91Ó°ÊÓ

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

(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 \(666.67{\rm{ N}}\).

(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 \(1.87{\rm{ gal}}\).