Scribe #2 (wow, it sure is a huge gap between the first and second scribes)

Very sorry about the tardiness of the scribe... I worked from 6:00-11:00 tonight and it was a long one today, so... read this before our test for a good review =D
Well, we started off the class with Mrs. K. going over the handout on "WORK: A DEFINITION". It reviews the concept of work (with Force and Displacement as its two essential elements).
an example of these elements are in the diagram: "F" being force, and "d" being displacement

Work ("W") is Calculated by the following equation with the same parameters from the diagram above.
The next topic was on "Work Done by a Variable Force". This means that while the work is being done over the displacement interval, the Force is not constant throughout and therefore, the equation of W = F • d will not work. In this case, the graph of the Force vs. Displacement graph would be that of a curve or line with a slope x<0as seen below) .
Now, one can imagine (should be easy for Calculus students) that as the number of these rectangles increase infinitesimally, the size of the intervals will begin to become smaller, and smaller, and smaller... and continue as such forever, infinitesimally close to zero, such that the rectangles could be said to have a width of the exact point on the graph. At this point in time, if the areas of all these rectangles are added together, we would get the Work done... why? because area of these rectangles as we said are A = b • h or A = ∂d • F and since ∂d • F = W, then A = W. So basically, the area between the graph and the x-axis is equal to the Work done.

WOW!!! That was a lot of explaining... anyways, next we took a look at Work being positive or negative. Now it can be either, depending on the direction of the Force relative to the displacement, if the Force is acting in the opposite direction of the displacement, the Work would be negative. So to support this, if the Force were to act in the same direction as the displacement, the Work would have to be positive. ***However, if the Force is acting (or has a component acting) neither in the opposite nor the same direction and is perpendicular to the displacement [eg. Force acting UP, displacement to the RIGHT], the Work is ZERO (0)***

The next section dealt with Work and Force at an Angle.
Basically, the Force acting upon the object in question is doing so at an angle (Ø). However, this angular force is not used to determine Work, only the Horizontal component is used in the calculation. (or the vertical one if it is parallel to the displacement in question). The way to calculate this horizontal component is given by Fx = F • cos(Ø). Therefore, to find the Work of an object moving as a result of an angular Force, we must use W = F • cos(Ø) • d.
Finally we briefly looked at the calculation of Kinetic and Gravitational Potential energies. Kinetic Energy (KE) is said to be: the work needed to accelerate a body of a given mass from rest to its current velocity. Gravitational Potential Energy (PE) is: the energy that an object of a given mass has by virtue of its position relative to an arbitrary "zero ground". The formulae for these are as follows:
Finally, we finished off class with a couple questions from the "SPORTS BOOK"!!! (Duck Book). We will be correcting those in class tomorrow (p. 330 q's: 1-6)

That's all for tonight, very, very, very, very sorry for the late scribe... Hope this will at least be useful for test review. Thanks, and tomorrow's scribe is...

Grey-M!!!