# Work energy efficiency car problem

## Homework Statement

Gasoline has an energy content of about 132.65 MJ/US Gallon. An average passenger car has a mass of 1500kg. The average vehicle experiences a drag force F(drag) = (0.176 kg/m)v^2, regardless of road conditions. A uniform and level road surface provides a 'rolling resistance' given by F(roll) = 0.018mg. Engine efficiency varies with the speed of the car, as shown in graph. A) calculate the percent difference in the number of miles traveled by the car per gallon of gasoline consumed by the engine (mpg) when driving at 65 miles per hour(mph) vs. Driving at 75mph. Assume a uniform and level road surface. Calculate the percent difference in travel time when driving at 65mph instead of 75mph

B) calculate the percent difference in the mpg when driving at 65mph with tires that are 9 pounds per square inch (psi) below recommended pressure. Under pressure tires increases the rolling resistance of the car by 1% per 3psi below the recommended level. Assume a uniform and level road surface.

C) calculate the percent difference in the mpg when driving at 65mph a uniform and level road surface vs. Driving at 65mph on a road surface in poor condition. Poor surface conditions increase the rolling of the vehicle to F(roll) = 0.022mg.

## Homework Equations

Newtons kinematic
Wncf = fd
KE = 1/2mv ^2

3. Attempt at solution

Part a) so the way I go about this I'm thinking I use the equation of sum of non conservative forces = kinetic energy. After doing some conversions, It would look something like work done by engine[132.65MJ * efficiency rate] - [drag force * d] - [roll force * d] = 1/2mv^2. Solve for d and use that to compare the distances. And would calculating the difference in travel time would it be just to compare the velocities once conversions are done?

Part b and c is similar to a but just make a couple of adjustments to some of the forces and stuff?

Am I on the right path on how to go about this or am I completely off?

I know this is kind of a long read but many thanks for any/all input. Definitely much appreciated

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haruspex
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Part a) so the way I go about this I'm thinking I use the equation of sum of non conservative forces = kinetic energy.
A force cannot equal energy.
After doing some conversions, It would look something like work done by engine[132.65MJ * efficiency rate] - [drag force * d] - [roll force * d] = 1/2mv^2.
You do not have a quantity 132.65MJ. You have a quantity 132.65MJ/Gal. You cannot go changing the dimensions of constants like that - it becomes meaningless.
Solve for d and use that to compare the distances.
What distances?
am I completely off?
Don't worry about kinetic energy here. The car is moving at constant speed, so that doesn't change. What force (thrust) is needed to keep the car going at 65mph? How much energy does it need for each mile? How much fuel does it need to produce that energy?

Oh I see, so first I need to find the thrust force that will allow the car to move at a constant acceleration against the forces such as the roll and drag, which looks like f(thrust) = f(drag) + (froll), to find the energy I multiply the thrust by the distance (65 or 75 miles convert this to meters), which will give me the number I can use to find out how many gallons it takes to produce this energy. Then I can compare the 2 numbers to find out the percent difference..

Am I on the right path so far?

haruspex
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Oh I see, so first I need to find the thrust force that will allow the car to move at a constant acceleration
No, constant speed.
against the forces such as the roll and drag, which looks like f(thrust) = f(drag) + (froll), to find the energy I multiply the thrust by the distance (65 or 75 miles convert this to meters), which will give me the number I can use to find out how many gallons it takes to produce this energy. Then I can compare the 2 numbers to find out the percent difference..

Am I on the right path so far?
Yes, much better.

Do i convert mph to m/s? And use that for the distance in the work equation?

haruspex
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Use whatever units you find convenient. My preference is always to work symbolically, only plugging in numbers and units at the end.

Or am I using the distance per 1 mile?

haruspex
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Again, pick what unit you like. The question asks in terms of mpg, so miles seems reasonable, but it won't affect the percentage.

ok so this is what I have so far,
I converted 65mph to m/s and used that for the velocity to find the drag. I found both F(thrust) for each speed and multiplied it by a mile, or 1609m and divided that number by the gasoline energy content to find how much of a gallon does it take to create that energy. And then calculated the percent difference for both. now for b & c it would be more or less the same right, except I just modify the roll forces?

haruspex
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You need to use different engine efficiencies at the two speeds, reading off the graph. I can't see where you've done that.

Oh I see, I made the mistake of using the one given in the problem statement. So what I do is I look at the graph, for example when driving at 65mph I see that it is about 0.199 efficiency, I multiply that by the gasoline energy content (132.65mj/g) and use that number to divide the energy found to find the gallons?

haruspex
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Quite so. (But it looks more like .197 to .198 to me.)

Cool thanks, now to find the time, do I use f=ma and solve for a? Where f is the sum of the forces(Fthrust - Fdrag - Droll) or would it be just the thrust force?

Then use newtons kinematic equation to find the time such as xf - xo = vot + .5at^2? Where the change of x is one mile?

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haruspex