### Video Transcript

In this video, we’ll learn how to
use Newton’s second law of motion to solve problems involving particles with
constant mass under the action of constant force. You’ve probably heard of Newton’s
first law of motion. That is, an object will remain at
rest or in uniform motion in a straight line unless acted upon by an external
force. This law does not stand alone,
however. In fact, it’s the first of three
laws that underpin classical mechanics as we know it today.

We’re interested in the second one,
so let’s find out what that is. Unlike the first law of motion,
which is interested in what happens when the forces are balanced, that is, the net
sum of the forces acting upon a body is zero, this one tells us what will happen
when these forces are not balanced. And it’s a bit of a mouthful.

Formally, it states that the
acceleration of an object as produced by a net force is directly proportional to the
magnitude of the net force in the same direction as that net force and inversely
proportional to the mass of the object. But what does that actually
mean?

Well, imagine you’re pushing a
wheelbarrow with a load of stones in it. The more force you apply to the
wheelbarrow in the form of pushing it, the more acceleration will occur. Your movement will speed up more
quickly. Double the force, and the
acceleration will double. The acceleration and the net force
are directly proportional to one another. Take some of the stones out of the
load, however, and the mass will decrease. And you’ll need to apply less force
to maintain that acceleration.

Now, in fact, we can simplify all
of this massively. We define 𝑎 to be equal to the
acceleration, the net force is equal to 𝐹, and the mass of the object is 𝑚. For the acceleration to be directly
proportional to the force and inversely proportional to the mass, we can say that 𝑎
must be equal to 𝐹 over 𝑚. We want to make 𝐹 the subject
really. So we’re going to multiply both
sides of this equation by 𝑚. And we get 𝑚𝑎 equals 𝐹 or 𝐹
equals 𝑚𝑎. And this is the formula we tend to
use, and it’s certainly the one you’ll want to learn.

Note that we tend to measure forces
in newtons. And for this to be the case, the
mass will be in kilograms and the acceleration in meters per square second. Now that we have the rule in some
context, let’s have a look at how we might apply it.

A constant force acted on a body of
mass nine kilograms such that its speed changed from 58 kilometers per hour to 66
kilometers per hour in half a second. Calculate the magnitude of the
force.

In order to link the motion of a
body and its mass and the force acting on that body, we need to recall Newton’s
second law of motion. This tells us that the net force on
an object is equal to the mass of that object multiplied by its acceleration. 𝐹 equals 𝑚𝑎. Now, we have the mass of the
object. It’s nine kilograms. And so to be able to calculate the
force, we’re going to need to first calculate the acceleration.

Now, of course, we’ve been given
the speed of the body. And so acceleration is change in
speed divided by time. To be able to calculate forces in
newtons though, we need the units for speed to be meters per second so that the unit
for acceleration is meters per square second. So let’s begin with 58 kilometers
per hour. This tells us that every one hour
the body must travel 58 kilometers. There are 1,000 meters in one
kilometer, so we multiply the numerator by 1,000. And we see that the body must
travel 58,000 meters in one hour.

We then know that there are 60
minutes in one hour and 60 seconds in a minute. So that must mean that there are
3,600 seconds in an hour. So we multiply the denominator by
3,600. And we find that the body will
travel 58,000 meters in 3,600 seconds. 58,000 divided by 3,600 is 145 over
nine. So the initial speed of the body is
145 over nine meters per second. In a similar way, we convert 66
kilometers per hour into meters per hour. That’s 66,000. And then we find that it travels
66,000 meters in 3,600 seconds. This fraction simplifies to 55 over
three. And so the end speed of the body
must be 55 over three meters per second.

Acceleration then in meters per
square second is 55 over three minus 145 over nine. That’s the change in speed. And then we divide that by one-half
since it changes the speed in half a second. That gives us 40 over nine. So we found the acceleration of the
body to be 40 over nine meters per square second. And that’s great. We have the acceleration and we
know the mass of the body. So we can now use the formula 𝐹
equals 𝑚𝑎.

We saw that the mass is nine
kilograms, so mass times acceleration is nine times 40 over nine. But of course, these nines are
going to cancel. And so the force is 40. We’ve been working in meters per
square second and kilograms, so the units for our force are newtons and the
magnitude or the size of the force acting on the body is 40 newtons.

In our next question, we’ll look at
how we can use Newton’s second law to calculate the mass of a body.

The diagram shows a body of mass 𝑚
kilograms, which moves with a constant acceleration of 0.8 meters per square second
under the action of three vertical forces. Given that the weight of the body
is 𝑤 and the forces are measured in newtons, find 𝑚. Use 𝑔 equals 9.8 meters per square
second.

We’re trying to find the value of
𝑚, but we’ve got an extra unknown 𝑤. And we’ll see how they link in a
moment. To begin with, we’re just going to
work out the weight of the body. Weight is a force, and so we’re
going to use Newton’s second law of motion. Net force is equal to mass times
acceleration. And so we’re going to find the net
force acting on this body and set that equal to mass times acceleration. And to do so, we need to find the
positive direction. Let’s define downwards to be
positive. Then we can say in the positive
direction, we have 𝑤. In the negative direction, we have
10 and 71. So the net force in our system is
𝑤 minus 10 minus 71, which is 𝑤 minus 81. Then that’s equal to mass times
acceleration.

Notice that the acceleration is
acting in the positive direction. So 𝑚𝑎 becomes 𝑚 times 0.8. And so we can say that 𝑤 minus 81
is 0.8𝑚. But what is 𝑤 and how does it link
with the mass 𝑚? Well, 𝑤 is known as the weight of
the body. And somewhat counterintuitively,
that is in fact a force. In fact, as a direct result of
Newton’s second law, weight is known as the mass of the object times its
acceleration due to gravity. So weight is 𝑚𝑔.

We can therefore rewrite our
equation as 𝑚𝑔 minus 81 equals 0.8𝑚. Remember, we’re trying to find the
value of 𝑚, so we need to make 𝑚 the subject. Let’s subtract 0.8𝑚 from both
sides. That gives us 𝑚𝑔 minus 81 minus
0.8𝑚 equals zero. Next, we add 81, and we then notice
that on the left-hand side of our equation we can factor 𝑚. So our equation becomes 𝑚 times 𝑔
minus 0.8 equals 81. But of course, 𝑔 is equal to
9.8. So 𝑔 minus 0.8 is 9.8 minus 0.8,
and 9.8 minus 0.8 is just nine. So we get nine 𝑚 equals 81. And to solve our equation for 𝑚,
we’re going to divide both sides by nine. 81 divided by nine is nine. And so the value of 𝑚 is nine. The body has a mass of nine
kilograms.

Up until this point, we’ve
considered forces acting in just one direction. It’s important to realize that we
can perform this process for forces acting perpendicular to one another, for
instance, where the force is a vector quantity. Let’s see what that might look
like.

A body of mass 478 grams has an
acceleration of negative four 𝐢 plus three 𝐣 meters per square second, where 𝐢
and 𝐣 are perpendicular unit vectors. What is the magnitude of the force
acting on the body?

Remember, to link mass,
acceleration, and net force, we can use Newton’s second law of motion. This says that force is equal to
mass times acceleration. Now, mass will always be a scalar
quantity. It will just have a magnitude. But force and acceleration can be
vector quantities. They will have a magnitude and a
direction. And so we can write this as the
vector sum of 𝐹 is equal to mass times the vector acceleration.

Now, since we are working in meters
per square second, we can’t use this formula until we convert from grams to
kilograms. We know that there are 1,000 grams
in a kilogram, and so we divide 478 by 1,000. And that tells us that it’s
equivalent to 0.478 kilograms. The vector sum of the forces acting
on the body then is equal to this mass times the vector acceleration. It’s 0.478 multiplied by negative
four 𝐢 plus three 𝐣.

We can of course distribute this
scalar quantity across our vector. 0.478 multiplied by negative four
is negative 1.912. And the 𝐣-component is 1.434. So we have the vector force, but we
need to find the magnitude. The magnitude of the vector is
found by finding the square root of the sum of the squares of each of its
components. So that’s the square root of
negative 1.912 squared plus 1.434 squared. That’s equal to 2.39. And so the magnitude of the force
that acts upon this body is 2.39 newtons.

Now, it is also worth noting that
we could have simply found the magnitude of the acceleration first and then
multiplied that by the mass. We would have achieved the same
result.

In our final example, we’ll look at
how we can use Newton’s second law of motion and the equations of motion to find a
resistive force.

A bullet of mass 63 grams was fired
toward a fixed barrier at 80 meters per second. Given that it penetrated five
centimeters into the barrier before it stopped, find the resistance of the barrier
to the bullet’s motion.

Let’s begin by sketching a
diagram. The bullet is traveling towards the
barrier at 80 meters per second. It then hits the barrier and
travels five centimeters in before it stops. We want to find the resistance of
the barrier to the bullet’s motion.

Now, of course, this resistance is
a force, and it’s acting in the opposite direction to that which the bullet is
traveling. We also know the mass of the body,
and we know that we can link mass and force by using Newton’s second law of
motion. That is, force is equal to mass
times acceleration. We’re trying to find the resistance
force, and we know the mass. So we’re going to need to work out
the acceleration of the body. To do so, we can use the laws of
constant acceleration, sometimes known as SUVAT equations.

We know the starting speed of the
bullet. It was traveling at 80 meters per
second on the moment of impact. It stops a distance five
centimeters in, and so its final speed was zero. It traveled five centimeters, but
of course we’re working in meters, so let’s convert this. Five divided by 100 is 0.05. And so it travels 0.05 meters
before stopping. We’re trying to find the
acceleration, so let’s let that be equal to 𝑎.

The equation that links these
unknowns is 𝑣 squared equals 𝑣 naught squared plus two 𝑎𝑠. So let’s substitute everything into
this equation and solve for 𝑎. That gives us zero squared equals
80 squared plus two times 𝑎 times 0.05. The right-hand side simplifies to
6,400 plus 0.1𝑎. We’re going to solve for 𝑎 by
subtracting 6,400 from both sides of our equation. And we get negative 6,400 equals
0.1𝑎. Then we divide by 0.1, remembering
that that’s the same as multiplying by 10. And we get our acceleration to be
negative 64,000, and that’s in meters per square second.

We now have everything we need in
order to be able to calculate the resistance force of the barrier. Before we can though, we are going
to convert the mass from grams into kilograms. And we do so by dividing by
1,000. So 63 grams is the same as 0.063
kilograms. The net force is negative 𝑅 since
that’s happening in the opposite direction to that which we’re modeling as
positive. That’s equal to mass times
acceleration, which is 0.063 times negative 64,000. That gives us negative 𝑅 equals
negative 4,032. And so 𝑅, the resistance of the
barrier to the bullet’s motion, must be 4,032 newtons.

We’ll now recap the key points from
this lesson. In this video, we learned that
Newton’s second law links force, mass, and acceleration. The net force is equal to the mass
of the object multiplied by the acceleration. So it’s 𝐹 equals 𝑚𝑎. And whilst mass is very much a
scalar quantity, force and acceleration can be vectors. So we can rewrite this as the
vector sum of 𝐅 is equal to mass times a vector acceleration. And of course, we’ve applied this
formula when the mass of the object is constant.