Question Video: Power of Motion on Inclined Planes with Resistive Forces Mathematics

A car of mass 3 metric tons was ascending a road inclined to the horizontal at an angle whose sine is 1/40, at its maximum speed of 54 km/h. Later on, the same car ascended another road which was inclined to the horizontal at an angle whose sine is 1/120. On this hill, its maximum speed was 72 km/h. Given that the resistance to the car’s motion was the same on both roads, determine the horsepower of the car’s engine 𝑃 and the resistance of the roads 𝑅.

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Video Transcript

A car of mass three metric tons was ascending a road inclined to the horizontal at an angle whose sin is one over 40 at its maximum speed of 54 kilometers per hour. Later on, the same car ascended another road which was inclined to the horizontal at an angle whose sin is one over 120. On this hill, its maximum speed was 72 kilometers per hour. Given that the resistance to the car’s motion was the same on both roads, determine the horsepower of the car’s engine 𝑃 and the resistance of the roads 𝑅.

So here we’re thinking about a car as it drives up two roads that are inclined at different angles. One of the first things we might notice about this question is we’re not told the actual angles that the roads are inclined at. The first road we’re told is inclined to the horizontal at an angle whose sin is one over 40. So we could say that the slope is inclined at an angle of πœƒ one and that the sin of πœƒ one is one over 40. This equation can be rearranged to show us that πœƒ one equals the inverse sin of one over 40, which our calculator will show us is approximately equal to 1.4 degrees.

We can do the same for the other angle. Let’s call this πœƒ two. The question tells us that this road is inclined to the horizontal at an angle whose sin is one over 120. This means that we can say the sin of πœƒ two is one over 120. So πœƒ two is equal to the inverse sin of one over 120, which our calculator tells us is approximately 0.48 degrees. So why doesn’t the question just tell us the angle of the slope in degrees? Well, as we’ll see later on, it actually makes our calculations easier if we know the sine of the angle rather than the angle itself. So for now, we can forget about the values of the angles in degrees. And we’ll just remember instead that the first slope is slightly steeper than the second slope, which we’ve exaggerated in our diagrams.

So this question asks us to calculate the resistance of the roads, which is a force 𝑅, and also the horsepower of the car’s engine, which we’re calling 𝑃. We can recall that horsepower is simply a unit of power and that generally power is equal to force times velocity. Since here we’re finding the power of the car’s engine, the force in question is specifically the force produced by the car’s engine. So since we’re interested in the forces that act on the car, let’s start by labeling our diagrams with all of the forces present.

Firstly, we have a force produced by the engine of the car in both situations. It’s not immediately clear whether the engine produces the same force on both inclines, so let’s call these forces 𝐹 one and 𝐹 two. We’re also told that the car experiences some resistance force and this resistance force is the same on both roads. So this resistance force has a value of 𝑅 in both cases. We also have the car’s weight, which of course, acts downwards. And this is equal to the car’s mass π‘š multiplied by the acceleration due to gravity 𝑔. Finally, in both cases, we have a normal reaction force of the road pushing on the car. Let’s call these 𝑁 one and 𝑁 two, respectively.

Now, in a situation like this, where we have multiple forces acting on a single object, we can use Newton’s second law to find out how these forces relate to each other. Newton’s second law tells us that the net force acting on an object is equal to that object’s mass multiplied by its acceleration. And we can recall that the net force is simply the vector sum of all of the forces acting on an object. Now when we’re summing vectors, we first need to resolve them into perpendicular components. Often in mechanics, we’ll find the vertical and horizontal components of our forces. But when we’re dealing with an inclined plane, it’s often easiest to find the components of each force that act parallel to the slope and perpendicular to the slope.

In this question, we’re only interested in the motion of the car parallel to the slope. This means that we can get away with only thinking about the components of each force that act parallel to the slope. So let’s start by calculating the net force parallel to the slope for the diagram on the left. First, we can spot that 𝐹 one and 𝑅 both act completely parallel to the slope. So let’s arbitrarily say that up the slope is the positive direction and down the slope is the negative direction. Then the component of 𝐹 one acting up the slope is simply 𝐹 one, and the component of 𝑅 acting up the slope is negative 𝑅.

If we look at the normal reaction force 𝑁 one, we can see that this force acts perpendicular to the slope. It, therefore, has no component parallel to the slope. So it doesn’t make any contribution to the net force parallel to the slope. Finally, we have the weight of the car π‘šπ‘”. This force does have a component parallel to the slope. We can visualize the components of this weight force using two arrows. This is the component of the weight force acting perpendicular to the slope. And this is the component that acts parallel to the slope. We can notice that these three arrows form a right-angle triangle, where the hypotenuse is the magnitude of the weight force π‘šπ‘”. This angle is the same as the angle of the slope β€” that’s πœƒ one β€” and the opposite side of the triangle is the component of the weight force that acts parallel to the slope.

Trigonometry tells us that the length of this arrow is π‘šπ‘” sin πœƒ one. And this is, therefore, the component of the weight force that acts parallel to the slope. Noticing that this arrow points in the negative direction, the net horizontal force acting on the car is 𝐹 one minus 𝑅 minus π‘šπ‘” sin πœƒ one.

And what about the right-hand side of this expression? We know the car’s mass π‘š, but it doesn’t look like we’ve been told anything about the car’s acceleration in the question. However, we have been told the maximum speed of the car on both slopes. On the first road, the maximum speed of the car is 54 kilometers per hour. And on the second road, the maximum speed is 72 kilometers per hour. So how does this help us? Well, if we consider the car when it’s traveling at its maximum speed, we know that the speed of the car is constant. And if the speed of the car is constant, that means it’s not accelerating. So we can say the acceleration of the car π‘Ž is equal to zero. If π‘Ž is zero, then π‘š times π‘Ž is also zero, which means Newton’s second law yields this equation for the first scenario.

Now, we do exactly the same thing for the scenario on the right side. Using Newton’s second law as guidance, we first calculate the net force that acts parallel to the slope. This is given by 𝐹 two, the driving force of the car, minus 𝑅, the force resisting the motion of the car, minus the component of the weight that acts parallel to the slope. This is π‘šπ‘” sin πœƒ two. Finally, we put this equal to mass times acceleration. And since the acceleration is zero when the car is traveling at its top speed, this means the right-hand side of this equation is zero.

So Newton’s second law has given us two equations, but we still have three unknowns, 𝐹 one, 𝑅, and 𝐹 two. But remember that the question doesn’t actually ask for the driving force, 𝐹 one and 𝐹 two, of the car on the different slopes. What we’ve been asked to calculate is the power of the engine, which is equal to the force produced by the engine multiplied by the speed of the car 𝑣.

Looking at the equation on the left, we can come up with an expression for the power of the engine by first rearranging to make 𝐹 one the subject. We can do this by adding 𝑅 to both sides and then adding π‘šπ‘” sin πœƒ one to both sides, giving us 𝐹 one equals π‘šπ‘” sin πœƒ one plus 𝑅. Next, because the power of the engine is equal to the force produced by the engine multiplied by the speed of the car, we can multiply 𝐹 one by the speed of the car to give us the power. Written as an equation, we can say that 𝑃, the power of the engine, equals 𝐹 one, that’s the force produced by the engine, multiplied by the speed of the car, 𝑣 one. And since we’ve shown that 𝐹 one is equal to π‘šπ‘” sin πœƒ one plus 𝑅, we can substitute all this in place of 𝐹 one, giving us this expression for the power of the engine.

Let’s now do the same for the expression on the right, first making 𝐹 two the subject and then recalling that the power of the engine is the force produced by the engine multiplied by the speed of the car, which for this scenario we’ll call 𝑣 two. Okay, so we now have these two equations. π‘š, the mass of the car, is given in the question. 𝑔 is a known constant. The inclines of the roads, πœƒ one and πœƒ two, are also given in the question, and so are the maximum speeds of the car, 𝑣 one and 𝑣 two. This means that the only unknown quantities in these equations are 𝑃, the power of the engine, and 𝑅, the resistance of the roads, which are the two quantities we’re asked to find in the question.

Since we have two unique equations and they each contain the same two unknown quantities, it’s possible to solve these equations simultaneously and find the answers to the question. So let’s clear some space on the screen and solve them. Now because 𝑃, the power of the engine of the car, is the subject of both equations, we can start by equating the right-hand side of each equation, effectively eliminating 𝑃 and enabling us to solve for 𝑅. Next, distributing 𝑣 across the parentheses on the left-hand side gives us 𝑣 one π‘šπ‘” sin πœƒ one plus 𝑣 one 𝑅. And doing the same on the right-hand side gives us 𝑣 two π‘šπ‘” sin πœƒ two plus 𝑣 two 𝑅.

Next, since we’re solving for 𝑅, we’ll gather all the terms with an 𝑅 in them on the right-hand side. So first, we subtract 𝑣 one 𝑅 from both sides and then we subtract 𝑣 two π‘šπ‘” sin πœƒ two from both sides. This gives us 𝑣 one π‘šπ‘” sin πœƒ one minus 𝑣 two π‘šπ‘” sin πœƒ two equals 𝑣 two 𝑅 minus 𝑣 one 𝑅. Let’s clear some space on the screen and move this line up to the top. And finally, if we look at the right-hand side of the equation, we can take out a factor of 𝑅, giving us 𝑅 times 𝑣 two minus 𝑣 one. And we can then divide both sides of this expression by 𝑣 two minus 𝑣 one to give us 𝑅 equals 𝑣 one π‘šπ‘” sin πœƒ one minus 𝑣 two π‘šπ‘” sin πœƒ two all over 𝑣 two minus 𝑣 one.

Okay, now we have an expression for 𝑅, all we need to do is substitute in the values of the known quantities, being careful to express them in standard units. We’re told that the mass of the car is three metric tons. And we know that the standard units for mass are kilograms. One metric ton is equal to 1000 kilograms. So three metric tons is equal to 3000 kilograms. We’re also told that the sin of the angle of the first slope is one over 40. Note that we’re able to substitute the sin of the angles straight into our equation, so we don’t need to worry about what the actual values of the angles are. Likewise, the sin of the angle of the second slope, πœƒ two, is one over 120.

Next, we’re told that the maximum speed of the car on the first slope, which we’ve called 𝑣 one, is equal to 54 kilometers per hour. The standard units for speed are meters per second. To convert this quantity into meters per second, we first multiply it by 1000, that’s the number of meters in a kilometer, and divide it by 3600, or the number of seconds in an hour. Overall, this is equivalent to dividing the quantity by 3.6. And 54 over 3.6 gives us 15 meters per second. The maximum speed of the car on the second slope, which we called 𝑣 two, is given as 72 kilometers per hour. Once again, we can divide this quantity by 3.6 to give us the value in meters per second, which in this case is 20.

Finally, we just need 𝑔, the gravitational acceleration on the surface of the Earth, which is equal to 9.8 meters per second squared. Using standard units for each of these quantities ensures that the value we calculate for 𝑅 will be expressed in newtons. Now, substituting all these values in gives us this. And carefully evaluating this expression on our calculator gives us a value of 1225 newtons. Since 𝑅 is a force, we could also express this in units of kilogram weight. To do this, we simply divide the value by 9.8, which is the value of 𝑔. 1225 divided by 9.8 gives us 125 kilograms weight.

Now that we found the value of 𝑅, we can use it along with either one of these equations to find 𝑃. So taking the equation on the left, we just need to substitute in the values of π‘š, 𝑔, sin πœƒ one, 𝑅 β€” and notice that we’re using the value expressed in the standard units of newtons β€” and finally 𝑣 one, which gives us a value of 29400.

Since we used quantities expressed in standard units in our calculation, this means that our answer is given in the standard units for power, which are watts. However, the question asks us to determine the horsepower of the car’s engine. To do this, we just need to recall that one metric horsepower is equal to 735 watts. So to convert this value in watts to metric horsepower, we just need to divide it by 735, giving us a final value of 40 horsepower.

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