Problem#1
Suppose the empty car described in Table 7.2 has a fuel economy of 6.40 km/liter (15 mi/gal) when traveling at 26.8 m/s (60 mi/h). Assuming constant efficiency, determine the fuel economy of the car if the total mass of passengers plus driver is 350 kg.Answer:
At a speed of 26.8 m/s (60.0 mph), the car described in Table 7.2 delivers a power of P1 = 18.3 kW to the wheels. If an additional load of 350 kg is added to the car, a larger output power of
P2 = P1 + (power input to move 350 kg at speed v)
will be required. The additional power output needed to move 350 kg at speed v is:
∆P = (∆f)v = µrmgv
Assuming a coefficient of rolling friction of µr = 0.0160, the power output now needed from the
engine is
P2 = 18.3 kW + 0.0160(350 kg)(9.80 m/s2)(26.8 m/s) = 19.8 kW
With the assumption of constant efficiency of the engine, the input power must increase by the
same factor as the output power. Thus, the fuel economy must decrease by this factor:
(fuel economy)2 = (P1/P2)(fuel economy)1
(fuel economy)2 = (18.3/19.8)(6.40 km/L) = 5.92 km/L
Problem#2
A compact car of mass 900 kg has an overall motor efficiency of 15.0%. (That is, 15% of the energy supplied by the fuel is delivered to the wheels of the car.) (a) If burning one gallon of gasoline supplies 1.34 x 108 J of energy, find the amount of gasoline used in accelerating the car from rest to 55.0 mi/h. Here you may ignore the effects of air resistance and rolling friction. (b) How many such accelerations will one gallon provide? (c) The mileage claimed for the car is 38.0 mi/gal at 55 mi/h.
What power is delivered to the wheels (to overcome frictional effects) when the car is driven at this speed?
Answer:
(a) the amount of gasoline used in accelerating the car from rest to 55.0 mi/h = 24.6 m/s is
Fuel needed = ∆K/(useful energy per gallon)
Fuel needed = ∆K/[eff. X energy content of fuel)
With
∆K = ½ mvf2 – 0 = ½ (900 kg)(24.6 m/s)2 = 2.72 x 106 J, then
Fuel needed = (2.72 x 106 J)/[0.150 x 1.34 x 108 J/gal) = 1.35 x 10-2 gal
(b) Many such accelerations will one gallon provide is
1/(1.35 x 10-2 gal) = 73.9/gal
(c) Power is delivered to the wheels (to overcome frictional effects) when the car is driven at this speed is
P = (15%)(1 gal/38.0 mi)(55.0 mi/1.00 h)(1.00 h/3,600 s)(1.34 x 108 J/1 gal)
P = 8.08 kW
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