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A vehicle of mass three metric tons was moving at 51 kilometers per hour along a horizontal section of road.
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When it reached the bottom of a hill inclined to the horizontal at an angle whose sine is 0.5, it continued moving at the same speed up the road.
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Given that the resistance of the two sections of road is constant, determine the increase in the vehicle’s power to the nearest horsepower.
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Take the acceleration due to gravity 𝑔 to equal 9.8 meters per second squared.
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All right, so in this example, we have this vehicle moving originally along a horizontal section of road.
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Its speed at that point is 51 kilometers per hour.
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And we’re told that eventually this vehicle reaches the base of a hill and then starts to move up the hill, maintaining the same speed.
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If we call the angle that this incline makes with the horizontal 𝜃, we’re told that the sin of 𝜃 equals 0.5 or one-half.
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We’re also told that the resistance of the horizontal stretch of road to the vehicle’s forward motion is equal to the resistance of the inclined stretch of road.
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In other words, the frictional forces opposing the vehicle’s motion are the same in either case.
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What is different is that once the vehicle is moving uphill, it now has to work against the force of gravity.
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To maintain its original speed, the vehicle will need to increase the power its engine generates.
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That’s the increase we want to calculate here.
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Before we clear some space on screen to work, let’s recall the fact that the mass of our vehicle, we’ll call it 𝑚, is given to us as three metric tons.
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We’ll abbreviate that three t.
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To get started on our solution, we can recall that when an object is subject to a force causing it to move at a speed 𝑣, the product of that force and 𝑣 is equal to the power applied to it.
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In our case, though, we really want to solve for an increase in power, we’ll call it Δ𝑃, that the car’s engine needs to supply.
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This increase in power is to supply an increase in force, Δ𝐹, needed for the car to maintain its same speed, 𝑣, as it moves onto this inclined plane.
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To get a clearer sense for the forces involved here, let’s look at this up-close sketch and draw a free body diagram.
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Recall that this involves sketching in all the forces that are acting on our vehicle.
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We know that our vehicle is subject to a gravitational force.
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Its magnitude is the vehicle’s mass times the acceleration due to gravity.
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It’s also acted on by a normal force and by a force that resists its motion across this road surface.
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Lastly, there’s the force applied to the vehicle, thanks to the power from its engine.
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This is what allows it to move up the incline.
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Our problem statement told us that what we’ve called 𝐹 sub r, the resistive force, hasn’t changed from when our vehicle was traveling on flat ground to this incline.
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What has changed is that we now need to work against gravity.
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Specifically, if we break the gravitational force into its two perpendicular components, we’re now opposed by this component of that force.
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To solve for the magnitude of this component, we recognize that this triangle we’ve sketched is a right triangle where this angle here, just like the angle of our incline, has a measure of 𝜃.
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That tells us that the component of the gravitational force we now need to work against is 𝑚𝑔 times the sin of 𝜃.
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In magnitude, this is equal to the increase in force that needs to be exerted on our vehicle to move it uphill.
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So we can replace Δ𝐹 in our equation with 𝑚 times 𝑔 times the sin of 𝜃.
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And note that we know 𝑚, that’s given to us, as is the acceleration due to gravity 𝑔, as is the sin of 𝜃 and the speed 𝑣.
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If we plug in all these values, though, we see that the units involved don’t match one another.
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That is, they don’t all come from the same system of units.
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And even those that do aren’t expressed in quite the same way.
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For example, our acceleration due to gravity has distance expressed in meters, while our speed has distance expressed in kilometers.
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To make sense of our calculation for Δ𝑃, we’ll want the units on the right-hand side to all be on the same basis.
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To do this, we can recall that one metric ton is equal to 1000 kilograms and that one kilometer per hour is equal to one over 3.6 meters per second.
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Since one metric ton equals 1000 kilograms, three metric tons is equal to 3000 kilograms.
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And then considering this unit conversion here, we find that if we divide 51 by 3.6, then we have effectively changed the units of this value to meters per second.
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Now all of our units are SI base units, kilograms for mass, meters for distance, and seconds for time.
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If we were to calculate Δ𝑃 now, we would get a result in units of watts, the SI base unit of power.
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Our question, though, wanted us to state this increase in power in units of horsepower.
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One metric horsepower is equal to 735 watts.
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So if we divide the right-hand side of our equation by 735 watts per horsepower, what we’ll find, and we can check it if we want, is that all of the SI base units will cancel out, while the units of horsepower will move into the numerator.
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All this to say, we’ll get the units that we’re after, horsepower.
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When we do go ahead and calculate this fraction, rounding to the nearest horsepower, we get a result of 283.
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This is the increase in power necessary from this vehicle’s engine to keep it moving at the same speed as it moves up the incline.