In physics, work is defined as exerting force over a certain distance. Work is a form of
energy, and therefore takes on it's unit, the Joule (J). There are many different types of
energy, but the one we will be dealing with in this series of lessons will be Work.
Seeing as how work is a force exerted over a certain distance, it is only logical that to
find the total amount of work done, you multiply the net force by the distance. In fact, that
is the equation for work.
W = F * d
Work is only determined by the force exerted by one object on another. For example, if a dog
picks up a bone, walks 10m, and then sets it down, it only does work for part of the trip. The
dog exerts a force upon the bone when picking it up and setting it down, and therefore does work;
but, there is no force exerted upon the bone as the dog carries it in his mouth across the room,
and therefore no work is performed.
Let's go ahead and do a sample problem.
A boy pulls a wagon with an 80N force directed 25° above the horizontal a distance of 30m.
a. How much work is done by the boy?
b. How much work does he do if the force is directed parallel to the ground?
First, for part a, we need to break the force down into it's x and y components.
80N cos 25° = 72.505N
80N sin 25° = 33.809N
Now we need to figure out the total amount of work done. Remembering that work is force
times distance, we can automatically cancel out any possibility of vertical work. Why? Because
the wagon only moved horizontally, it didn't move up and down at all. Therefore, all we need
to worry about is horizontal work.
W = 72.505N * 30m
W = 2,175.15J
Now, for part b it said what is the work if the force is parallel to the ground. All that
means is that we don't have to break the force into components. Since it is traveling
horizontally, the entire 80N is the x component. All we need to do is plug-n-chug.
W = 80N * 30m
W = 2,400J
You probably noticed that when I wrote the formula for work at the beginning of the lesson,
I didn't put any subscript after the F in the formula. The reason is because you want to use
the F that corresponds with the type of work you want to find. For example, if you want WNET,
then you use FNET; or, if you want to find the amount of work friction did, you use FK as your
force. The neat thing about this equation, like Now that you know a little about work, we will discuss power next.
