### Video Transcript

In this lesson, weβll learn how to
calculate the work done by a constant force vector acting on a body over a
displacement vector using the dot product.

Remember, work is done when a force
causes a body to move over some distance. And in its most simplest form, it
can be calculated by finding the product of those two measures, force times
displacement. When the force πΉ is constant and
the angle between the force and the displacement π is π, then work done is
alternatively given by πΉ times π cos π. And whilst work is a scalar
quantity, we can alternatively represent the force and displacement as vectors
rather than as the magnitude of vectors.

Then the work done by a force
vector π
with displacement vector π is the dot product of the force vector and the
displacement vector. This can be alternatively
calculated using the magnitude of the force vector times the magnitude of the
displacement vector times cos π. And of course since the dot product
or scalar product of two vectors is a scalar quantity, then this definition holds
with our understanding of how work is represented. So, given these definitions, weβll
consider a demonstration of how to calculate work done.

A particle moves in a plane in
which π’ and π£ are perpendicular unit vectors. A force π
equals nine π’ plus π£
newtons acts on the particle. The particle moves from the origin
to the point with position vector negative nine π’ plus six π£ meters. Find the work done by the
force.

Remember, the work done by a force
vector π
with displacement vector π is the dot product of π
and π. The force acting on our particle is
nine π’ plus π£ newtons. But what is its displacement? Well, weβre told it moves from the
origin to the point with position vector β letβs call that π© β negative nine π’
plus six π£. This means that the displacement is
simply the vector π¨π©. Thatβs negative nine π’ plus six π£
minus zero π’ plus zero π£. And of course, we simply subtract
the components. In this case, of course, weβre
subtracting the vector zero π’ plus zero π£. So, our original vector remains
unchanged. And so the displacement, which we
can give in meters, is negative nine π’ plus six π£. So, work done is the dot product or
the scalar product of π
and π. Thatβs nine π’ plus π£ dot negative
nine π’ plus six π£.

Then, of course, we find the scalar
product by finding the sum of the products of the components. So, we multiply the π’-components,
nine times negative nine, and then we add the product of the π£-components. Well, thatβs one times six. Thatβs negative 81 plus six, which
is, of course, negative 75. Since the force is in newtons and
the displacement is in meters, the work done is going to be in joules. Now, we notice that the work done
is negative. Well, if the energy of the particle
is conserved, then the kinetic energy of the particle will decrease. If the energy of the particle is
not conserved, the work done might instead increase the potential energy of the
particle. Either way, the work done by the
force in this question is negative 75 joules.

In this question, we considered a
single force acting upon a body. But there will be times where
multiple forces are acting on the same body. In these cases, weβll consider the
resultant of these forces. And that will help us to calculate
the work done. So, in our second example, weβre
going to demonstrate what that looks like.

A body moves in a plane in which π’
and π£ are perpendicular unit vectors. Two forces π
sub one equals nine
π’ minus two π£ newtons and π
sub two equals nine π’ minus seven π£ newtons act on
the body. The particle moves from the point
with position vector negative six π’ plus two π£ meters to the point two π’ plus
three π£ meters. Find the work done by the resultant
of the forces.

We begin by recalling that we can
find the work done by a force vector π
over some displacement vector π by finding
the dot product of these two vectors. We do need to be a bit careful here
though, because weβre working with two forces and we want to find the work done by
their resultant. Now, of course, the resultant of
two forces is simply their sum. So, the resultant force π
is the
sum of π
sub one and π
sub two. In this case, thatβs nine π’ minus
two π£ plus nine π’ minus seven π£. Now, of course, we simply add the
components, so weβll add the π’-components first. Nine plus nine is 18. Then, we add the π£-components. Negative two minus seven is
negative nine. And so, the resultant of the forces
in this case is 18π’ minus nine π£ newtons.

Now, we need to find the
displacement. Now, of course, the particle moves
from the point with position vector negative six π’ plus two π£ to the point two π’
plus three π£. Letβs call the first point π΄ and
the second point π΅. Then, the displacement is the
vector ππ. Well, thatβs the difference between
vector ππ and ππ. So, itβs two π’ plus three π£ minus
negative six π’ plus two π£. As before, we simply subtract the
individual components, beginning with the π’-components. Two minus negative six is eight,
and three minus two is one. So, we find the displacement of the
body is eight π’ plus π£. And of course, thatβs in
meters.

And we have everything we need. The work done is simply the dot
product or the scalar product of these two vectors. Thatβs 18 times eight plus negative
nine times one, which is 144 minus nine, or 135. The work done by the resultant of
the two forces then is 135 joules.

In our next example, weβll have a
look at how to use our understanding of the features of vectors to find work done by
a force.

A particle moved from point π΄
seven, negative three to point π΅ negative nine, two along a straight line under the
action of a force π
of magnitude eight root 10 newtons acting in the same direction
as the vector π equals negative three π’ minus π£. Calculate the work done by the
force, given that the magnitude of the displacement is measured in meters.

Remember, the work done by a force
of vector π
over some displacement vector π is the dot product or scalar product
of those two vectors. In this question, weβre told that
the particle moves from point π΄ to point π΅, so we can calculate the displacement
of the particle fairly easily. In vector form, weβll say that the
displacement π is the vector ππ, and thatβs the vector ππ minus the vector
ππ. Now, since we know the position
from the origin, thatβs the point zero, zero, then the vector ππ is simply
negative nine π’ plus two π£ and the vector ππ is seven π’ minus three π£. So, the vector π is the difference
of these, and we can find that by subtracting the individual components. Negative nine minus seven is
negative 16, and two minus negative three is five π£. So, we have the displacement
vector. Itβs negative 16π’ plus five
π£.

How do we calculate the force
though? We know it has a magnitude of eight
root 10, but we donβt know its vector. We do, however, know it acts in the
same direction as this vector here, negative three π’ minus π£. So, letβs sketch this out. Vector π looks a little something
like this. For every three units left, we move
one unit down. Vector π
acts in the same
direction as this but has a magnitude of eight root 10. Once again, for every three units
left, we still move one unit down. So, we can say that if we define π
sub π as being the π’-component of our force and π
sub π as being the
π£-component, then π
sub π is equal to three times π
sub π.

And, of course, since the
π’-components and π£-components of the vectors are perpendicular to one another, we
can form a right triangle and apply the Pythagorean theorem to find the value of π
sub π and π
sub π. That is, the magnitude of π
sub π
squared plus the magnitude of π
sub π squared is equal to eight root 10 all
squared. Now, of course, π
sub π is three
times π
sub π. So, we can replace the magnitude of
π
sub π with three times the magnitude of π
sub π. Similarly, eight root 10 all
squared is 640. Three squared is nine, so we have a
total of 10 lots of π
sub π squared. Then, we divide through by 10. And finally, weβll take the square
root of both sides. Now, we only need to take the
positive square root of 64 because weβre looking at the magnitude, which is simply a
length. And so, the magnitude of π
sub π
is equal to eight. Since the magnitude of π
sub π is
three times the magnitude of π
sub π, itβs three times eight, which is equal to
24. We know the direction in which our
force is acting. So we can define the vector force
to be negative 24π’ minus eight π£ newtons.

Letβs clear some space and weβre
ready to take the dot product of our force and our displacement. The work done is the dot or scalar
product of negative 24π’ minus eight π£ and negative 16π’ plus five π£. Thatβs negative 24 times negative
16 plus negative eight times five, and thatβs equal to 344. The work done is in joules. So, we find the work done by our
force is 344 joules.

In our previous examples, weβve
looked at finding the work done when both displacement and force are vectors, where
each component is simply a number. In our final example, weβll
demonstrate how work done can also be calculated when the vectors for displacement
and/or force are given as expressions in terms of time.

A particle moves in a plane in
which π’ and π£ are perpendicular unit vectors. Its displacement from the origin at
time π‘ seconds is given by π« equals two π‘ squared plus seven π’ plus π‘ plus
seven π£ meters. And it is acted on by a force π
equals six π’ plus three π£ newtons. How much work does the force do
between π‘ equals two seconds and π‘ equals three seconds?

Since work done is the dot product
of the vector force and the vector for displacement, weβre going to need to
calculate the displacement of this particle. Weβre given its displacement at
some time π‘. But weβre also asked how much work
the force does between π‘ equals two seconds and π‘ equals three seconds. So, letβs substitute π‘ equals two
and π‘ equals three into our expressions for the displacement. Beginning with π‘ equals three
seconds, we get two times three squared plus seven π’ plus three plus seven π£. Thatβs 25π’ plus 10π£. And weβre working in meters. So the displacement from the origin
and, hence, the position of the particle at three seconds is 25π’ plus 10π£
meters. In a similar way, we can find the
position of the particle at two seconds by substituting π‘ equals two into our
expression. And we get 15π’ plus nine π£
meters.

The displacement then of the
particle between these two times is the difference here. Itβs π sub three minus π sub
two. Thatβs 25π’ plus 10π£ minus 15π’
plus nine π£. And of course, we simply subtract
the individual components. 25 minus 15 is 10, and 10 minus
nine is one. So, the displacement is simply 10π’
plus π£, and thatβs in meters. Since the work done is the dot
product of the force and displacement, we need to find the dot product of six π’
plus three π£ and the displacement we just calculated. Itβs the dot or scalar product of
six π’ plus three π£ and 10π’ plus π£. Thatβs six times 10 plus three
times one, which is, of course, equal to 63. The work done then is equal to 63
joules.

Weβll now recap the key points from
this lesson. In this lesson, we learned the work
done by a constant force vector π
over a displacement vector π is equal to the dot
product of π
and π. Alternatively, if π is the angle
between π
and π, itβs the magnitude of the force times the magnitude of the
displacement times cos π. And if we express π
and π in
component form, we donβt actually need to use this second form. We can simplify the dot product to
find the work done.