figure the car went 100km. Using the force of friction, force of wind drag, with distance,that is the work the car did on friction.
Because of efficiency, you used 5 times this amount of energy of fuel. Now figure the amount of fuel you used.
and you then have km/liter.
A car experiences rolling (tractive) friction with coe�cient of friction mut = 0.0200, and turbulent-flow (bv^2) drag, with drag coefficient CD = 0.300. The car has mass (including
fuel, which is a negligible fraction of the total) M = 1000. kg and frontal cross-sectional area S = 2.00 m^2. The density of air is approximately 1.20 kg/m^3. The heat of combustion
of gasoline is approximately 36.0 MJ per liter, but when thermal and mechanical losses are considered only 20.0 percetn of this is available for the work of propulsion. Calculate the gas mileage (km/liter will do) of this car traveling on a straight, level road at a constant speed
of 30.0 m/s.
2 answers
I don't understand very well what I have to do...What I did before was
v= P/W
P= (mu(mg)+(1/2)(CD)(rho)Sv^3)
w=36.0 MJ*20%
V=((mu(mg)+(1/2)(CD)(rho)Sv^3)/ (36x10^6 *0.2)
I plugged in the values and got .00217 L/s and then to get the gas mileage=(0.00218^-1 x30)/1000
13.84 km/L
v= P/W
P= (mu(mg)+(1/2)(CD)(rho)Sv^3)
w=36.0 MJ*20%
V=((mu(mg)+(1/2)(CD)(rho)Sv^3)/ (36x10^6 *0.2)
I plugged in the values and got .00217 L/s and then to get the gas mileage=(0.00218^-1 x30)/1000
13.84 km/L
2 answers