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

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SUMMARY

The discussion focuses on calculating the average force exerted on a 2000 kg car that travels 40 km per gallon of gasoline, which has a chemical potential energy (PE) of 1.2 x 10^8 J. The average force can be derived from the work-energy principle, where work (W) equals force (F) multiplied by distance (d). Given the energy conversion from gasoline to work, the coefficient of friction can also be determined based on the force required to maintain constant speed against frictional forces.

PREREQUISITES
  • Understanding of work-energy principle
  • Basic physics concepts of force and friction
  • Knowledge of chemical potential energy
  • Ability to perform unit conversions (e.g., km to meters)
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  • Calculate the average force using the formula W = F * d
  • Determine the coefficient of friction using the relationship between force and friction
  • Explore the implications of energy conversion efficiency in automotive applications
  • Investigate the impact of vehicle weight on fuel efficiency and friction
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Physics students, automotive engineers, and anyone interested in the mechanics of vehicle motion and energy efficiency.

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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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