### Video Transcript

A car of mass ๐ tons was initially at rest on a hill inclined at an angle ๐ to the horizontal, where sine of ๐ equals one-half. After 100 seconds, its velocity was 21 metres per second. Calculate the resistance per ton of the carโs mass. Take ๐ to equal 9.8 metres per second squared.

We can call the elapsed time during which the car is speeding up 100 seconds ๐ก. And the speed it achieves after that time 21 metres per second weโll call ๐ฃ sub ๐. Weโre told that the sine of the angle of inclination of the hill on which the car rests is one-half. We want to calculate the resistance per ton of the carโs mass. Weโll call that value ๐น sub ๐ over ๐.

Letโs begin our solution by drawing a sketch of this scenario. Weโre told that we begin with a car at rest on a hillside inclined at an angle ๐ from the horizontal. At ๐ก equals zero, the car is released and begins to slide downhill. After a time of 100 seconds, the car has achieved a particular speed 21 metres per second downhill. Knowing this about the carโs descent, we want to solve for the resistive force โ weโve called it ๐น sub ๐ โ applied to the car per ton of its mass ๐.

To start on our solution, letโs draw in the forces that are acting on the car while it descends. First, there is the force of gravity acting down. Thatโs equal to the mass of the car times the acceleration due to gravity ๐, which weโre told to treat as exactly 9.8 metres per second squared. There is also a normal force pushing the car perpendicularly away from the plane of the hill. We call that ๐น sub ๐. And finally, there is a resistive force due to friction, which weโve called ๐น sub ๐, acting on the car and opposing its downward motion.

To solve for that force per ton of the carโs mass, letโs begin by recalling Newtonโs second law of motion. The second law says that the net force acting on an object is equal to that objectโs mass multiplied by its acceleration. If we place a pair of coordinate axes on our diagram so that ๐ฅ is acting up the hill and ๐ฆ is positive perpendicularly to the plane, then we become able to apply the second law to both directions.

Letโs consider the forces in the ๐ฅ-direction. Two of our three forces in our diagram act along this axis ๐น sub ๐ and a component of the weight force ๐๐. If we divide our weight force ๐๐ into ๐ฆ- and ๐ฅ-components, then the right triangle formed has an angle of ๐ in its upper corner. This means when we apply Newtonโs second law to forces in the ๐ฅ-direction, we can write ๐น sub ๐ minus ๐๐ times the sine of ๐ is equal to ๐ times ๐.

If we then divide both sides of our equation by the mass of the car ๐, that term cancels out of two of our three terms and we now have ๐น sub ๐ over ๐ in the far left, which is what we want to solve for. We can rearrange our equation. So it reads ๐น sub ๐ over ๐ is equal to ๐ sine ๐ plus ๐.

We know the gravitational acceleration ๐ and we also know the sine of ๐. But what about the acceleration ๐? We can recall that acceleration is defined as a change in velocity over change in time. Weโre told that our car starts out from rest. So our initial velocity is zero and that it attains a final velocity ๐ฃ sub ๐ of 21 metres per second. All this happens over the course of a time of 100 seconds.

So when we plug in ๐ฃ sub ๐ and ๐ก, we see that ๐ is negative 21 metres per second divided by 100 seconds or negative 0.21 metres per second squared. Weโre now ready to plug this value in for ๐. And we know ๐ and the sine of ๐.

When we enter these values on our calculator, we find that ๐น sub ๐ over ๐ is equal to 4.69 newtons per ton. Thatโs the resistive force of the hill on the car per ton of the carโs mass.