Question

Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 1.90 gallons of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (The energy content of gasoline is 1.30 ✕ 108 J per gallon.)
(a) What is the force (in N) exerted to keep the car moving at constant speed? 2287.0 N
(b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s? gallons.

Answer 1

Answer:

686.11 N

1.7733 gallons

Explanation:

$\eta$ = Efficiency = 30%

V = Volume of gasoline

E = Energy content of gasoline = $1.3\times 10^8\ J/gal$

F = Force

s = Displacement = 108000 m

v = Velocity

Work done is given by

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{VE\eta}{s}\\\Rightarrow F=\frac{1.9\times 0.3\times 1.3\times 10^8}{108000}\\\Rightarrow F=686.11\ N$

The force required to keep the car moving at a constant speed is 686.11 N

Here the force is directly proportional to speed

$\\\Rightarrow F=v$

$\\\Rightarrow \frac{F_1}{v_1}=\frac{F_2}{v_2}\\\Rightarrow F_2=\frac{F_1\times v_2}{v_1}\\\Rightarrow F_2=\frac{686.11\times 28}{30}\\\Rightarrow F_2=640.36\ N$

$W=F\times s\\\Rightarrow 0.3\times 1.3\times 10^8\times V=640.36\times 108000\\\Rightarrow V=\frac{640.36\times 108000}{0.3\times 1.3\times 10^8}\\\Rightarrow V=1.7733\ gal$

The gallons that will be used is 1.7733 gallons