### Video Transcript

A frog of mass 30 grams jumping
into the air from a resting position accelerates to a speed of 12 centimeters
per second in 0.025 seconds. How many newtons of force did
the frog’s legs exert in the jump?

Okay, so here’s our frog in its
initial resting position. And we’ve been told that it has
a mass of 30 grams. And as well as this, it’s just
about to jump. In fact, here he is a few
moments later, having just jumped. We’ve been told that, in this
position, the frog is moving at a speed of 12 centimeters per second. So since the frog jumped from
here to here, let’s say that the frog is moving in this direction at 12
centimeters per second. The direction at the moment is
arbitrary. But it is important that we
label it just for the sake of drawing a complete diagram.

Now, as well as this, we’ve
been told that the time taken between the frog being at rest and the frog
traveling at 12 centimeters per second is 0.025 seconds. So that’s how long it took for
the frog to accelerate from zero meters per second here to traveling at 12
meters per second here.

We’ve been asked to find how
many newtons of force the frog’s legs exerted in this jump. Now this is a really useful
sentence because not only is it telling us that we need to find the force
exerted by the frog’s legs, but we also need to find it in newtons. And, as we know, newtons is the
base unit of force. So regardless of whatever
equations we have to use to find the force, if we want to find it in its base
unit, we need to convert everything else to base units as well.

So the mass of the frog given
to us as 30 grams needs to be converted to kilograms. And similarly, the speed of 12
centimeters per second needs to be converted to meters per second. However, the 0.025 seconds are
already in their own base unit, seconds. So we don’t need to worry about
that one. So let’s focus on converting
the mass to kilograms and the speed to meters per second. we can start by
recalling that one gram is equivalent to one thousandth of a kilogram. And therefore, 30 grams is
equivalent to thirty thousandths of a kilogram, or in other words 0.03
kilograms. So let’s replace our label here
on the diagram with 0.03 kilograms.

Then we can focus on converting
the speed to meters per second. To do this, we can first recall
that one centimeter is equivalent to one hundredth of a meter. And therefore, if we divide
both sides of the equation by the unit seconds, we can see that one centimeter
per second is equivalent to one hundredth meters per second. Then we multiply both sides of
the equation by 12 to see that 12 centimeters per second is equivalent to twelve
one hundredth meters per second, or in other words 0.12 meters per second. Therefore, on our diagram, we
can replace the final speed of the frog with 0.12 meters per second.

Now at this point, we’ve
converted all of the quantities we’ve been given into their base units. So let’s now actually look at
finding the number of newtons of force exerted by the frog’s legs to get it from
here to here and moving at 0.12 meters per second. We can recall that the force
exerted on an object is given by finding the change in momentum of that object,
Δ𝑝, divided by the amount of time taken for that change in momentum to
occur. We’ll call that time interval
Δ𝑡.

Now the change in momentum of
an object is given by finding the final momentum of the object, which we’ll call
𝑝 subscript 𝑓, minus the initial momentum, which we’ll call 𝑝 subscript
𝑖. In this case, the object is the
frog. So all we need to do to find
the change in momentum of the frog is to find the final momentum of the frog and
subtract from this the initial momentum of the frog.

To do this, we can further
recall that the momentum of an object, 𝑝, is given by multiplying the mass of
that object by the velocity with which it’s traveling at that point in time. So the momentum of the frog
here in its initial position is going to be given by multiplying the mass of the
frog, 0.03 kilograms, by its velocity at that point in time.

But then, at that point in
time, the frog is in a resting position. That’s what we’ve been told in
the question. Therefore, its velocity is
zero. And hence, its momentum —
that’s its initial momentum 𝑝 subscript 𝑖 — is also equal to zero.

Then we can move on to working
out the momentum of the frog when it’s already jumped and it’s moving at 0.12
meters per second. We can say that the final
momentum of the frog, 𝑝 subscript 𝑓, which is the momentum at this point, is
given by multiplying the mass of the frog, which is 0.03 kilograms still, by its
velocity, which is 0.12 meters per second. And of course, the velocity is
in this direction. But that’s not really relevant
to us right now, cause all we’re trying to find is the magnitude or size of this
momentum.

So evaluating the right-hand
side of this equation, we find that it’s 0.0036 kilograms meters per second, at
which point we can go back to finding the change in momentum of the frog,
Δ𝑝. We can see that this is equal
to the final momentum, 0.0036 kilograms meters per second, minus the initial
momentum, which is zero kilograms meters per second, which then simplifies to
0.0036 kilograms meters per second. That’s how much the momentum of
the frog has increased because of it jumping.

And we’ve also been told that
this momentum increase occurs over a time period of 0.025 seconds. Therefore, the question already
gives us the value of Δ𝑡. So now we can sub in the values
of Δ𝑝 — that’s 0.0036 kilograms meters per second — and Δ𝑡 — that’s 0.025
seconds — to give us a value for the force exerted by the frog’s legs to get it
from here to here. And that value for force is
going to be in newtons by the way. This is because we earlier
converted everything into base units, which means our answer for the force is
going to be in its own base unit, which is the newton.

So when we evaluate the
right-hand side, we find that it’s equal to 0.144 newtons. And hence, we have our final
answer to the question. The frog’s legs exert 0.144
newtons of force in the jump.