A car with mass of $1120 \mathrm{~kg}$ 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 $22.0 \%$ efficiency, which means that $22.0 \%$ of the energy released by the burning gasoline is converted into mechanical energy.
What is the average mechanical power output of the engine?
$\mathrm{KW}$

Step 1 ：Given that the mass of the car, \(m = 1120 \, \text{kg}\), the final velocity of the car, \(v = 40.0 \, \text{m/s}\), the time taken, \(t = 10.0 \, \text{s}\), and the efficiency of the engine, \(\eta = 0.22\).

Step 2 ：First, we calculate the final kinetic energy of the car using the formula \(\frac{1}{2}mv^2\). Since the car starts from rest, the initial kinetic energy is 0. Therefore, the work done by the car, which is equal to the change in kinetic energy, is also the final kinetic energy.

Step 3 ：Substituting the given values into the formula, we get \(\frac{1}{2} \times 1120 \, \text{kg} \times (40.0 \, \text{m/s})^2 = 896000.0 \, \text{J}\).

Step 4 ：The average power output of the engine is the work done by the car divided by the time taken. Therefore, the average power output is \(\frac{896000.0 \, \text{J}}{10.0 \, \text{s}} = 89600.0 \, \text{W}\) or \(89.6 \, \text{KW}\).

Step 5 ：However, only 22% of the energy released by the burning gasoline is converted into mechanical energy. Therefore, the total energy released by the burning gasoline is the work done by the car divided by 0.22.

Step 6 ：Substituting the given values into the formula, we get \(\frac{896000.0 \, \text{J}}{0.22} = 4072727.272727273 \, \text{J}\) or \(4072.727272727273 \, \text{KW}\).

Step 7 ：Final Answer: The average mechanical power output of the engine is \(\boxed{407.27 \, \text{KW}}\).