# Question Video: Calculating the Gravitational Force on a Car Driving up a Hill Physics • 9th Grade

A car is driven at a constant speed up a hill that slopes at 30°. The car is driven by a force of 6750 N from its engine. A friction force of 4790 N 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?

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### 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.