Suppose a car travels $108 \mathrm{~km}$ at a speed of $35.0 \mathrm{~m} / \mathrm{s}$, 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 $1.30 \times 10^{8} \mathrm{~J}$ per gallon.)
(b) If the required force is directly proportional to speed, how many gallons will be used to drive $108 \mathrm{~km}$ at a speed of $28.0 \mathrm{~m} / \mathrm{s}$ ? (Enter a number.) gallons

Step 1 ：Given that a car travels $108 \mathrm{~km}$ at a speed of $35.0 \mathrm{~m} / \mathrm{s}$, 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 $1.30 \times 10^{8} \mathrm{~J}$ per gallon.

Step 2 ：First, we need to find the force required to move the car at the initial speed. We know that work done is equal to force times distance. We also know that the work done is 30% of the energy content of the gasoline used.

Step 3 ：We can set up the equation: \(Force = \frac{Work}{Distance}\)

Step 4 ：Substituting the given values, we get \(Force = \frac{0.3 \times Energy content of gasoline \times Amount of gasoline used}{Distance}\)

Step 5 ：Substituting the given values, we get \(Force = \frac{0.3 \times 130000000.0 \times 1.9}{108000}\)

Step 6 ：Solving the above equation, we get \(Force = 686.1111111111111\)

Step 7 ：Since the force required is directly proportional to speed, we can set up a proportion to find the new amount of gasoline used at the new speed.

Step 8 ：We can set up the equation: \(\frac{Force1}{Force2} = \frac{Speed1}{Speed2}\)

Step 9 ：Substituting the given values, we get \(\frac{686.1111111111111}{0.3 \times Energy content of gasoline \times Amount of gasoline used at new speed / 108000} = \frac{35.0}{28.0}\)

Step 10 ：Solving the above equation for the amount of gasoline used at the new speed, we get \(Amount of gasoline used at new speed = 2.375\)

Step 11 ：Final Answer: The car will use \(\boxed{2.375}\) gallons of gasoline to drive $108 \mathrm{~km}$ at a speed of $28.0 \mathrm{~m} / \mathrm{s}$.