A car advertisement claims that its \(900 - {\rm{kg}}\) car accelerated from rest to \(30.0{\rm{ m}}/{\rm{s}}\) and drove \(100{\rm{ km}}\), gaining \(3.00{\rm{ km}}\) in altitude, on \(1.0{\rm{ gal}}\) of gasoline. The average force of friction including air resistance was \(700{\rm{ N}}\). Assume all values are known to three significant figures.

(a) Calculate the car’s efficiency.

(b) What is unreasonable about the result?

(c) Which premise is unreasonable, or which premises are inconsistent?

Efficiency: The term efficiency is defined as the ratio of useful output energy to the total input energy.

Mathematically,

\(\eta \% = \frac{{{\rm{Useful output energy}}}}{{{\rm{Total input energy}}}} \times 100\% \)

(a)

The total energy used by car is,

\({E_{out}} = \frac{1}{2}m{v^2} + mgh + fd\)

Here, \(m\) is the mass of the car \(\left( {m = 900{\rm{ kg}}} \right)\), \(v\) is the velocity of car \(\left( {v = 30.0{\rm{ m}}/{\rm{s}}} \right)\), \(h\) is the gain in altitude \(\left( {h = 3.00{\rm{ km}}} \right)\), \(g\) is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), \(f\) is the average force of friction \(\left( {f = 700{\rm{ N}}} \right)\), and \(d\) is the distance traveled by the car \(\left( {d = 100{\rm{ km}}} \right)\).

Putting all known values,

\(\begin{array}{c}{E_{out}} = \left[ \begin{array}{l}\frac{1}{2} \times \left( {900{\rm{ kg}}} \right) \times {\left( {30{\rm{ m}}/{\rm{s}}} \right)^2} + \left( {900{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {3{\rm{ km}}} \right)\\ + \left( {700{\rm{ N}}} \right) \times \left( {100{\rm{ km}}} \right)\end{array} \right]\\ = \left[ \begin{array}{l}\frac{1}{2} \times \left( {900{\rm{ kg}}} \right) \times {\left( {30{\rm{ m}}/{\rm{s}}} \right)^2} + \left( {900{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {3{\rm{ km}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right)\\ + \left( {700{\rm{ N}}} \right) \times \left( {100{\rm{ km}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right)\end{array} \right]\\ = 9.6865 \times {10^7}{\rm{ J}}\end{array}\)

The energy input of \(1{\rm{ gal}}\) of gasoline is,

\({E_{in}} = 1.2 \times {10^8}{\rm{ J}}\)

The efficiency of the car is,

\(\eta \% = \frac{{{E_{out}}}}{{{E_{in}}}} \times 100\% \)

Putting all known values,

\(\begin{array}{c}\eta = \frac{{9.6865 \times {{10}^7}{\rm{ J}}}}{{1.2 \times {{10}^8}{\rm{ J}}}} \times 100\% \\ = 80.72\% \end{array}\)

Therefore, the required efficiency of the car is \(80.72\).

(b)

From table 7.2: Energy of various objects and Phenomenon, the efficiency of a gasoline engine is about \(30\% \) only, but the efficiency calculated in part (a) is \(80.72\).

Hence, the efficiency is unreasonably high.

(c)

\(1{\rm{ gal}}\) of gasoline is inconsistent with \(100{\rm{ km}}\) traveled, especially when the car is going to an altitude of \(3{\rm{ km}}\) because as gaining altitude will cause in increase in energy consumption.