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While driving down the highway, your car must overcome the forces of friction and air resistance in order to keep the car moving. Experiments have indicated that a force of 560N is necessary to keep a car moving at a constant speed. While driving the car a distance of 100,000m , the car burns up 2.8gal of gasoline. If gasoline contains 100,000,000 Joules of potential energy per gallon, what is the efficiency of the car's engine?

User VinsanityL
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Answer:

Efficiency of the engine equals 20%

Step-by-step explanation:

We know that when the car moves it must do work against the resisting forces to keep moving and this work is spend as energy by the engine to keep the car moving.

we know that


Work=Force* Displacement

Thus to keep the car moving for 100,000 meters the theoretical work that requires to be done equals


Work=560N* 100,000m=56* 10^(6)Joules

Now the actual energy spend by the car equals the energy spend by burning 2.8 gallons of gasoline.

Thus the energy produced by burning 2.8 gallons of gasoline equals


2.8* 100* 10^(6)=280* 10^(6)Joules

Thus the efficiency is calculated as


\eta =(56* 10^(6))/(280* 10^(6))* 100\\\\\therefore \eta =20percent

User Soenke
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