Let's go ahead a try a one dimensional problem for Newton's second law.
Suppose I asked you if you accelerate a big 1.000 kilogram spaceship
but you only applied one Newton of force,
which by the way is approximately the weight of an apple.
Ironically enough, it's a very small force,
and that object, the spaceship is starting from rest in space,
starting from rest meaning it has no initial velocity.
We put it in space just so we know
there's no gravity or anything else that we have to worry about,
just the force on an object.
The question is, how far will the spaceship go in 30 days?
So in about a month's time was a tiny amount of force acting on it.
So first pause, see if you can solve this,
see if you can find out how far the spaceship has travelled
in 30 days with a tiny, tiny force acting on it.
Now if you solve this problem,
you can see an example of how to do so here.
So we're gonna run through this very quickly
with an object moving under the influenced
of a very tiny force of one Newton.
The first thing that we wanna do
is recognize that if we're going to do
any equation of motion type analysis,
then we could write down an equations of motion.
For example, this one for the motion of an object
under some acceleration,
the problem of course in this problem
is that we're not given an acceleration
so if we look at this equations of motion,
the distance, the final position minus the initial position
which will be exactly this distance here
where you see X and X0,
is equal to the initial velocity times the time
where the initial velocity is zero,
so we can carry with that term
plus one half the acceleration times the time square.
So we are already, almost done with this problem.
We need to put in the time which we know is 30 days.
We also need to put in the acceleration so we have to find that one still.
The time is not so bad.
We have 30 days and the first thing we have to do here
is some unit analysis because we do want all of our units
to be in our standard units, our kilograms,
our meters and our seconds,
and so we need to get days converted into seconds,
so this is a great practice in unit analysis
which I recommend you try doing first
before you see me do it here.
To convert the units of days into seconds,
we will keep multiplying by what our essential units of one.
In other words, there are 24 hours in one day,
so this term here is actually just the number one
because 24 hours and one day are the same thing,
so I'm not changing the number 30 days.
This is the way to convert it to units by not changing the number
but multiplying it by different sorts-of the number one.
So multiply by 24 hours per one day.
We can multiply by 60 minutes in one hour
and then we can multiply by 60 seconds in one minute.
And the reason this works so well is that I cancel days,
I cancel my units of hours, I cancel my units of minutes
so that if I take 30 times 24 times 60 times 60,
I will have my time in units of seconds.
So if we do this and you can do this on your own
using a calculator or anything else we have,
something that is approximately, 2.6 times 10 to the 6th seconds.
So that's a lot of seconds,
so eventhough we're applying a small force,
we do get a long time over which this force is applied
so maybe we will get this spaceship to move pretty well after all.
So let's see, the second thing we have to do is find the acceleration.
Here's what we're going to need to apply Newton's second law,
we know that F the force on our object
which is one Newton is equal to the mass of the object
times its acceleration, so we can find the acceleration quite easily.
We just take the force and divide by the mass.
In our case, our force is one Newton
and our mass is 1.000 kilograms.
Remember our units of Newtons,
what we have is kilograms meters per second squared
divided by kilograms so this is equal to one divided by 1.000
which is 10 to the minus 3 using our scientific notation,
meters per second squared which are the units for acceleration.
So now that we have these two things,
our acceleration from this sort of analysis
and our time from this unit analysis,
we can plug in and find the distance
of the change in position of our object.
Doing that we now have the change in position, X minus X0,
where the initial position is zero
equals 1/2 times the acceleration 10 to the minus 3,
times the time which is this complicated expression
2.6 times 10 to the 6th seconds squared.
So we have to do one last thing here
which is to square this quantity 2.6 times 10 to the 6th
so this is 1/2 times 10 to the minus 3
and 2.6 when you square it, is approximately 6.760
but as I've already discussed,
sometimes you're going to want to just make
some approximations on your own,
to work through a problem very quickly.
So for example in this problem,
we could've taken this 2.6 and just try to put in 3
because we know very quickly 3 squared is 9.
So just by doing something very quickly and simply
we can sometimes simplify our analysis
and find out what the correct answer, in an exam setting
especially could be without having to deal
with any tricky numbers like this ones.
So the 6.760 times, and then we have 10 to the 6th squared
which will be 10 to the 12th since we multiplied these exponents.
Now we do what we've always done with our unit analysis,
which is to treat our coefficients as we call them,
the numbers in front, on their own,
so we have 6.760 and then divide it by 2
and then we treat the powers of 10 on their own,
so 10 to the 12th times 10 to the minus 3,
subtract 3 from 12 giving us times 10 to the 9th
then all we have to do is simplify this slightly
and get 6.760 divided by 2 which is about 3.380
times 10 to the 9th power, and this is in meters.
So 3 times 10 to the 9th, 10 to the 9th
or changing our units, it'll be 3.380
times 10 to the 6th kilometers if we wanted to write it in kilometers.
Either one of these will be fine
so be careful in an exam setting to find out
which options have meters and which ones have kilometers
and make sure you get your powers of 10 correct.
And so we see just from this simple example
that even a tiny force if applied over a long enough period of time,
these accelerations will build up more and more and more
until you in fact get great distances from very, very small forces.
The equations of motion which we just used in this problem,
are often referred to simply as kinematics, how things move.
On the other hand, forces
and we have things under the influence of different forces.
We call dynamics because things are dynamic, they're moving.
In the problem we just solved,
we saw that the link between these two
and the way to solve each one was acceleration.
We want something to do with the equations of motion.
We didn't know the accelerations,
so we appeal to Newton's second law
F equals ma to find what the acceleration was
and to plug that acceleration into the equations of motion.
In some problems, you will know the forces
and be able to solve for the acceleration
as we just did and put that into the equations of motion.
In other problems it will be exactly the opposite.
You will know how something move
or be told how something moves
and you will need to deduce the forces on those objects
based on the equations of motion going to the dynamics,
using again the link of acceleration.
So be aware that this is very common for problems
that you are able to find the acceleration
and use the acceleration to solve for the other,
either the dynamics or the kinematics
the equations of motion or Newton's second law, using the acceleration.