College Physics by Openstax Chapter 7 Problem 4

Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (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?

Solution:

Part A

According to Table 7.1, the energy in 1 gallon of gasoline is 1.2×108 J1.2 \times 10^{8}\ \text{J}. Since only 30% of the gasoline goes into useful work, the work done by the friction WfW_{f} is

Wf=0.30(2.0 gal)(1.2×108 J/gal)Wf=72×106 J\begin{align*} W_{f} & =0.30 \left( 2.0\ \text{gal} \right)\left( 1.2 \times 10^{8} \ \text{J/gal}\right) \\ W_{f} & = 72 \times 10^{6}\ \text{J} \end{align*}

Now, the work done by the friction can also be calculated using the formula below, where FfF_{f} is the magnitude of the friction force that keeps the car moving at constant speed, and dd is the distance traveled by the car.

Wf=Ffd\begin{align*} W_{f}=F_{f}d \end{align*}

We can solve for FfF_{f} in terms of the other variables.

Ff=WfdF_{f} = \frac{W_{f}}{d}

Substituting the given values, we can now solve for the unknown magnitude of the force exerted to keep the car moving at constant speed.

Ff=WfdFf=72×106 J108 kmFf=72×106 N⋅m108×103 mFf=666.6667 NFf=6.7×102 N  (Answer)\begin{align*} F_{f} & = \frac{W_{f}}{d} \\ F_{f} & = \frac{72 \times 10^{6}\ \text{J}}{108\ \text{km}} \\ F_{f} & = \frac{72 \times 10^{6}\ \text{N}\cdot \text{m}}{108 \times 10^{3}\ \text{m}} \\ F_{f} & = 666.6667\ \text{N} \\ F_{f} & = 6.7 \times 10^{2}\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

If the required force is directly proportional to speed, then there must be a linear relationship between the required force and speed. In this situation, we can just simply used ratio and proportion to compute for the number of gallons.

2.0 gal30.0 m/s=x28.0 m/sx=(2.0 gal)(28.0 m/s)30.0 m/sx=1.8667 galx=1.9 gal  (Answer)\begin{align*} \frac{2.0\ \text{gal}}{30.0\ \text{m/s}} & = \frac{x}{28.0\ \text{m/s}} \\ x & = \frac{\left( 2.0\ \text{gal} \right)\left( 28.0\ \text{m/s} \right)}{30.0\ \text{m/s}} \\ x & = 1.8667\ \text{gal} \\ x & = 1.9\ \text{gal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}