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.

[tex]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[/tex]

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

Here the force is directly proportional to speed

[tex]\\\Rightarrow F=v[/tex]

[tex]\\\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[/tex]

[tex]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[/tex]

The gallons that will be used is 1.7733 gallons

samuelmeier187 samuelmeier187 · Answer 2 · 2021-11-15T17:40:04+01:00

Answers and Explanation:

a)

As 30% of the gasoline goes into useful work by the force against the friction, the work done by the force against the friction is equal to 30% of the energy of the 2 gallon gasoline. this energy is also equal to work done by friction of the care.

One gallon of gasoline contains 1.2*10^8J energy. Therefore, The work done by the friction force is:

Wf = (2 gal)(1.2*10^8 J/gal)(30%)

= 7.2*10^7J

Now, work done by the friction force is:

Wf = Ffd

Rewrite the above equation for Ff:

Ff = Wf/d

Then you substitute the values for the knowns:

= 7.2*10^7J/108 km(10^3m/1km)

= 666.7 N

b)

As the car travels 108 km at a speed of 30.0 m/s, and uses 2 gallons of gasoline and the force is directly proportional to the speed. SO, the amount of gasoline required to drive the car 108 km at a speed of 28 m/s is:

(28 m/s/30 m/s)2 gal = 1.867 gal

OR

To drive the same distance 108 km in 28 m/s:

V2 = (u2/u1)V1

Here u1 and u2 are the velocities and V1 is the initial volume of gasoline when driving at the speed of 30 m/s and V2 is the volume of gasoline when driving at the speed of 28 m/s. And so we substitute the knowns:

V2 = (28 m/s/30 m/s)2 gal = 1.867 gal

If you have any questions please comment below.