# Implications of Work-Energy Theorem

1. Nov 2, 2014

### Impulse

Wnet = ΔKE

By this equation, if I lift a 1 kilogram book at rest from the ground and place it to be at rest on a table 10 meters above the ground, no net work has been done on the book. (Its kinetic energy before and after is zero.)

However, its potential energy has changed by mgh or 1kg * 9.8m/s2 * 10m = 98J. Therefore its total mechanical energy (KE + PE) has increased but no net work has been done.

By this reasoning the work-energy theorem implies that no net work needs to be done to increase or decrease the mechanical energy of an object.

This is counter-intuitive to me. Is not 98J of work required to increase the total mechanical energy of a system by 98J? Is there a law that relates work to the change in mechanical energy of a system?

2. Nov 2, 2014

### OldEngr63

The equation that you opened with does not apply to the example you cited for the very reasons you see a problem with it.

3. Nov 2, 2014

### PPERERA

-W = ΔU
The negative of the work done on a system is also equal to the negative change in potential energy of the system. When an object is lifted to a new height, it gains potential energy because Earth does work against it through gravity to oppose its vertical displacement. (We say it gains potential energy because it is lifted from a lower potential to a higher potential.) And this change in potential energy from the work done on the object is what causes the increase in mechanical energy. Furthermore, the process described is where this equation comes from:

ΔUg = -Wg = -FgΔy = -mgΔy

4. Nov 2, 2014

### Simon Bridge

... from the definition of gravitational potential energy, an object gains PE because work has to be done on it to get it there.

But what everyone is saying is that the initial form of the work-energy theorem commonly taught is not the whole story.
Well done for noticing. It would be better to note that work is the total change in mechanical energy. $W=\Delta U + \Delta K$
As you advance you will come across other ways of looking at work, and you'll end up just using conservation of energy directly.

5. Nov 6, 2014

### Impulse

Is the most robust definition of work:

Wnet = ΔEsystem

?

Also, under what conditions does Wnet = ΔKE not apply?

Later in the chapter the text makes a distinction between "conservative" and "nonconservative" forces and defines:

Wnet = Wc + Wnc = -ΔU + Wnc

Therefore by substituting Wnet for ΔKE we can say:

Wnc - ΔPE = ΔKE

Is the above formula always applicable?

If it is, because Wnc = ΔKE + ΔPE = 0J + 98J, there must be 98J of nonconservative forces acting on the book. What would those nonconservative forces be?

6. Nov 6, 2014

### Simon Bridge

Pretty much - though different situations will use a slightly different definitions. The most common variation is the sign convention. YOu need to look out for that or just make your own and state it clearly at the start of your work.

Well, clearly when not all the work goes into changing kinetic energy.
You gave an example in post #1. A mass m climbs a hill at constant speed ... the kinetic energy does not change, but work is done.

No ... because $W_{net}=\Delta KE$ does not always apply, and the formula you quoted from the book does not always apply. You need to be careful of the context and try not to mix up different situations too much at this stage. Try to understand where the formula comes from: it's not a definition - it's a description of a particular class of situation.

Usually if you have some energy left over after accounting for changes in kinetic and potential energy, it means you have failed to include some energy transformations.

You need to take a close look at what you included in the kinetic and potential parts.

... friction.

The thing to remember is that the total energy is always conserved.
With work you are usually interested in the energy that gets used for a particular task (or is needed to perform a particular task), so you are not considering the whole system. As long as you focus on the energy transformations you should be fine.