### Video Transcript

A car is driven at a constant speed
up a hill that slopes at 30 degrees. The car is driven by a force of
6750 newtons from its engine. A friction force of 4790 newtons
also acts on the car. The direction of the force from the
car’s engine is considered the positive direction. What is the value of the
gravitational force that acts on the car in the direction of the slope of the
hill?

Let’s begin by sketching out the
problem with the forces mentioned in the question. The driving force of the car is
labeled as 𝐹 𝐷, and the friction force that acts on the car is labeled as 𝐹
𝑓. The gravitational force on the car
is called the weight of the car and is labeled as 𝑊. Since we want to find the magnitude
of the gravitational force acting in the direction of the slope, we can resolve this
force into two independent components: one that acts parallel to the slope, which we
will label as 𝑊 parallel, and the other component that acts perpendicular to the
slope, which we will label as 𝑊 perp.

We are told in the question that
the car drives up the hill at a constant speed. Since the car is at a constant
speed, this means that the car’s acceleration must be zero. Let’s now recall Newton’s second
law. The acceleration of an object as
produced by a net force is directly proportional to the magnitude of the net, or
total force, in the same direction as the net force, and inversely proportional to
the mass of the object. In equation form, this is written
as 𝑎 equals 𝐹 total divided by 𝑚.

Since the car’s acceleration is
zero, the net force in the direction of the hill must also be equal to zero due to
Newton’s second law. If we now sum up the forces in the
direction of the hill, we get 𝐹 𝐷 minus 𝐹 𝑓 minus 𝑊 parallel equals zero. Here, we note that the direction of
the force from the car’s engine is considered the positive direction. So 𝐹 𝐷 is positive, while 𝐹 𝑓
and 𝑊 parallel are negative, since they point in the opposite direction to the
force from the car’s engine.

We can make 𝑊 parallel the subject
by adding 𝑊 parallel to both sides of the equation, leaving us with 𝑊 parallel
equals 𝐹 𝐷 minus 𝐹 𝑓. We’re told in the question that the
force from the car’s engine is equal to 6750 newtons and the friction force is equal
to 4790 newtons. So if we substitute these values
into our equation, we find that 𝑊 parallel is equal to 6750 newtons minus 4790
newtons, which is equal to 1960 newtons. And so, we have calculated that the
magnitude of the gravitational force that acts on the car in the direction of the
slope of the hill is equal to 1960 newtons. This force acts in the opposite
direction to the direction of the force from the car’s engine.

Since we want to give the value of
the gravitational force that acts on the car in the direction of the slope, this
value must be a negative number. So the value of the gravitational
force that acts on the car in the direction of the slope of the hill is equal to
negative 1960 newtons.