# Work and Energy

1. Dec 6, 2012

### benrocks1

Can someone explain to me when you use 1/2mv^2-mgh as opposed to 1/2mv^2 or mgh

2. Dec 6, 2012

### K^2

In what context?

3. Dec 6, 2012

### the_emi_guy

So we can know were you are coming from, are you able to answer the following -

What is 1/2mv^2 ?

What is mgh ?

4. Dec 6, 2012

### benrocks1

I know that 1/2mv^2 is Kinetic energy and mgh is potential energy. My question is in some example problems they use 1/2mv^2 and in others they use 1/2mv^2-mgh.

5. Dec 6, 2012

### CWatters

You will have to tell us what problems.

In some case the total energy of a system is KE + PE. It's possible that the direction of "h" in your question has been defined such that PE = mg(-h) making your equation really...

1/2mv^2 + mg(-h)

or it might be nothing of the sort. Tell us about the problem where you saw this equation.

6. Dec 6, 2012

### the_emi_guy

Good,
So if a problem asks for the kinetic energy of an object, which formula would I use?

If a problem asks for the potential energy of an object, which formula would I use?

Do you know about the concept of conservation of energy? When a ball is falling what is happening to its kinetic energy? Its potential energy?

See if you can figure out from these clues where 1/2mv^2 - mgh would be useful.

EDIT to fix sign: should be 1/2mv^2 + mgh

Last edited: Dec 6, 2012
7. Dec 6, 2012

### benrocks1

While the kinetic energy increases, the potential energy will decrease. Is that correct? I've thought about it, but i still do not understand why I would use 1/2mv^2-mgh

8. Dec 6, 2012

### the_emi_guy

Good, yes that is correct.

Conservation of energy means that the *total* energy remains the same (this is what conservation laws are, some quantity remains unchanged even though its constituent parts are changing).

So if the kinetic energy increase by some amount (say x), now much will the potential energy have to decrease?

What will happen to the quantity 1/2mv^2 - mgh is this case?

EDIT: should be 1/2mv^2 + mgh

Last edited: Dec 6, 2012
9. Dec 6, 2012

### benrocks1

So 1/2mv^2-mgh would also increase by x?

10. Dec 6, 2012

### the_emi_guy

No, think about it carefully (don't get frustrated, you are very warm).

Start by answering: If the kinetic energy increase by some amount (say x), now much will the potential energy have to decrease?

11. Dec 6, 2012

### benrocks1

the potential energy will decrease by x. does that mean that 1/2mv^2-mgh also decreases by x?

12. Dec 6, 2012

### the_emi_guy

No,

Here is what you have so far:

Kinetic energy is 1/2mv^2

Potential energy is mgh

Kinetic energy increases by x

Potential energy decreases by x

What happens to 1/2mv^2 - mgh?

EDIT: should be 1/2mv^2 + mgh

Last edited: Dec 6, 2012
13. Dec 6, 2012

### benrocks1

would 1/2mv^2-mgh remain the same?

14. Dec 6, 2012

### the_emi_guy

Bingo!

1/2mv^2 - mgh is the total energy and it is conserved. In other words when a ball is falling, the quantity (1/2mv^2 - mgh) will remain constant no matter where that ball is in its flight.

(of course we are ignoring air resistance and other complicating stuff).

*So*, consider this problem, I a drop a ball from 10meters and want to know how fast it was going when it hit the ground.

Calculate (1/2mv^2 - mgh) before the drop (1/2mv^2 = 0).

Calculate (1/2mv^2 - mgh) at the ground (mgh = 0)

And trust that the difference will not change.

15. Dec 6, 2012

### benrocks1

I understand that mgh would = 0 when it hits the ground, but why is it 0 before the drop?

16. Dec 6, 2012

### the_emi_guy

Look at my previous post carefully, what did I say was = 0 before the drop?

17. Dec 6, 2012

### benrocks1

Oh so the kinetic energy is 0 because it is not in motion?

18. Dec 6, 2012

### the_emi_guy

Exactly!

I am just realizing that we have a sign error here, perhaps this was the source of your confusion.

It should be 1/2mv^2 + mgh = constant right? Sorry I did not notice this before.

19. Dec 6, 2012

### benrocks1

20. Dec 6, 2012

### CWatters

ONLY for certain problems.

Consider a pendulum swinging back and forth with no air resistance or friction.

At the top of the swing:

PE = Maximum = mghmax where h is measured from the bottom of the swing
KE = Zero because it stops momentarily

At the bottom of the swing:

PE = Zero
KE = Maximium (= 0.5mV^2)

At any point in between:

PE + KE = Constant = mghmax