Question

Suppose a car travels\(108{\rm{ km}}\)at a speed of\(30.0\,{\rm{m}}/{\rm{s}}\), and uses\(2.0{\rm{ gal}}\)of gasoline. Only\(30\% \)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\(108{\rm{ km}}\) at a speed of \(28.0{\rm{ m}}/{\rm{s}}\)?

Short answer

(a) The magnitude of force exerted to keep the car moving is\(666.67{\rm{ N}}\).

(b) The fuel used will be \(1.87{\rm{ gal}}\).

Step-by-step solution

Definition of Concept

Efficiency: Efficiency of any system is the ratio of total useful output energy to the total input energy.

Calculating the energy produced by burning gasoline

(a)

The amount of energy released by burning\(1.0{\rm{ gal}}\)of gasoline is\(1.2 \times {10^8}{\rm{ J}}\). Therefore, the produced by burning\(2.0{\rm{ gal}}\)of gasoline is,

\(\begin{aligned}{}E &= 2 \times \left( {1.2 \times {{10}^8}{\rm{ J}}} \right)\\ &= 2.4 \times {10^8}{\rm{ J}}\end{aligned}\)

Since only\(30\% \)of this energy is used in moving the car. Therefore, the amount of work done in moving the car is,

\(\begin{aligned}{}W &= \left( {2.4 \times {{10}^8}{\rm{ J}}} \right) \times \frac{{30}}{{100}}\\ &= 7.2 \times {10^7}{\rm{ J}}\end{aligned}\)

The work done is,

\(W = Fd\)

Here,\(F\)is the force and\(d\)is the displacement.

The expression for the force is,

\(F = \frac{W}{d}\)

Putting all known values,

\(\begin{aligned}{}F &= \frac{{7.2 \times {{10}^7}{\rm{ J}}}}{{108{\rm{ km}}}}\\ &= \frac{{7.2 \times {{10}^7}{\rm{ J}}}}{{108{\rm{ km}} \times \left( {\frac{{{{10}^3}{\rm{ m}}}}{{1{\rm{ km}}}}} \right)}}\\ &= 666.67{\rm{ N}}\end{aligned}\)

Therefore, the required magnitude of force exerted to keep the car moving is \(666.67{\rm{ N}}\).

Calculating the amount of fuel used

(b)

Since, the force is directly proportional to speed i.e.,

\(F \propto v\)

Here,\(v\)is the speed of the car.

The force is directly proportional for the work done or energy used i.e.,

\(E \propto F\)

The energy generated by the burning gasoline is directly proportional to the amount of the gasoline burnt i.e.,

\(m \propto E\)

Here, m is the amount of gasoline burnt.

From equations (1.1), (1.2), and (1.3) we get,

\(m \propto v\)

Here, the amount of the gasoline burnt is directly proportional to the speed of the car.

As a result,

\(\frac{{m'}}{m} = \frac{{v'}}{v}\)

Here,\(m\)is the amount of gasoline burnt to keep car moving with speed\(v = 30{\rm{ m}}/{\rm{s}}\)and\(m'\)is the amount of gasoline burnt to keep car moving with speed\(v' = 28{\rm{ m}}/{\rm{s}}\).

As a result, the amount of gasoline\(m'\)required to keep car moving with speed\(v' = 28{\rm{ m}}/{\rm{s}}\)is,

\(\begin{aligned}{}m' &= \frac{{mv'}}{v}\\ &= \frac{{\left( {2{\rm{ gal}}} \right) \times \left( {28{\rm{ m}}/{\rm{s}}} \right)}}{{\left( {30{\rm{ m}}/{\rm{s}}} \right)}}\\ &= 1.87{\rm{ gal}}\end{aligned}\)

Therefore, the required fuel used will be \(1.87{\rm{ gal}}\).