220k views
0 votes
Suppose a vehicle travels 150 km at a speed of 30.0 m/s, and uses 3.0 gal of gasoline. only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. if the energy content of the fuel is 1.2 x 108 j/gal. what is the magnitude of the force exerted to keep the car moving at constant speed?

1 Answer

4 votes

Only 30% of gasoline is used for useful work in that engine

Now it is given that 3 Gal gasoline is used in the car

So the actual gasoline used by engine will be


Q = 0.30 * 3 = 0.9 Gal

energy content of gasoline will be


E = 1.2 * 10^8 J/gal

Energy consumed by the engine will be given by


E = 1.2 * 10^8 * 0.9 = 1.08 * 10^8 J

now let say engine applied F force against the air drag to move the car

So work done by the engine = energy consumed by it


W = F*d


1.08 * 10^8 = F * 150*10^3


1.08 * 10^8 = F* 1.50* 10^5


F = (1.08* 10^8)/(1.50* 10^5)


F = 720 N

so engine must have to apply 720 N force

User Kinokijuf
by
5.5k points