In this video, we’re going to learn
about work. We’ll learn its physics definition,
how under that definition it can be positive or negative, and we’ll also get some
practice using it in examples.
To get started, imagine at one day
you and a couple of friends are out walking through the neighborhood. The three of you come across an
active construction site. And you see a sign saying that help
is wanted with pay given out by the net work performed. Interested in making some extra
spending money, the three of you decide to sign up for a day’s work.
Once hired on, each of the three of
you are assigned different tasks. Your first friend is assigned the
task of using a pulley system to lift tools and supplies up to workers on the top
level of the building being constructed. Once those workers are done with
the tools, your friend then returns them to their original location. Your other friend is given the job
of moving pallets of brick around to the back side of the building to be counted and
then to bring them back to where they were. And your job throughout the day is
to move all of the leftover building materials from the first floor of the building
up to the second floor, where that installation will begin.
At the end of a long day, the three
of you go to get paid for your work. To your friends’ great surprise,
they don’t get any money for the day, but you do. To understand why that it is, we’ll
want to understand a bit about work. We’ve all heard the word “work”
before and probably have a sense for what it means. But in physics, there’s a precise
meaning to the word “work.”
In physics, we define the work done
on an object as the scalar or dot product of the force on that object and its
displacement. As a symbol for work, we often use
a capital 𝑊. And because work is calculated by
taking a dot product, it itself is not a vector but a scalar. Another way to write out work is as
the product of the magnitudes of force and displacement multiplied by the cosine of
the angle between those two vectors. This cos 𝜃 term helps us
understand why one of your friends didn’t get paid at the work site.
For the friend who was carrying
pallets of material, the force they exerted on those objects to carry them was up to
oppose gravity, but the displacement was perpendicular to that direction. The cos of 90 degrees is zero. So they did no work. Going further, if we consider the
graph of the cosine function, we see that there are areas where this function goes
negative, which means that it’s possible for work to be negative.
Consider this example of a book
lying flat on a tabletop. Say that you pick the book up,
transfer it across to the other side of the table, and then set it down. We can draw in arrows representing
the book’s displacement at each of these three legs and then also draw in arrows
representing the force you apply to the book as it moves.
For the first part of the book’s
journey, force and displacement are aligned. So work is positive. In the second part, these two
vectors are at right angles to one another. So no work is done. Then, as you set the book down, the
force you exert on the book is up. But the book’s displacement is
down. For this part of the book’s
journey, the work you do on it is negative.
Our graph of the cosine function
supports this conclusion. We see that when the angle between
the force and displacement is 𝜋 radians or 180 degrees, we’ll have a negative
result. This helps explain why your first
friend didn’t get paid that day. This friend did positive work
lifting supplies up to the top of the building and then canceled it out with
negative work, lowering them back down. Just like with this book, the net
work done on the supplies was zero. So the work done on an object by a
force can be positive, negative, or can be zero. Let’s get some practice calculating
work through a couple of examples.
A toy cart is pulled for 4.4 meters
in a straight line across a floor. The force applied to the cart has a
magnitude of 17 newtons and is aligned at 28 degrees above the horizontal. How much work does the applied
If we draw a picture of this
process, we have a cart being pulled along by an applied force, 𝐹, where the force
is applied at an angle we’ve called 𝜃 above the horizontal. If we call motion to the right
motion in the positive direction, we’re told that the cart is pulled 4.4 meters that
way. We can call this 𝑑. We want to know the work done on
the cart by this applied force, 𝐹.
To solve for this value, we can
recall that work is not only equal to the dot product of force and displacement. But it’s also equal to the product
of their magnitudes times the cosine of the angle between them. In our case, we know the force
magnitude, 𝐹, as well as the magnitude of the displacement, 𝑑. We’re given the angle 𝜃 that
separates displacement from force and so are ready to plug in and solve for 𝑊.
When we calculate this product, we
find that, to two significant figures, 𝑊 is 66 newton meters or 66 joules. That’s how much work is done on the
cart over this distance by the applied force.
Now let’s look at an example that
will help us see work from a geometric perspective.
A crane is lifting construction
materials from the ground to an elevation of 60 meters. Over the first 10 meters, the motor
linearly increases the force it exerts from zero kilonewtons to 10 kilonewtons. It exerts that constant force for
the next 40 meters then winds down to zero newtons again over the last 10 meters, as
shown in the figure. What is the total work done on the
Calling the net work done on these
materials capital 𝑊, that work is related to the force the materials experience
over the distance they experience that force. We can recall that work is equal to
the dot product of force and displacement. And in our case, since the force
and the displacement are in the same direction, we can write this expression as a
simple product. Since work equals force times
distance, looking at our graph, we can see that this will be equal to the area under
To make solving for that area a bit
easier, we can divide it up into three pieces. Piece number one has an area of
one-half its base, 10 meters, times its height of 10 kilonewtons. Piece number two has an area of 50
minus 10 meters, or 40 meters, multiplied by 10 kilonewtons. And the third piece has an area
like the first of one-half 10 meters multiplied by a height of 10 kilonewtons. When we add the areas of these
three sections together to get the total area, we find it’s equal to 500
kilojoules. That’s the total work done on the
construction materials over this movement.
Let’s summarize what we’ve learned
so far about work. We’ve seen that work is equal to
the scalar product of force and displacement. For a given force 𝐹, the work that
force does on an object is equal to the dot product of that force and the object’s
displacement. We’ve also seen that work tells us
a bit about the angle between the vectors of force and displacement. This cosine term means that work
can be positive and it can also be negative, or of course it can be zero. And finally, we’ve seen that work,
which has units of joules, is related to energy, which is expressed in the same