# Heigh Ho, Heigh Ho, Problems with Work, Y'Know?

1. Feb 16, 2004

### holly

Okay, in a different thread I asked a question about a block of ice and an incline plane and needed to know how to figure out the work and the force to move a 500N block up a 6m ramp, and also directly to the top of the ramp, 3m high.

I couldn't figure it out because I didn't know how to use the W=Fxd formula...the book gives only one little example, and it doesn't qualify the distance thing, to wit:

Could it be that the DISTANCE part of the formula refers to VERTICAL distance? I can't get it to make sense otherwise...but what about when a crate is pushed along the floor, and isn't raised? Is it still Vertical Distance or is it now flat distance? How can W=Fxd be valid for flat distance one time, and vertical the next? Isn't distance distance? I mean, the ramp was 6 m long. That's more distance than only 3 m up...

2. Feb 16, 2004

### deltabourne

The distance you push it is the ACTUAL length of the ramp.. which looks to be 6m in this case

3. Feb 17, 2004

### jamesrc

True. To add to the explanation:

For a constant applied force, as in the problem you have, the work done in moving an object between two points is given by the product of the component of the force along the path taken and the length of the path. This is because computation of work done by a force to move an object is a dot product of vectors. (The general definition for work involves a line integral and looks something like this:
$$W_{ab} = \int_a^b \vec F\cdot d\vec{r}$$
but I have a feeling that you're not interested in that right now.)

Anyway, if we compute the work done in moving the object up the (frictionless) ramp, we should see that we have to multiply the applied force by the length of the ramp. The applied force required to move the object is found with a free body diagram to be the component of the weight along the ramp, namely: mgsin(&theta;).

(We are given the geometry of the ramp so we know that &theta; is 30 degrees (sin&theta; = .5 = 3/6).)

So the work W = mgsin&theta;*d = (500N/2)*6m = 1500 J

In order to pick the ice up vertically to the top of the ramp, the applied force will be equal to the weight, mg = 500 N. The distance that the applied force acts through is 3 m (the force and the displacement vector both point up). So the work is simply W = Fd = 500N*3m = 1500 J

Note that the answer is the same as before. That is because the work needed to move an object between two points in a conservative force field (such as the one provided by gravity here) is path-independent.

4. Feb 17, 2004

### holly

Lovely and elegant solutions to my problem, nothing of which I understood. I never had geometry, and yes, it's too late now.

How can the work for pushing the ice up the 6m ramp be the same number as the work for picking it up directly and setting it on top the ramp? When I put 3m in for the d in W=F d, that doesn't give me the same number as when I put 6m in for d. I am totally turned around. Is 500N the F in the equation? I am going to flunk.

5. Feb 17, 2004

### jamesrc

Hmm,

I don't know what to tell you about the geometry. Haven't you ever done a free body diagram for a block on a ramp in your class? Maybe you could look online for trigonometry tutorials.

As far as the work calculation goes, W=Fd does not tell the whole story. How do you decide what F is and what d is? Let's answer this question for the special case where the applied force is constant (does not change with time) and the path is a straight line between two points. (I believe these are the only situations you will run into and I don't want anyone to jump on me for not explaining this for a more general case.) d is the distance traveled along the path. In part a, the distance traveled is 6m and in part b, the distance traveled is 3 m. Now the applied force that counts when discussing work is the force that points in the same direction as the path. In part a, it's the force that pushes up the ramp: F = 500N*(1/2) = 250 N. (Again you're going to need some bacground in trigonometry/working with vectors to understand that more). In part b, it's the force that points up: this force is simply equal to the weight, 500N.
Another way to look at this is with the work-energy theorem. There are no kinetic energies in this problem (everything has 0 velocity) so the work done on the object is equal to its change in gravitational potential energy. Look at the first case: initially the block is on the ground, so we set the potential energy equal to 0. At the top of the ramp, the ice is at a height of 3 m, so it has potential energy given by (mg)*h = (500N)*3m = 1500 J

If you follow that formulation for the second case (where you pick the block straight up) what is different? The answer is nothing, so you get the same work. The only thing that affects the potential energy (and therefore the work) is the height off the ground. In both cases, the block goes from 0 to 3m high.

Last edited: Feb 17, 2004
6. Feb 24, 2004

### Mac13

whats up with not being able to read james thread..its all black

7. Feb 24, 2004

### jamesrc

8. Feb 25, 2004

### turin

There is something called the "work energy theorem." It roughly states that the amount of work (W) required to do something to a system is equal to the amount that the mechanical energy of the system increases (&Delta;U) plus the amount of wasted energy (Q) in the transaction (which is unsually in the form of heat and usually due to friction). This is also the first law of thermodynamics, but that probably doesn't matter to you.

Anyway:

W = &Delta;U + Q.

The sign conventions are arbitrary but important (quite often, you may see W + &Delta;U = Q in thermo, but the way I defined the variables is consistent with the first way I wrought the equation; again, probably not important to you). Fortunately, they make sense in terms of conservation of energy.

How does this apply to your problem?

Well, since its ice, you are probably supposed to assume (I would) that there is no friction. Basically, that amounts to Q = 0 (not because it's could, but because it's slippery). Then, assume that the U is entirely gravitational potential energy (which is conservative as jamesc hinted). This leads to:

W = mg&Delta;h

regardless of the path.

Crate problems on flat ground will probably have &Delta;U = 0 while Q /= 0. In these cases, it is usually easiest to use that integral with force and distance.

Last edited: Feb 25, 2004