Question Video: Calculating the Resistance of an Inclined Plane on a Moving Body given Its Velocity after a Certain Time | Nagwa Question Video: Calculating the Resistance of an Inclined Plane on a Moving Body given Its Velocity after a Certain Time | Nagwa

# Question Video: Calculating the Resistance of an Inclined Plane on a Moving Body given Its Velocity after a Certain Time Mathematics • Third Year of Secondary School

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A car of mass π tons was initially at rest on a hill inclined at an angle π to the horizontal, where sin π = 1/2. After 100 seconds, its velocity was 21 m/s. Calculate the resistance per ton of the carβs mass. Take π = 9.8 m/sΒ².

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### Video Transcript

A car of mass π tons was initially at rest on a hill inclined at an angle π to the horizontal, where sine of π equals one-half. After 100 seconds, its velocity was 21 metres per second. Calculate the resistance per ton of the carβs mass. Take π to equal 9.8 metres per second squared.

We can call the elapsed time during which the car is speeding up 100 seconds π‘. And the speed it achieves after that time 21 metres per second weβll call π£ sub π. Weβ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 carβs mass. Weβll call that value πΉ sub π over π.

Letβs begin our solution by drawing a sketch of this scenario. Weβre told that we begin with a car at rest on a hillside inclined at an angle π from the horizontal. At π‘ 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 carβs descent, we want to solve for the resistive force β weβve called it πΉ sub π β applied to the car per ton of its mass π.

To start on our solution, letβs draw in the forces that are acting on the car while it descends. First, there is the force of gravity acting down. Thatβs equal to the mass of the car times the acceleration due to gravity π, which weβ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 πΉ sub π. And finally, there is a resistive force due to friction, which weβve called πΉ sub π, acting on the car and opposing its downward motion.

To solve for that force per ton of the carβs mass, letβs begin by recalling Newtonβs second law of motion. The second law says that the net force acting on an object is equal to that objectβs mass multiplied by its acceleration. If we place a pair of coordinate axes on our diagram so that π₯ is acting up the hill and π¦ is positive perpendicularly to the plane, then we become able to apply the second law to both directions.

Letβs consider the forces in the π₯-direction. Two of our three forces in our diagram act along this axis πΉ sub π and a component of the weight force ππ. If we divide our weight force ππ into π¦- and π₯-components, then the right triangle formed has an angle of π in its upper corner. This means when we apply Newtonβs second law to forces in the π₯-direction, we can write πΉ sub π minus ππ times the sine of π is equal to π times π.

If we then divide both sides of our equation by the mass of the car π, that term cancels out of two of our three terms and we now have πΉ sub π over π in the far left, which is what we want to solve for. We can rearrange our equation. So it reads πΉ sub π over π is equal to π sine π plus π.

We know the gravitational acceleration π and we also know the sine of π. But what about the acceleration π? We can recall that acceleration is defined as a change in velocity over change in time. Weβre told that our car starts out from rest. So our initial velocity is zero and that it attains a final velocity π£ sub π of 21 metres per second. All this happens over the course of a time of 100 seconds.

So when we plug in π£ sub π and π‘, we see that π is negative 21 metres per second divided by 100 seconds or negative 0.21 metres per second squared. Weβre now ready to plug this value in for π. And we know π and the sine of π.

When we enter these values on our calculator, we find that πΉ sub π over π is equal to 4.69 newtons per ton. Thatβs the resistive force of the hill on the car per ton of the carβs mass.

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