Calculate Force & Coefficient of Friction for 2000kg Car & 1.2 x 10^8 J/Gallon

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To calculate the average force on a 2000 kg car that travels 40 km per gallon of gasoline with a chemical potential energy of 1.2 x 10^8 J, the energy must be converted into work done over the distance. The work done is expressed as F*d, where F is the force and d is the distance traveled. Since the car is moving at constant speed, the force opposing motion is equal to the force produced by the combustion of the gasoline. The coefficient of friction can be derived from the relationship between the force and the normal force acting on the car. Understanding the distinction between chemical potential energy and gravitational potential energy is crucial for accurate calculations.
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A gallon of gasoline has a chemical PE of 1.2 x 10^8 J. If a 2000 kg car gets 40 km/gallon, what is the average force on the car? Assume constant speed and remember the force on the car is applied over the total distance traveled. What is the coefficient of friction?

I tried PE = mgx, using m = 2000 and solving for x (which was 6122.45). I'm not really sure where to go from there though.
 
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Not quite.

The PE of the gasoline is not gravitational PE. Completely different.

This PE gets converted into work over the distance of 40 km.

What you do is convert the energy they gave you into the work to move the car. Work is F*d.

At constant speed then the force retarding motion is consumed by the combustion of the gas.
 
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