This video is going to be about the idea of work in physics. Work has a very specific definition
and equation that we're going to go into. Work is basically a force applied on an object multiplied
by the distance or displacement that the force is applied for. The specific definition is that
it's an object's displacement multiplied by the component of force parallel to the displacement.
The variable for work is W and the unit is J, which stands for a Joule, and the equation for
work is: work is equal to the force times the displacement s times the cosine of the angle
between them. I'll explain why that equation looks the way it does using this picture. Let's say that
you want to push this box along this displacement s, and to do that you apply this force in this
upward and rightward direction. Here we can imagine maybe this is a rope or something dragging
the box along. And there's a certain angle between the force and the displacement that you're moving
the box, so the displacement and movement looks something like this. So our definition says that
work is equal to the component of the force that's parallel to the displacement multiplied
by the displacement, and not all of that force is parallel to the displacement. So that force has
two components: an X component and a Y component that look like this, and you can see when I draw
those out that only one of those components is parallel to the displacement, which is the force
times the cosine. That's the X component in this case, so you can see that the force times
the cosine of the angle is equal to the force component that's adjacent to the displacement,
which according to our definition is the component that we need to calculate the work. So the work
here is going to be equal to the component that's adjacent to the displacement times the
displacement, so that's where we get the equation force times the displacement times the cosine of
the angle between them. Some patterns show up here. You'll notice for example that if a force
is perfectly parallel with the displacement, the angle between them is equal to 0 and cosine
of 0 is just 1, so that means that the force times the displacement times the cosine is just
equal to the force times the displacement, which makes sense because the whole force is parallel,
so we don't really have to worry about angles or components at all if the force is perfectly
perpendicular to the displacement. We know that the cosine of 90 degrees is 0, so that actually
means that the whole work equation is going to 0. So if you have a force applying at a perfectly
perpendicular direction to the displacement of the object, that actually means that the force is
not doing any work on the object as we understand work in physics. Work can be positive or negative,
but the rules for when it's positive and negative are different from when vectors are positive and
negative, and in fact work is a scalar. It has a magnitude but it does not have a direction.
I'm gonna give you a reason why. Let's say that I define right as being positive and left as being
negative. In this case I have a force pointing to the right and a displacement pointing to the
right, so we can see that the force is positive and the displacement is positive. So that means
that the work is also positive. If I flip these around and I consider the rightward direction to
be negative, now I now have a negative force because right is negative, and a negative
displacement because right is negative, but when I multiply those two things together I still get
a positive work, so whether work is positive or negative has nothing to do with the specific
direction that the force and displacement are pointing. Whether they're pointing positive or
negative by themselves doesn't affect the sign of the work, so work does not depend on a particular
direction. We can see that work is always positive if the parallel component of the force points in
the same direction as the displacement, so what work actually depends on and whether work is
considered to be positive or negative is whether the force and displacement point in the same
direction as each other, or in the opposite direction as each other. So as an example,
if the force points in the negative direction and the displacement points in the positive direction,
I can see that the work is going to be negative here, and if I switch these around now the force
points positive and the displacement points negative, but the work altogether is still
negative. So what's determining whether the work is positive or negative is not the direction of
just the force or just the displacement, it's how those 2 things relate to each other. So we can't
say that work by itself has a direction in the real world, because it's based on other things
that have directions, but it itself does not. So work is always negative if the parallel component
of the force points in the opposite direction of the displacement. We can talk about the work done
on the object by specific forces or the total work done on the object. So we use language for
both. There are two ways to find the total work done on an object. We can either add the work
done by every force on the object (if we add all the individual works of all the individual
forces we get an overall total work done on the object) or we can just use the net force on the
object as F in our work equation. If we just add up all the forces to find the net force,
that's another way of finding the total work done. I'll give you an example of what I mean.
Let's say that we want to find the work done by every individual force on this object, and then
find the total work done. So I'm just gonna go through one at a time and look at the work done by
each force as this box is moved five meters to the right, starting with this 10 Newton force. I can
see that because this is perfectly perpendicular to the displacement, you'll remember that that
means that the total work done is 0 because the angle is 90 degrees and cosine of 90 degrees is 0.
The work done by the 8 Newton force is gonna be the force times the displacement times the angle
between them, and here the angle is just zero, so that means that cosine is just 1, so the force
times the displacement is 8 times 5 which is 40, and that's measured in joules J and again this
downward force here of 10 is perpendicular to the displacement, so that means that it's going
to have a total work of 0, and this 3 Newton force is pointing in the opposite direction of
the displacement, and you'll remember from the previous slide that if the force is pointing
in the opposite direction of the displacement the work is always considered to be negative,
so this is going to be negative 3 times 5 which is equal to negative 15 Joules, so the total work
done on this object over those 5 meters I can see is 25 Joules. if I add up the work done by each
individual force it adds together to make 25, and like I said the other way you can calculate the
total work is by finding the net force on the object, and I can see here that the net force,
if I cancel out the 10 going up the 10 going down and the 3 takes away from the 8 I get a total net
force of 5 going to the right, so here the net force is 5 and the displacement is 5. There's no
angle between the net force and the displacement, so the total work is equal to 25. So those are 2
different ways of finding the total work done on the object. Here's an example of calculating work:
a person who pulls a 10 kilogram box 10 meters along a horizontal floor with a rope that applies
100 Newtons angled up at 30 degrees. If the coefficient of kinetic friction between the
floor and box is 0.5, what work is done by the rope, by gravity, and in total on the box? It's
always helpful to start these by drawing a free body diagram, so that's what I'm gonna do. This
says that there's a force going up and to the right of 100 Newtons at an angle of 30 degrees,
so if I break this into its X component and Y component this is what it looks like. There's
also a force of gravity going straight down, which I can find just by multiplying the mass
times gravity, so I know that this object is not moving up or down at all, so that means that
the total force in the Y direction has to add together to make 0, so I can see that I have an
upward force of tension of 50 as one component of my force of tension, so if I put that on my force
table above my head and add that to the force of gravity going down which I'll consider negative,
I know that in total these forces have to add together to make 0. So that means that the normal
force to add to those things to make 0 has to be equal to positive 48 Newtons, so that's going up.
To figure out the force of kinetic friction I can just say that the force of kinetic friction is
equal to the coefficient times the normal force. That's my equation for the force of friction:
multiplying the coefficient by the normal force gets me 24 Newtons, so that force is going to the
left because it always opposes the motion of the object. Now that I have this complete free body
diagram I'm just going to enter in information about the X component of the net force above
my head. So we can see that I have a force of tension of 86.6 Newtons and a force of friction of
negative 24 Newtons, so altogether the net force is 62.4 Newtons to the right so the work done by
the rope is going to be equal to its force times cosine times the displacement, and I already know
the force of tension times cosine of 30 degrees is just 86.6 Newtons and that's pointing in the same
direction as the displacement, so I'm going to consider that to be positive work. So I multiply
that force times the displacement of 10 to get 866 joules, so that's the work done by the rope.
I can see that gravity is pointing at a 90-degree angle to the displacement,
so because it's at a 90 degree angle the cosine of that is just 0, so the total work done by gravity
is 0 and to find the total work done on the box I can just use my net force in the work equation and
I know that the net force is just pointing in the horizontal direction. I can see above my head that
it does not have a vertical component, so because it's just pointing in the horizontal direction in
the same direction as the displacement, the angle between it and the displacement is zero, so that
means that the cosine is 1, so we just have to multiply the net force by the displacement,
and when I do that I get 624 Joules of work done in total on the box. The reason why this is lower
than the work done by the rope is that the force of friction is taking away, it's adding negative
work to the box, so if I add the negative work from friction to the positive work from the
rope I get a total work done on the box of 624 Joules. Here's another example: a person carries
a 10 kilogram bag of groceries up a 6 meter high flight of stairs to their apartment. How much work
do they do on the bag? So this is what it looks like: they're moving up at a constant velocity up
the stairs. I'm gonna start by making a free body diagram for this bag. I know automatically that
there's a force of gravity of 98 Newtons going down, and if this bag is moving up the stairs at
a constant velocity that actually means that the forces only need to balance out. There doesn't
need to be any kind of special force pointing diagonally up the stairs, because if it's moving
at a constant velocity we just know that the net force has to make 0. So if the person applies an
upward force that's equal to the downward force of gravity applied on the bag, that will be enough to
keep it moving at a constant velocity, so if the bag pulls down with 98 Newtons the person pulls
up with 98 Newtons, so that upward force is what we want to focus on as they move up because we're
asking how much work does the person do on the bag, not how much work is done in total on the
bag. I can see that this force applied by the person is parallel to the vertical displacement
and perpendicular to their horizontal displacement as they go up the stairs, so that actually means
that the horizontal displacement isn't going to come into my equation at all because the
work done over the horizontal displacement by this force is just 0, because it's perpendicular to the
horizontal displacement, so it's only the vertical displacement that matters when calculating the
work done by this person. So here the work is 98 times 6 which is equal to 588 Joules, so for this
person to move the bag up the flight of stairs they had to do 588 Joules of work on the bag. I'll
do one more quick example: we can imagine we have the same situation but now we're given the full
length of the stairs of ten meters, and the angle of the stairs so they go up stairs that are angled
at thirty degrees and they walk for ten meters up those stairs, and they're applying the same force
to the same bag. So in this situation we just have to solve for what their horizontal and vertical
displacements are. When I do that using trig I find that they went 5 meters up, and it doesn't
actually matter how far they went to the right because that rightward component is perpendicular
to the force, so it's not contributing to the work done at all, because they're going up 5
meters and the force that they're applying is 98 Newtons. This is going to be 98 times 5 which is
equal to 490 Joules, so that's the basics of how you use the equation for work and the concept
of work to solve for the work done. In the rest of this unit we're going to explain why we care
about work so much. It's actually one of the big foundational problem solving tools in physics,
and that doesn't entirely become clear until we talk about energy in the next lecture.