- [Instructor] There's a hamburger sitting right in
front of you, but you don't want it.
Maybe it's because you're a vegetarian,
maybe you're already full, maybe you found someone that
needs it more than you do, so you're gonna push this
hamburger to the right, with a force of four newtons
and you're gonna do this and you're gonna push this
a distance of five meters to the right.
So a question we get asked would be
how much work did we do in passing
this hamburger to the person to our right?
Now since this is a pretty simple problem we can just plug
straight into the work formula, but I'm not gonna do that.
I'm gonna show you an alternate way to think about this,
because what we learn from this alternate
approach is gonna help us in more
challenging, complicated work examples.
So I ask you to imagine this.
Instead of just plugging straight in,
consider the fact that we exerted a force of
four newtons for the entire five meters
that we pushed this hamburger to the right,
so if we were to plot what the force was
on our hamburger as a function of its position,
it would look something like this.
So if it started at zero it moved five meters to the right
and we exerted a constant force of four newtons,
That's why this is a horizontal line.
It's horizontal, that means it was a constant amount
of force, and the reason I'm doing this is because
there's gonna be a geometrical significance to this work.
Check it out.
We can just use the work formula.
We know work is the force in the direction of motion
times the displacement or you could say the entire force
times displacement times cosine theta.
If you don't like this cosine theta, this just makes sure
that you're singling out the force in the direction
of motion and only that component of force that's
directed in the direction of motion but for our hamburger
example, the force already was in the direction
of motion so we don't really need this.
In other words the angle here would be zero.
Whenever you take cosine of zero, you just get one
and one times anything is just that thing so whenever
your force is already in the direction of the
motion of the object you don't really need that
cosine theta and we don't need it here.
So I'm gonna say that the force here was four newtons
so we can calculate four newtons was the force for
the entire displacement of five meters
and we get the work that we did on the hamburger
was positive 20 joules of work.
Now if you're clever you might be like
wait, four newtons times five meters, look it,
that's just the area of this rectangle.
So notice that this line is a straight line in this
force graph just forms a rectangle in here
and all we did, we took four newtons, that was just
the height of this rectangle and we multiplied by
five meters and that was just the width of this rectangle
and if you multiply a height times width,
you know what you get, you get the area of a rectangle,
so what we found was when the force is constant
one way to find the work done is by using the work formula
but another way to find the work done is just to find
the area enclosed by the graph.
So from this line that determines the force
down to this x-axis, if you find the area,
that's gonna equal the work done.
Now you might be like, alright, well,
fat lot of good that does us, we can
already find it with this formula.
Why would I ever want to know that the work is
equal to the area under a force graph?
I'll show you why.
Because in this case, yes, it was easy,
we had a formula, we could just plug the force in here,
we could just plug the displacement in there,
we get our value, but think about this.
What if our force was not constant?
What if we had a varying force?
In that case, what force would we plug in here?
It'd be changing.
One way to handle that scenario is
using calculus, but if you don't know
calculus that doesn't do you any good.
Fortunately, there's another way to determine
the work done by a varying force
and that's to take this idea seriously.
That's why I showed you this.
It turns out the area underneath any force versus
position graph is gonna equal the work,
not just ones where the force is constant,
even where the force is varying,
if you can find the area underneath that graph,
that's a quick and easy way to
get the work done by that force.
So what I mean is this.
Instead of just pushing with a constant four newtons
for the entire duration of this trip,
let's say you started pushing with four newtons
but you were getting tired and your force was diminishing
and you were pushing with a weaker and weaker force
until your force became zero newtons.
This would be zero newtons of force when it hit this axis.
Let's say the hamburger still made it five meters.
You might be confused.
How could it make it five meters
if we're pushing with less force?
Well it may have taken more time to get there,
but let's say it still made it five meters.
How do we find the work done by our force now?
Well I'm still gonna make the claim that
the work done is gonna be equal to the area underneath
this force versus position graph,
but you might be skeptical, you might be like wait a minute,
all we really showed in that previous
example was that the work done for a constant force
was equal to the area underneath.
How do I know for this varying force that the area
underneath this graph is still gonna give me the work done?
Well physicists and mathmeticians are clever.
What they did is they said this,
alright, they were like, the only thing we know
is that for a constant force the area underneath
is equal to the work, so let's just say this,
instead of just considering this case where I diminish
my force continuously, let's say I pushed with a whole
bunch of constant forces but for a small displacement,
so I started with four newtons but I pushed for like only
10 centimeters with four newtons, and then I dropped that
down to maybe 3.9 newtons and I pushed for another
10 centimeters, and I keep doing this over and over,
reducing my force but keeping it constant,
and then dropping it again and here's why that's useful.
Because the work done for this rectangular case
is just gonna equal the area of all of these rectangles
added up, right, because they're each constant forces
so I know the work done is just equal to the area
underneath, so if I add up the area of all these
rectangles I get the work done during that rectangular
process and here's the really clever idea.
If we made these rectangles infinitesimally small
they would add up to the total area just underneath
this triangle, it'd be the exact same area.
Now the area under this triangle would be exactly
the same as the area under all of these rectangles
and it'd be the exact same process we had before,
because think about it, you'd be pushing with four newtons
for like a millimeter but if it's infinitesimal I mean it's
even smaller than that, but for the sake of just
conceptually thinking about it,
let's say you push with four newtons for a millimeter
and then for the next millimeter you pushed
with 3.99 newtons and then for the next
millimeter you pushed with 3.98 newtons.
That's basically just you diminishing your force
continuously and going through this exact
process that we described earlier.
But now we know the area under all these rectangles
is gonna just equal the work done,
but that's the same area and process
that's just the area underneath this triangle.
Now, you might not buy that.
You might be like, wait a minute, look it,
these edges are throwing everything off,
look at this little edge, this shoots out
a little too far, that's a little overestimating the area,
and then you gotta little hole in here,
that's underestimating the area,
you got these all over the place.
Is that really gonna be equal to the area?
And it will be if you let these widths of the rectangles
become infinitesimally small, the error you're incurring
from all these little overshoots and undershoots
are also gonna get infinitesimally small,
and if the error is getting infinitesimally small
that means that the area under all these
infinitesimal rectangles is gonna exactly equal
the area under this triangle which is great news for us,
because that means that the area
under any force versus position graph,
it didn't have to be a triangle,
we could use infinitesimally small rectangles to
represent the area under any graph.
That means the area under any force versus position
graph is gonna equal the work done by that force
and now we have a powerful tool to determine the work
done for cases where the force is not constant.
This formula here where you have work equals
Fd cosine theta, this only gives you true
answers if you use a constant force.
If the force is constant you can use this equation,
but if the force is varying you couldn't just use this,
but now we derived this.
This is great.
This showed me that if I can find the area under
my force graph, my force versus position graph,
I can find the work done.
So, in other words, for this case right here,
if I push with four newtons on my hamburger
and then I reduce that force to zero
and it went five meters, how much work was done?
Well now I know how to find it.
I just need to get the area under this force versus
position graph and it's a triangle
and I know how to find the area of a triangle.
So the area under this triangle is
1/2 base times height, well so I'll have 1/2,
the base here is gonna be five because this is five meters,
so I have five meters and the height of this triangle
is four newtons and so I've got four newtons times five
meters times a 1/2 is gonna give me positive ten joules
of work done during this process where my force diminished
continuously, and that's why this idea is so powerful,
because if your force versus position graph happens
to take a shape that you know how to find the area of,
like a rectangle or a triangle or some combination
of rectangles and triangles, then you can quickly
get the work done without having to know anything
about calculus because you just get the area underneath.
Let me just clear up any confusion.
You don't actually have to draw any rectangles.
That was just to prove this relationship.
Once you know this relationship you don't need to draw
or think about those little infinitesimal rectangles.
That was just how we were proving to ourselves
that even for cases where the force was varying
the area underneath, it's still gonna equal the work.
So there's one more thing you gotta be careful about.
When we say the area underneath the graph
is equal to the work done, we mean the area
from the line that represents the force to that x-axis,
and you might be like, well of course
that's what you mean, what else could you mean?
Well sometimes it gets a little unclear when you
have a case like this.
So let's say your force started down here.
So we'll say we first started pushing on our hamburger
to the left and then we're like oh there's
no one there, awkward, we'll start pushing
to the right, so our force starts off
negative and then it becomes positive.
Now if we wanted to find the total work done during
this entire process, how would we do it?
We can still use this idea that the work done
is gonna equal the area underneath the curve,
but for this first portion what do I do?
How do I find the area underneath this portion?
Some people are like underneath, alright, I'm gonna
go underneath, we'll keep going this way,
how much area's down here, but that's just crazy.
That goes on infinitely.
I did not do infinite work on this cheeseburger.
No.
The area I'm talking about is gonna be this area right here,
so I gotta take all this area right here.
That area is finite.
It's from this line that represents
the force to the x-axis.
That's the area we're talking about,
just like over here the area we'd be talking about
is from this line that represents the force to the x-axis.
It's all of this area.
So how would I find these areas?
They're both triangles.
I can do this, so this one's gonna be 1/2 base times height,
so it's gonna be 1/2 the base, this base right here is
one meter, so it'd be one meter times the height.
I've got negative height.
Look it, it's down here.
It's negative two newtons.
That means I'm gonna have a negative area.
Is that okay?
Yep, that just means I did negative work
on this cheeseburger during that time
so I'm negative one joule.
During the first meter of me exerting force
on that hamburger, I did negative one joule of work,
and then for this portion of the
trip I'll find the area underneath.
It's also a triangle, so 1/2 base times height.
I'll have 1/2, the base this time is not three,
the base of this triangle is only two, because it goes from
one to three, so that's two meters times the height,
well that is four newtons, so I'll multiply
by four newtons and then I get for that
portion of the trip, I did four joules of work.
So, positive four joules of work for this portion,
negative one joule of work for that portion.
The total work done for the entire three meters
would be four joules minus one joule
and that would be positive three joules.
So, recapping, if your force is constant you can
just use this formula to find the work done.
It's just Fd cosine theta, but you can also find
the work done by determining the area underneath
a force versus position graph and this is useful
because this works even when the force is varying
which would render this equation somewhat unusable
but if the shape of the graph is one that you know
how to find the area of, you can still find the work
done by determining the area underneath
the force versus position graph.