### Video Transcript

In this lesson, we’ll learn how to
calculate the kinetic energy of a moving particle of mass 𝑚 that moves with a
velocity of magnitude 𝑣.

An object that is moving is said to
have kinetic energy. The faster that object moves, and
the more kinetic energy it has. It’s worth noting that the kinetic
energy can be increased or decreased by doing work on the object. If a force acts on an object as
that object moves along the path, we say the force does work. And if that force causes the object
to accelerate, then the speed of the object increases. So, its kinetic energy will also
increase.

Similarly, if the force that acts
on the object acts in the opposite direction to its motion, the object will
decelerate. The speed of the object will
decrease and so will its kinetic energy. Imagine, for instance, throwing a
bowling ball down the lane of a bowling alley. When you throw the ball, you exert
a force on it, that force causes the ball to accelerate. Hence, it does work on the ball and
increases its kinetic energy.

So, what is the formula we use to
calculate kinetic energy? Suppose we have an object with mass
𝑚 moving at velocity vector 𝐯, where the magnitude or algebraic measure of this
vector is simply 𝑣. Then, its kinetic energy, which can
be denoted using the letter 𝑇, is a half 𝑚 times the magnitude of 𝐯 squared, or
simply a half 𝑚𝑣 squared. Since the magnitude of the velocity
squared is equivalent to the dot product of the velocity vector with itself, then
the kinetic energy can be alternatively expressed as a half 𝑚 times 𝐯 dot 𝐯.

And from this definition, it’s
quite clear that the kinetic energy of a particle is a nonnegative scalar
quantity. And if the velocity is zero, then
the kinetic energy is also zero. According to the definition, the
kinetic energy of a particle may change from one instant to another during the
particle’s motion in regard to the magnitude of its velocity. So, with this in mind, let’s
demonstrate how to calculate the kinetic energy of a body whose velocity is given as
a vector quantity.

A body of mass 500 grams is
moving at a constant velocity 𝐯 equals two 𝐢 minus three 𝐣 centimeters per
second, where 𝐢 and 𝐣 are two perpendicular unit vectors. Find its kinetic energy.

Let’s begin by reminding
ourselves of the formula for kinetic energy of a body whose mass is 𝑚 and whose
velocity vector is 𝐯. The kinetic energy of this body
is given by 𝑇 equals a half 𝑚 times 𝐯 dot 𝐯. Now, we generally tend to
calculate kinetic energy in joules. In these cases, the mass of the
object is given in kilograms, whilst its speed or velocity would be given in
meters per second.

Here, we’ve been told that the
mass 𝑚 is 500 grams, whilst the velocity is in centimeters per second; it’s two
𝐢 minus three 𝐣. And that’s absolutely fine. What it does mean is that the
kinetic energy of the body will be in ergs. With that in mind, let’s
substitute everything we know about this body into the formula. When we do, we find that
kinetic energy is a half times 500 times the dot product of two 𝐢 minus three
𝐣 with itself. So, what is the dot product of
this pair of vectors?

In fact, finding the dot
product of a vector with itself is actually equivalent to finding the square of
its magnitude. Here, it’s two times two. And then, we find the sum of
negative three times negative three. Two times two is four, and
negative three times negative three is positive nine. Four plus nine is 13. And so, we find the dot product
of 𝐯 with itself is equal to 13.

Hence, the kinetic energy of
this body is given by a half times 500 times 13, and that’s 3250. And remember, we said the units
here were ergs. So, the kinetic energy of the
body is 3250 ergs. Now, had we converted our
original values into kilograms and meters per second or considered a conversion
directly from ergs to joules, we could equivalently have calculated the kinetic
energy of the object to be 3.25 times 10 to the power of negative four
joules.

Let’s now demonstrate how to find
the kinetic energy of a particle when given its speed, in other words, the magnitude
of its velocity rather than its velocity in vector form.

A body of mass 940 grams is
moving at four meters per second. Determine its kinetic
energy.

Remember, the kinetic energy of
the body is linked to both its mass and velocity. Specifically, let’s assume we
have a body with mass 𝑚 and the magnitude of its velocity, in other words, its
speed, is 𝑣. In this case, the kinetic
energy is a half 𝑚𝑣 squared. And before we move any further,
let’s investigate in a little more detail the units in this question.

The body has a mass given in
grams, whilst the speed is given in meters per second. What we’ll do is convert the
mass into the SI unit kilograms. We do so by dividing by
1000. So, the mass of the object is
in fact 0.94 kilograms. If we’re working in kilograms
and meters per second, we can then give the SI unit joules for our energy. So, let’s define 𝑚 to be 0.94
kilograms and 𝑣 to be four meters per second.

Then, we can substitute
everything we know about this body into the formula for kinetic energy. We get a half times 0.94 times
four squared, in other words, a half times 0.94 times 16. And that’s 7.52. Now, of course, the units here
are in fact kilogram square meters per square second. And that’s the same, as we
said, as the standard unit for energy joules, which has the symbol J. So, the kinetic energy of this
object is 7.52 joules.

Now, it follows that whilst we can
use this equation quite easily to calculate the kinetic energy, we can also use it
to calculate the speed or mass of an object given its kinetic energy. And if the quantity is a scalar,
like in this question, it’s simply a matter of substituting and solving the
resulting equation. We’re going to demonstrate what we
would do if presented with vector quantities.

A body is moving at a constant
velocity 𝐯 equals 250𝐢 minus 250𝐣 centimeters per second, where 𝐢 and 𝐣 are
two perpendicular unit vectors. Given that the kinetic energy
of the body is 4.8 joules, find the mass of the body.

Suppose we have a general body
with mass 𝑚 and velocity vector 𝐯. The kinetic energy, which we
can denote 𝑇, is given by a half 𝑚 times the dot product of 𝐯 with
itself. This is in fact equivalent to
finding a half 𝑚 times the magnitude of 𝐯 squared. So, let’s substitute what we
know about our vector into this formula.

We have 𝐯 is 250𝐢 minus 250𝐣
centimeters per second, and the kinetic energy is 4.8 joules. But in fact if we’re working in
joules, we want to work with meters per second. So, we’ll divide each component
in the vector for 𝐯 by 100 to achieve this. Then, the velocity vector is
2.5𝐢 minus 2.5𝐣 meters per second. And now that we’re working in
SI units, we know that the mass will be in kilograms. Then, substituting everything
we know into the formula, we get 4.8 equals a half 𝑚 times the square of the
magnitude of 𝐯.

Now the magnitude of a vector
is the square root of the sum of the squares of its components. So, the magnitude of 𝐯 here is
root 2.5 squared plus negative 2.5 squared. And if we square a square root,
we just get the expression inside the square root. Inside our root, we have 2.5
squared plus negative 2.5 squared, which is 12.5. And so, we have 4.8 is a half
𝑚 times 12.5. To solve for 𝑚, we’ll multiply
by two and divide by 12.5. That gives us 0.768. And so, we find the mass of the
body is 0.768 kilograms, which is equivalent, of course, to 768 grams.

Now, in practice, we might actually
encounter systems involving more than one moving object. When working with such systems,
it’s actually helpful to instead use relative velocities. And in this case, the formula for
kinetic energy behaves as expected. In certain cases, it may even cut
down on calculations, as finding the kinetic energy of one moving body relative to
another moving body can give a clearer picture of the energy that can be transferred
if two objects were to collide. Let’s have a look at one such
example.

A cannon fired a shell of mass
16 kilograms at 285 meters per second toward a tank that was moving at 72
kilometers per hour in a straight line directly toward the cannon. Determine the kinetic energy of
the shell relative to the motion of the tank.

Remember when we’re calculating
kinetic energy, we’re interested in the mass and velocity of our object. In this case though, we’re
going to have to begin by finding the relative speed of the shell. The shell and the tank are
moving toward each other. In this case then, the
magnitude of the relative velocity will actually be equal to the sum of the
magnitudes of their velocities. But of course, the velocities
here are given in different units. They are meters per second and
kilometers per hour. Let’s convert the motion of the
tank, 72 kilometers per hour, into meters per second.

Since there are 1000 meters in
a kilometer, we can begin by simply multiplying by 1000. And we find that the speed of
the tank is 72000 meters per hour. Then, we can divide by 60 times
60, and that will convert from meters per hour into meters per second. 60 times 60 is 3600. So, this is equivalent to 20
meters per second. And so, we’ve determined that
the relative speed of the shell to the tank must be 285 plus 20, which is 305
meters per second.

With that in mind, we can just
use the kinetic energy formula for the magnitude of the velocity to find the
kinetic energy of the shell relative to the tank. Suppose we have an object with
mass 𝑚 whose speed, in other words, the magnitude of its velocity, is 𝑣. Then, the kinetic energy,
denoted 𝑇, is a half 𝑚𝑣 squared.

Now, of course, we’re finding
the kinetic energy of the shell relative to the tank. So, we need the mass of that
shell; it’s 16 kilograms. The relative speed is 305
meters per second. Now, since we’re working in
kilograms and meters per second, the energy is in fact going to be in
joules. The kinetic energy is a half
times 16 times 305 squared. This gives us a value of
744200. And so, we’ve calculated the
kinetic energy of the shell relative to the motion of the tank. It’s 744200 joules.

Now that we’ve demonstrated the
various formulae that can be used to calculate the kinetic energy of a moving object
and considered that in a number of examples, let’s look at the key concepts from
this lesson.

In this lesson, we learned that
kinetic energy is a nonnegative scalar quantity. And it vanishes — it’s zero — if
the velocity of the object is zero. We saw that an object with mass 𝑚
and velocity vector 𝐯 has kinetic energy a half 𝑚 times the magnitude of the
velocity squared. And we saw that this can
equivalently be represented as a half 𝑚 times the dot product of 𝐯 with
itself.

We also saw that if we’re working
purely with scalar quantities where the magnitude of the velocity is simply 𝑣, then
the kinetic energy is a half 𝑚𝑣 squared. And finally, we saw that given the
kinetic energy and velocity of an object, we can find its mass. But given the kinetic energy and
mass, we can find its speed. However, we cannot find the
velocity since we only are able to calculate scalar quantities and we can’t obtain
the direction of motion.