# Quantifying Energy

1. Feb 2, 2005

### pierre45

Hello All,
I am new to this forum and am most definately NOT a physcis guy....but I have a question that I thought someone here could help me with.

I want some way to quantify what I would call energy, but that is probably not the right word. But basically, if I gave you the following information:

Starting at 0 velocity. Mass is say 40kg

after 21.3 seconds has traveled 1000 ft.
after 22 more seconds has traveled another 1000 ft.
after 22.7 more seconds has traveled another 1000 ft.

how can I quantify how much energy (whatever) was expended, in a way that I come up with a different answer than for this case:

after 22 seconds has traveled 1000ft.
after 22 more seconds has traveled another 1000 ft.
after 22 more seconds has traveled another 1000 ft.

It took the same amount of total time to cover the same distance, but the energy of the first must have been higher since a higher velocity was reached? Can someone please help me to understand this?

Thanks,
Pierre

2. Feb 2, 2005

### dextercioby

What u have described is the method of computing MEAN/AVERAGE VELOCITY.However,this has little to do with the concept of Kinetic Energy that comes out of your words as the one u'd like to discuss/clarify.
If a body of mass "m" has a velocity $\vec{v}$,then

$$KE=:\frac{1}{2}m\vec{v}^{2}$$

Daniel.

3. Feb 2, 2005

### pierre45

I am very confused as you can tell by my question.

I know the formula you post for kinetic energy, how can I use that to tell me how much energy was expended over the entire course of events?

I know each of those bodies had such and such kinetic energy at three different points in time...I want to know the total amount of energy put into that object to make it move that way from start to finish, assumming in both examples they have the same amount of force impeding them.

I apologize if this doesn't make much sense, but say a car for example, with a fixed rolling resistance and zero wind and given cross section etc., given the data I gave how can I quantify (within a ballpark #) how much energy was expended in the two cases I gave? In watts, calories, newtons, BTUs...I don't know what the correct measure is, I just want to compare one set of events to another to see which used more energy.

4. Feb 2, 2005

### dextercioby

In the case i've given you,to be able to compute the KE you must know bth the mass and the velocity.If the velocity is constant (by virtue of Newton's laws,it means that all forces acting on the body (if any) are 0 when added as vectors),then the KE is constant.
If not,then u have ot use the theorem of Leibniz

$$(\Delta KE)_{1\rightarrow 2}=W_{1\rightarrow 2}$$

which asserts that the net chenge in KE (between 2 states labeled with 1 & 2) is equal to the work done by the forces which act on the body...

Daniel.

5. Feb 2, 2005

### kanato

If you have a drag force, the work done by that force is -F*x, where F is the force applied and x is the distance travelled. For the second case with constant velocity, whatever force is fighting the drag force would have to do the same amount of work (but positive sign) to keep the kinetic energy constant. For the first case, the object is slowing down, because it's taking more time to cover the same distance, so whatever force fighting the drag force is smaller, so it is doing less work over the same distance.

Keep in mind that if there was no drag force or friction, in the object maintains constant velocity (as in the second example), then there are no net forces on the object.

6. Feb 3, 2005

### pierre45

that's what I don't know...what I do with my force numbers.

So I've got a + force and a - force direction wise. I want to know how much power was needed for the + force....but everything says net force...I know what the net force is, and I have a model for what the - force is...can I sum them (changing the sign of the - force) and call that my +force, and calculate power based on that, just fix it as if there is no drag? Calculat ehow much work would have been done if there was no drag?

seems reasonable...

7. Feb 3, 2005

### kanato

Hmm.. if you notice your second example, the object is moving at constant velocity, so there is no net force, and hence no net work done on it. So the work done by the drag is exactly opposite the work done by whatever force is keeping it moving. If there were no drag, then the other force (if present) would have accelerated the object.