Chapter 6, Problem 90P

(a)

To determine

The average mechanical output power of the engine.

Answer to Problem 90P

The average mechanical output power of the engine is $80.0\text{kW}$.

Explanation of Solution

Write the expression from work energy theorem.

  $W=\Delta K$        (I)

Here, $W$ is the work done, $\Delta K$ is the change in energy of the system.

The change in energy of the system is,

  $\Delta K=K_f-K_i$        (II)

Here, $K_f$ is the final kinetic energy of the car and $K_i$ is the initial kinetic energy of the car.

Rewrite the above equation.

  $\begin{array}{c}\Delta K=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2\\ =\frac{1}{2}m\left(v_f^2-v_i^2\right)\end{array}$        (III)

Write the expression for mechanical power output of the engine.

  $P=\frac{W}{t}$        (IV)

Here, $P$ is the mechanical power output of the engine and $t$ is the time period.

Conclusion:

Substitute $1000\text{kg}$ for $m$, $40\text{m/s}$ for $v_f$, and $0\text{m/s}$ for $v_i$ in equation (III) to find $\Delta K$.

  $\begin{array}{c}\Delta K=\frac{1}{2}\left(1000\text{kg}\right)\left[{\left(40\text{m/s}\right)}^2-{\left(0\text{m/s}\right)}^2\right]\\ =8.0\times {10}^5\text{J}\end{array}$

Substitute $8.0\times {10}^5\text{J}$ for $\Delta K$ in equation (I) to find $W$

  $W=8.0\times {10}^5\text{J}$

Substitute $8.0\times {10}^5\text{J}$ for $W$ and $10.0\text{s}$ for $t$ in equation (IV) to find $P$

  $\begin{array}{c}P=\frac{8.0\times {10}^5\text{J}}{10.0\text{s}}\\ =80.0\times {10}^3\text{W}\left(\frac{{10}^{-3}\text{kW}}{1\text{W}}\right)\\ =80.0\text{kW}\end{array}$

Therefore, the average mechanical output power of the engine is $80.0\text{kW}$.

(b)

To determine

The volume of the gasoline consumed.

Answer to Problem 90P

The volume of the gasoline consumed is $79\text{mL}$.

Explanation of Solution

Write the expression for total energy needed for the consumption of 22% .

  $E=\frac{\Delta K}{22\%}$        (V)

Here, $E$ is the total energy.

The volume of the gasoline consumed is,

  $V^{\prime }=V\frac{E}{E^{\prime }}$        (VI)

Here, $V^{\prime }$ is the volume of the gasoline consumed, $V$ is the volume of the burning fuel, and $E^{\prime }$ is the energy releases from the burning of gasoline.

Conclusion:

Substitute $8.0\times {10}^5\text{J}$ for $\Delta K$ in equation (V) to find $E$.

  $\begin{array}{c}E=\frac{\left(8.0\times {10}^5\text{J}\right)}{\frac{22}{100}}\\ =\frac{8.0\times {10}^5\text{J}}{0.22}\\ =36.4\times {10}^5\text{J}\end{array}$

Substitute $36.4\times {10}^5\text{J}$ for $E$, $1.0\text{L}$ for $V$, and $46\text{MJ}$ for $E^{\prime }$ in equation (VI) to find $V^{\prime }$.

  $\begin{array}{c}{V}^{\prime }=\left(1.0\text{L}\right)\frac{36.4\times {10}^5\text{J}}{46\text{MJ}\left(\frac{{10}^6\text{J}}{1\text{MJ}}\right)}\\ =\left(1.0\text{L}\right)\frac{36.4\times {10}^5\text{J}}{46\times {10}^6\text{J}}\\ =79\times {10}^{-3}\text{L}\left(\frac{{10}^3\text{mL}}{1\text{L}}\right)\\ \approx 79\text{mL}\end{array}$

Therefore, the volume of the gasoline consumed is $79\text{mL}$.