### Video Transcript

A block of mass five kilograms is
initially at rest on a smooth surface. A constant force of magnitude 39
newtons acting parallel to the surface is applied to the block. Determine the speed of the block
nine seconds after the force was initially applied.

Let’s begin by drawing a free-body
diagram of this scenario. We have a block that is at rest on
a horizontal surface. Since it’s at rest, we can say its
initial velocity 𝑢, which is sometimes represented as 𝑣 sub zero, is equal to
zero. It has a mass of five kilograms,
which means it exerts a downwards force of its weight on the surface on which it
rests. We can use Newton’s second law of
motion to calculate the exact value of this force. Its force is equal to mass times
acceleration. Well, the block itself has a mass
of five kilograms, and the acceleration due to gravity we’ll say is 𝑔. And that’s equal to 9.8 meters per
square second.

By Newton’s third law of motion,
since the body itself exerts a downwards force on the surface, the surface must
exert a force equal in magnitude and opposite in direction on the block. We’ll call that force 𝑅. Next, we’re told a constant force
of magnitude 39 newtons acts parallel to the surface. And since the surface is smooth,
this is the only force acting in this direction; there’s no frictional force. Now we want to find the speed of
the block, let’s call that 𝑣, nine seconds after the force was initially
applied. And so we have three values
here. We have the initial speed is zero,
the final velocity 𝑣 is what we’re trying to find, and 𝑡, time, is nine
seconds.

Now, it follows that if we can find
the value of 𝑎, acceleration, or 𝑠, displacement, we’ll be able to use one of our
equations of constant acceleration to calculate 𝑣. In fact, we can use Newton’s second
law of motion to calculate the acceleration of the block. Thinking about the forces that act
on the block in the horizontal direction, we know that the sum of these forces is
simply 39. And we’re taking the direction in
which the block is moving to the right to be positive. The mass of the block is five, so
mass times acceleration is five 𝑎. And we have an equation for 𝑎;
it’s 39 equals five 𝑎. Dividing both sides of this
equation by five and we find 𝑎 is equal to 39 over five meters per square
second.

And so we now have everything we
need to be able to apply one of our equations of constant acceleration. The equation that we’ll use is 𝑣
equals 𝑢 plus 𝑎𝑡. Since the initial speed is zero,
the acceleration is 39 over five, and 𝑡 is nine, 𝑣 is zero plus 39 over five times
nine, which is equal to 351 over five or equivalently 70.2.

The speed of the block then nine
seconds after the force was initially applied is 70.2 meters per second.