied different in the Work-Energy Theorem?

[daniellionyang]

Assuming you are lifting a block up 1 meter from rest to rest with constant work. You know that the work is -deltaU or 10. However, you also know W=deltaKE which is 0. You finally know that W=Fx=10*F. How do you explain why the numbers are different? Thanks!

 

[Ken G]

It's the difference between work by a single force, versus net work. The work-energy theorem only applies to net work, which is zero. There is positive work by the lifting force, and negative work by gravity, so the net work is zero.

 

[daniellionyang]

Thanks for your reply! I still have questions however. If this is true, when does W=deltaEmech, W=deltaKE, work=-deltaU? Also, since W=deltaEmech, shouldn't the total be deltaKE+deltaU=deltaU=-10 not 0?

 

[Ken G]

W doesn't equal delta E mech, it equals delta KE. But you need to use the net work-- the sum of all the work done by all the forces on the object. There would be no point giving delta E mech its own name, it is always zero unless there is heat generated (and when that happens, life can get complicated, because you have friction and inelastic collisions and all that jazz). So if you would give delta E mech a name, it would just be the negative of the heat dissipated.

 

[daniellionyang]

Thanks again! However, my textbook states that work done on a system, assuming no friction is deltaEmech. Also, if net W=deltaKE=-deltaU, why is deltaKE 0 and why is -deltaU 10?

 

[nasu]

Assuming no friction (or other non-conservative forces) the mechanical energy is conserved so the change is zero.
However the work doesn't have to be zero. Maybe you misunderstand what the book actually says.
It seems this is a common matter of confusion. It was discussed here several times.
What book is that? Can you show an image?

 

[daniellionyang]

I am using Halliday Fundamentals of physics 6th edition. nasu or someone else, can you explain why I am getting different numbers though for the problem? And, yes, this concept is confusing me! A more specific question, I guess, is when is W=deltaEmech, when is W=deltaKE, and when is work=-deltaU? Thanks!

 

[Ken G]

It would be quite unusual to say that work is delta E mech, unless they are not counting work done by the forces associated with potential energy functions (called "conservative forces") as work! But it normally would be called work, even the conservative forces. If you don't count the work done by conservative forces in the work-energy theorem, then the work done equals the change in mechanical energy (if there's no heat loss). I can't recommend that conceptual approach however, work is too valuable a concept to waste on purely nonconservative forces. Just count all forces as able to do work, and then if there is no heat loss, work is always equal to the change in kinetic energy, you can derive that from F=mA.

So in summary:
work by a single conservative force = - delta U
is only for work done by conservative forces (forces with a potential energy associated with them). This isn't a statement about reality, it is just the definition of potential energy-- it's bean counting.

net work = delta KE
is the work-energy theorem, equivalent to F=mA. This is a statement about how reality works, given the more general definition that net work is the sum of all the work by all the forces (including friction).

work = delta mechanical energy
is only true if you don't count work done by conservative forces as work on the left-hand-side (their - delta U is pulled to the other side of the equation, and absorbed into mechanical energy, just a different way to count those beans). So the meaning of "work" here is external forces applied manually (like a person pulls on a rope or something), or frictional forces. If you include work done by frictional forces, you are not including heat that is dissipated in the mechanical energy, that's just KE+U.

One final note-- when you do inelastic collisions, like objects that stick together, it is generally hard to include the work done there, so you usually drop the attempt to bean count using work, unless you are told the heat dissipated in the elastic collision. Usually you just don't use conservation of energy at all for inelastic collisions, but you can if there is friction, if you can track how much work the friction does.

 

[daniellionyang]

Actually, the textbook says this formula is true for nonconservative forces. It says the work done by an external force for conservative forces is W=deltaEmech and for nonconservative, it is W=deltaEmech+deltaEtherm

EDIT Usign another textbook (college Physics by Serway), it seems that it is a little clearer. It explicitly state that non-conservative work is deltaEtherm. Now, can some one help me with the individual problem? Why do I get different numbers by doing W=deltaKE=0 and W=-deltaU=10? Thanks!

 

[Ken G]

It's because the first answer uses net work, which is zero, and the second answer only uses the work done by the person lifting the object, which is positive. There is negative work done by gravity, which makes the net work zero, but that's a conservative force rather than an "external" force, so one of the books is counting that work separately in its bean counting. I don't recommend that at all.

What you can take to the bank is the work-energy theorem, which says, count all the work done by all the forces (including friction), add it up, and it equals change in KE. If any of the works that are done cannot be tracked, you will need to be just told what they are, or not use conservation of energy at all.

 

[daniellionyang]

Wait, so W=deltaKE gives total work and work=-deltaU gives work by a specific force?

 

[Ken G]

Yes, if U refers to a given potential energy term (that's the single force). If U adds them all, then it refers to all their work added up.

 

[daniellionyang]

Oh... since Ui would be 0 and Uf =Ug+UF=0. Ok! So for W=KE, if we knew the individual deltaKE for the force, it would be 10. I have one last question. W=Fx also. Thus, assuming F is constant, W done by the force would be W=F*1=F. Thus, work depends on F. However, W=-deltaU=10 gives us that it is always 10. Why are these yielding different results?

 

[Ken G]

They aren't-- you are forgetting that if you would lift something against gravity, you don't get to use whatever F you like, it must balance gravity. If you use less F, it will drop instead of lift, and if you use more F, you get some excess KE by the end.

 

[daniellionyang]

Yes. However, what if the final velocity is not 0. Then the extra KE would just cause a bigger final velocity which won't influence the PE. Thus, even if you have extra KE, the work still should be 10.

 

[Ken G]

The work is only 10 if the force is 1, and that barely balances gravity, so there'll be no excess KE. The principles work.

 

[daniellionyang]

Firstly, since the work is 10, W=F(gravity)*x+F(force)*x=>10=-10+F, so isn't F=20.
If this is so, isn't it possible to choose the magnitude of the force you apply to the object? Why does the force HAVE to be g+10?
Sorry if I am being rally naive about this topic.

 

[Ken G]

I should have said the force is 10, I forgot it is the distance that is 1. Still, there's no force that is 20 here, if you lift something up against gravity, such that it acquires no significant velocity but rises up a distance of 1 (let's not bother with units, it's metric), the work looks like this:
work by external force = 10*1 = 10
work by gravity = -10*1 = -10
net work = 0
change in kinetic energy = 0
which all checks with no significant velocity, verifying that we did indeed need an external force of +10 to get this done.

 

[nasu]

You are mixing the work done by different forces. And using units won't hurt either.:)

If you lift a 1kg block (you never mentioned the mass of the block, I think) to a height of 1 m "from rest to rest" the net work is zero. The change in KE is zero. This is the work-energy theorem.
The work done by gravity is -10J (assuming g=10 m/s²). The force of gravity is opposite to the displacement so the work is negative.
You can associate this work with a change in gravitational potential energy, ΔPE=-W=10J . The gravitational PE has increased by lifting the body. The KE did not change. So the change in mechanical (PE+KE) is +10J.

Because the net work is zero, the work done by the force lifting the body should be +10J, to cancel the -10J done by gravity.
So the work done by this force is equal to the change in mechanical energy.

As long as you use the definition consistently there is no contradiction.

 

[daniellionyang]

Oh wait... Yeah! I forgot that potential is for a specific force. Can you explain to me specifically in what cases -deltaU and deltaKE gives you the work of a specific force and when it gives you total work? Also, why must F be 10. Can't you just apply any force to the object you are lifting? Sorry if I am bothering you too much.

 

[nasu]

I would if I knew that you really read the posts trying to explain these questions.
Your questions were already explained in detail by Ken G.
Read post 8 and see if you can answer which work is equal to the change in KE.
Or even better, read the work-energy theorem in your book or on wiki.

 

[daniellionyang]

In actuality, I did read the other post. I guess I need to be a little for clear with my question.Although Ken G said that -deltaU is for a single conservative force, consider a push to the right 1 meter. of an object resting on the table. There is no change in potential energy, but there is work done. However, in this case, deltaKE gives you the formula for the work done by the force. I guess I just need little clarification or an example for something. Why does -deltaU work for the push up and why does -deltaU not work for the push sideways? Also, if I am understanding this correctly, deltaKE always gives net work, on the object, right?

 

[nasu]

OK, you got the part with KE. :)
If you are sometimes confused, remember that you can always solve the problem just by this. (work-energy theorem).

Now, if you want to get into PE, the point to understand is that only some forces have a PE associated. For the problems you are likely to encounter in elementary mechanics, only gravity and elastic forces may be associated with PE. If you have work done by gravity then you can talk about change in PE. If not, forget about it.
But even if you have work done by gravity, only the work done by gravity is associated with PE. The work done by other forces during the same motion don't associate with changes in PE.

Now, can you see why when you move something horizontally, even some work is done there is no change in PE associated?

 

[daniellionyang]

Oh... I see! Thanks! Now, can you answer the question of why F has to always equal 10? Can't you simply apply any force you want while lifting the object? (the math is in post 18.

 

[nasu]

First you have to write the full text of the problem. Without that, you question has no sense.
What force?

 

[Ken G]

Perhaps a lot would fall in place if we told you why the force needs to be 10 N. It's because you said there is no gain in KE-- the mass is lifted at nearly zero speed. If you use a force larger than 10 N, you will accelerate the mass, right? (Assuming, as nasu said, the mass is 1 kg.) To see if you understand, try the problem with an upward force of 20 N for 1 m, and calculate the work done by the 20 N force, the work done by gravity (and only that work can be accounted for as "coming from" the gravitational potential energy, though remember it's negative), and the change in kinetic energy.

By the way, a useful analogy for energy is money. Think of KE like the money the object has "in its wallet", and U like a "bank account" (if negative, that's like taking a loan from the bank), and external work like "money being paid to the object." The work-energy theorem then says the money in your wallet increases by the amount you are paid, minus what is getting deposited in your bank account. Each different force over the distance displaced is like a different way that money is changing hands.

 

[daniellionyang]

Ah... I see... Well, thank you all for clearly up my conceptuals regarding work!

 

[Mark Harder]

The notion of energy dissipation has been mentioned in several posts above. Just to be clear, when we consider heat and other forms of energy lost or gained by a mechanical system, we are in the realm of thermodynamics. There, changes in energy are not equivalent to work done (i.e. force * distance.) If you want to work with energy in a thermodynamic system and assume its energy is conserved, you have to include in the system the environment that absorbs or supplies the energy to the mechanical, work-doing part of the system. Only then is the total energy of the system conserved. That's the essence of the First Law.

 

[daniellionyang]

Ok! Thanks! All of you helped a lot to allow me to understand a question that had been driving me crazy for a while!