A car of mass 𝑚 tons was initially at rest on a hill inclined at an angle 𝜃 to the horizontal, where sine of 𝜃 equals one-half. After 100 seconds, its velocity was 21 metres per second. Calculate the resistance per ton of the car’s mass. Take 𝑔 to equal 9.8 metres per second squared.
We can call the elapsed time during which the car is speeding up 100 seconds 𝑡. And the speed it achieves after that time 21 metres per second we’ll call 𝑣 sub 𝑓. We’re 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 car’s mass. We’ll call that value 𝐹 sub 𝑓 over 𝑚.
Let’s begin our solution by drawing a sketch of this scenario. We’re told that we begin with a car at rest on a hillside inclined at an angle 𝜃 from the horizontal. At 𝑡 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 car’s descent, we want to solve for the resistive force — we’ve called it 𝐹 sub 𝑓 — applied to the car per ton of its mass 𝑚.
To start on our solution, let’s draw in the forces that are acting on the car while it descends. First, there is the force of gravity acting down. That’s equal to the mass of the car times the acceleration due to gravity 𝑔, which we’re 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 𝐹 sub 𝑁. And finally, there is a resistive force due to friction, which we’ve called 𝐹 sub 𝑓, acting on the car and opposing its downward motion.
To solve for that force per ton of the car’s mass, let’s begin by recalling Newton’s second law of motion. The second law says that the net force acting on an object is equal to that object’s mass multiplied by its acceleration. If we place a pair of coordinate axes on our diagram so that 𝑥 is acting up the hill and 𝑦 is positive perpendicularly to the plane, then we become able to apply the second law to both directions.
Let’s consider the forces in the 𝑥-direction. Two of our three forces in our diagram act along this axis 𝐹 sub 𝑓 and a component of the weight force 𝑚𝑔. If we divide our weight force 𝑚𝑔 into 𝑦- and 𝑥-components, then the right triangle formed has an angle of 𝜃 in its upper corner. This means when we apply Newton’s second law to forces in the 𝑥-direction, we can write 𝐹 sub 𝑓 minus 𝑚𝑔 times the sine of 𝜃 is equal to 𝑚 times 𝑎.
If we then divide both sides of our equation by the mass of the car 𝑚, that term cancels out of two of our three terms and we now have 𝐹 sub 𝑓 over 𝑚 in the far left, which is what we want to solve for. We can rearrange our equation. So it reads 𝐹 sub 𝑓 over 𝑚 is equal to 𝑔 sine 𝜃 plus 𝑎.
We know the gravitational acceleration 𝑔 and we also know the sine of 𝜃. But what about the acceleration 𝑎? We can recall that acceleration is defined as a change in velocity over change in time. We’re told that our car starts out from rest. So our initial velocity is zero and that it attains a final velocity 𝑣 sub 𝑓 of 21 metres per second. All this happens over the course of a time of 100 seconds.
So when we plug in 𝑣 sub 𝑓 and 𝑡, we see that 𝑎 is negative 21 metres per second divided by 100 seconds or negative 0.21 metres per second squared. We’re now ready to plug this value in for 𝑎. And we know 𝑔 and the sine of 𝜃.
When we enter these values on our calculator, we find that 𝐹 sub 𝑓 over 𝑚 is equal to 4.69 newtons per ton. That’s the resistive force of the hill on the car per ton of the car’s mass.