### Video Transcript

A sprinter of mass 60 kilograms
accelerates from 2.00 meters per second to 8.00 meters per second over a 25.0-meter
distance of racetrack. The sprinter is running into the
wind which exerts an average force of 30.0 newtons. What average force does the
sprinter exert on the track?

Let’s highlight some of the
important information given to us. The sprinter’s mass is given as 60
kilograms. We’ll record that value as 𝑚. The sprinter accelerates from an
initial speed of 2.00 meters per second to a final speed of 8.00 meters per
second. We’ll call the sprinter’s initial
speed 𝑣 sub 𝑖 and the sprinter’s final speed 𝑣 sub 𝑓. We’re told that this acceleration
occurs over a distance of 25.0 meters, a value we’ll call 𝑑. The sprinter faces resistance due
to wind at an average force of 30.0 newtons. We’ll call this 𝐹 sub 𝑤 for the
force exerted by the wind. We want to solve for the average
force the sprinter exerts on the track. We’ll call that 𝐹 sub 𝑠.

Let’s begin our solution by drawing
a diagram of the situation. The sprinter running across the
level track experiences two forces in the horizontal direction. There is the force of the wind
pushing against the sprinter and the force of the track pushing the sprinter
forward. We’ll define motion down the track
as motion in the positive direction.

To begin solving for 𝐹 sub 𝑠,
let’s recall Newton’s third law. Newton’s third law says that for
every action, there exists an equal and opposite reaction. As we look at the diagram of the
sprinter, Newton’s third law tells us that the force that the track exerts on the
sprinter is equal and opposite to the force that the sprinter exerts on the track,
𝐹 sub 𝑠. So if we can solve for 𝐹 sub 𝑡,
the force of the track on the sprinter, then we’ll have solved for 𝐹 sub 𝑠 in
magnitude.

Now let’s recall another of
Newton’s laws, his second law. This second law states that the net
force on an object is equal to its mass multiplied by its acceleration. If we consider only the forces in
the horizontal direction that are acting on the sprinter, we can write that 𝐹 sub
𝑡 minus 𝐹 sub 𝑤 is equal to the sprinter’s mass times their acceleration. We are given 𝐹 sub 𝑤 and the
sprinter’s mass 𝑚. But what about acceleration? Because the sprinter’s acceleration
is constant over this path, we can use the kinematic equations of motion to solve
for it. Let’s recall those equations.

As we consider the information
given to us in the problem, with an initial and final speed and a distance over
which the sprinter moves, and then look over the kinematic equations, we see that
the second one listed is a good match for the information we have and the
information we want to solve for. In this equation, 𝑣 sub 𝑓 is
given to us as 8.00 meters per second. 𝑣 zero, we’ve called 𝑣 sub 𝑖;
that’s 2.00 meters per second. The acceleration 𝑎 is what we want
to solve for and 𝑑 is given as 25.0 meters.

Let’s rearrange this equation to
solve for the acceleration 𝑎. But before we rearrange the
equation, let’s replace 𝑣 sub zero with 𝑣 sub 𝑖. We can solve for the acceleration
𝑎 in two steps. First, we’ll subtract 𝑣 sub 𝑖
squared from both sides, cancelling that term on the right side. Next, we’ll divide both sides of
the equation by two times 𝑑, cancelling out both those terms on the right side of
the equation, leaving us with an equation that reads: acceleration 𝑎 equals 𝑣 sub
𝑓 squared minus 𝑣 sub 𝑖 squared divided by two 𝑑.

We can plug in the values given for
𝑣 sub 𝑓, 𝑣 sub 𝑖, and 𝑑. When we enter these values on our
calculator, we find a value for 𝑎 of six-fifths meters per second squared.

Now that we know the acceleration
𝑎, we can solve for 𝐹 sub 𝑡, the force of the track on the sprinter, remembering
that that will equal 𝐹 sub 𝑠, the force of the sprinter on the track. First, let’s add 𝐹 sub 𝑤 to both
sides of the equation which cancels that term out of the left-hand side, leaving us
with an equation that reads: 𝐹 sub 𝑠 equals the sprinter’s mass 𝑚 times their
acceleration 𝑎 plus 𝐹 sub 𝑤, the force of the wind against the sprinter.

We plug in 60 kilograms for 𝑚,
six-fifths of a meter per second squared for 𝑎, and 30.0 newtons for 𝐹 sub 𝑤. When we enter these values on our
calculator, we find that 𝐹 sub 𝑠 is equal to 102 newtons. This is the force the sprinter
exerts on the track.