A car with mass of \(1000.0 \mathrm{kg}\) accelerates from $0 \mathrm{m} / \mathrm{s}\( to \)40.0 \mathrm{m} / \mathrm{s}\( in \)10.0 \mathrm{s} .$ Ignore air resistance. The engine has a \(22 \%\) efficiency, which means that \(22 \%\) of the energy released by the burning gasoline is converted into mechanical energy. (a) What is the average mechanical power output of the engine? (b) What volume of gasoline is consumed? Assume that the burning of \(1.0 \mathrm{L}\) of gasoline releases 46 MJ of energy.

First, we need to determine the kinetic energy gained by the car during its acceleration. To do this, we will use the formula for kinetic energy, \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the car and \(v\) is its final speed. \(KE = \frac{1}{2}(1000.0\,\mathrm{kg})(40.0\,\mathrm{m/s})^2 = 800,\!000\,\mathrm{J}\) The work done by the engine on the car is equal to the change in kinetic energy, which is the final kinetic energy minus the initial kinetic energy. Since the car starts from rest, its initial kinetic energy is 0. \(W = KE - 0 = 800,\!000\,\mathrm{J}\)

Now we will find the average mechanical power output of the engine. Power is defined as the work done per unit time. The given time for acceleration is 10 seconds. \(P_{avg} = \frac{W}{t} = \frac{800,\!000\,\mathrm{J}}{10.0\,\mathrm{s}} = 80,\!000\,\mathrm{W}\) So, the average mechanical power output of the engine is \(80,\!000\,\mathrm{W}\).

The engine has a \(22 \%\) efficiency, which means that \(22 \%\) of the energy from burning gasoline is converted into mechanical energy (work done). Therefore, we will find the total energy released from gasoline by dividing the work done by the percentage efficiency. \(E_{total} = \frac{W}{0.22} = \frac{800,\!000\,\mathrm{J}}{0.22} = 3,\!636,\!363.64\,\mathrm{J}\)

Finally, we will calculate the volume of gasoline consumed by the car. We know that burning \(1.0\,\mathrm{L}\) of gasoline releases \(46\,\mathrm{MJ}\) of energy or \(46\,\mathrm{x}\,10^6\,\mathrm{J}\). We can calculate the volume of gasoline by using the energy per liter and the total energy released. \(V_{gasoline} = \frac{E_{total}}{46\,\mathrm{x}\,10^6\,\mathrm{J/L}} = \frac{3,\!636,\!363.64\,\mathrm{J}}{46\,\mathrm{x}\,10^6\,\mathrm{J/L}} = 0.0079\,\mathrm{L}\) The volume of gasoline consumed during the acceleration is approximately \(0.0079\,\mathrm{L}\) or \(7.9\,\mathrm{mL}\). Therefore, the answers are: (a) the average mechanical power output of the engine is \(80,\!000\,\mathrm{W}\), and (b) the volume of gasoline consumed is \(0.0079\,\mathrm{L}\) or \(7.9\,\mathrm{mL}\).