Newton's Laws Problems (part 1)

Examples of exercises using Newton's laws.

Welcome back.

Now that we've hopefully,
learned a little bit about

Newton's law, let's apply them
to solve some problems.

Let's say I have a-- I don't
know-- some kind of vehicle, a

car, a motorcycle
or something.

And let's say its mass is 500
grams. And let's say that I

can accelerate this vehicle at
an acceleration of-- I don't

know-- 3 centimeters
per second squared.

What's the force that I need
to apply to this mass to

accelerate it at this speed?

And we want the answer
in Newton's.

So what's the force?

So we're just going to use
Newton's second law and

Newton's second law tells us
force is equal to mass times

acceleration.

So you might want to, or you
might be tempted just to

multiply mass times acceleration
and you'd get

force is equal to-- let's see,
500 grams times 3 centimeters

per second squared.

And you would get force
is equal to 1,500 gram

centimeters per second
squared.

And if you did this,
you would be right.

Although your answer would not
be in Newtons and now you

would have to somehow,
try to convert this

set of units to Newtons.

And what are Newtons?

Well we learned when we did
Newton's laws that a Newton,

so 1 Newton is equal
to 1 kilogram

meter per second squared.

So somehow we have to convert
the gram to kilograms and we

have to convert the centimeters
to meters.

We could do it after the fact
here, or what I find it easier

to do is actually to convert the
mass and the acceleration

units first and then just
do the F equals ma.

So what's 500 grams
in kilograms?

Well 500 grams is half-- well,
a kilogram as a thousand

grams. So 500 is going to
be half a kilogram.

1 kilogram is a thousand grams,
so 0.5 kilograms is 500

grams.

Similarly, 3 centimeters
is how many meters?

Well, 1 meter is 300-- sorry.

I think I'm dehydrated.

1 meter is a hundred
centimeters, right?

So 3 centimeters is 0.03 meters
per second squared.

Hopefully this make
sense to you.

3 centimeters is 0.03 meters.

And now we already.

We have our mass in kilograms
and we have our acceleration

in meters per second squared.

And if this is confusing, you
should watch the unit videos

because this is all I'm doing;
I'm just doing unit

conversion.

So let's go back to force
equals mass times

acceleration.

So the force was equal to
0.5 kilograms times the

acceleration, which is 0.03
meters per second squared.

And this equals--

What's 0.5 times 0.03?

I'll do it down here just
because multiplying decimals

seems to be a problem for a
lot of people, including

myself, many times.

So what you do, you just
multiply the numbers.

5 times 03 or 5 times 3 is 15.

And then, how many points do
we have behind the decimal?

How many digits behind
the decimal?

Let's see.

We have 1, 2, 3.

So 1, 2, 3.

We have to add the 0 because
we need three spaces behind

the decimal point.

So we get the force is equal to
0.015 kilogram meters per

second squared.

And this is a Newton.

So the force is equal
to 0.015 Newtons.

Let's do another problem.

And this one's going to be-- and
actually, I think you'll

find most of the difficult
Newton's laws problems or

force problems, they're just
some combination of making

sure you get the units right
when we're talking about in

one dimension.

The difficult part is usually
getting the units right or

just the math, just
the algebra.

So if you have trouble with this
it's usually because you

have to just brush up a little
bit on the algebra.

The physics itself is just
force equals mass times

acceleration as we will
see in this problem.

So let's say that when I apply
some force, some particular

force, I use this
little 0 here.

So I call that force F sub 0.

So this means a particular
force.

This is some value.

When I apply that force to some
mass, let's call that m

sub 0, I get some
acceleration.

I get acceleration a sub 0.

We could've put numbers here.

We could've said, well, if I
apply a force of 10 Newtons to

a mass of-- I don't know--
to a mass of let's say 2

kilograms, I have an
acceleration of 5 meters per

second squared.

But I'm just doing this because
this could be any

relationship.

And let's say the problem
tells us that when I put

another mass with this first
mass, so let's say that, you

know-- let me draw
this diagram.

So here's my mass, m sub 0.

When I apply a force of f sub 0
to it, I get an acceleration

of a sub 0.

Now the problem tell us when I
add another mass-- so let's

says I stack it up and
we're like in an ice

skating ring or something.

And I stack another mass
up here, and let's

call this mass m1.

When I stack another mass on
here-- so let me redraw it

actually down here.

Because it's a different case.

And I apply the same force,
and now I have

this new mass on here.

I'll do it in red.

This is m1.

The problem tells us that my new
acceleration is 1/5 of the

original acceleration.

So it's 1/5 of whatever
this was.

So it's 1/5 a sub 0.

So the question is, what
is the ratio of m

sub 0 to m sub 1?

So m sub 0 to m sub 1
is equal to what?

And I'm going to keep it in
abstract variables just to

confuse you.

And then I'll show you that you
can actually substitute

numbers and the problem becomes
a little easier.

And you might want to pause it
and try it for yourself.

So let's work it through.

So we know we have
this relationship

to start off with.

And just for simplicity, let's
write what m sub 0 is in terms

of F and a.

So we just divide both sides by
a sub 0 and you get F sub 0

divided by a sub 0 is
equal to m sub 0.

Good.

So let's just put that
aside for a second.

And let's do that same
relationship here with this.

So here, this relationship tells
us that F sub 0 is equal

to m1 plus m0 times this
new acceleration, which

is 1/5 a sub 0.

And so, if we divide both sides
by this term right here,

we get-- dividing by 1/5
is the same thing as

multiplying by 5.

So you get 5 F sub 0 over a sub
0 is equal to m1 plus m0.

I just divided both sides
by this term right here.

Well, what's F sub 0?

What's this?

Let me switch colors again.

What's F sub 0 divided
by a sub 0?

Well it's here's.

It's what we solved for
in the beginning.

We just got it from
this relationship.

So we could substitute.

5 times F sub 0 divided
by a sub 0 is the same

thing as 5 times m0.

Draw a line here.

So we have a new relationship.

5m0 is equal to m1 plus m0.

All I did is I substituted
this for

this, or this for that.

And I used this relationship,
which we got in the beginning

to do that.

And now what do we have?

We have 5 m sub 0 is equal
to m1 plus m0.

We could subtract m0
from both sides.

You get 4 m0 is equal to m1.

You could divide both sides by
m0 and you got 4 is equal to

m1 over m0.

And you could invert this
relationship and you can get

m0 over m sub 1 is
equal to 1/4.

So what we learned is the ratio
of the old mass to the

new mass is 1 to 4.

And that's a problem.

And actually, I will leave it
for you as an exercise to

figure out-- to just
do the same

problem using the numbers.

I will do that actually, in the
next video just to show

you that that actually
would've been a

simpler way to do it.

But it's good to get used to
this just so you can solve

things in general terms. I'll
see you in the next video.