- Calculate the change in kinetic energy: \(\Delta K = \frac{1}{2}mv_f^2 = 0.5 \times 2000 \text{ kg} \times (12 \text{ m/s})^2 = 144,000 \text{ J}\).
- Use the efficiency formula to find input energy: \(E_{in} = \frac{\Delta K}{e} = \frac{144,000 \text{ J}}{0.20} = 720,000 \text{ J}\).
- Convert to kilojoules: \(720,000 \text{ J} = 720 \text{ kJ}\).
- Final Answer: \(\boxed{720 \text{ kJ}}\).
Explanation
- Problem Analysis and Physical Principles
To determine the total energy transferred to the engine by burning gasoline, we begin by identifying the physical parameters provided: the mass of the car is \(m = 2000 \text{ kg}\), the initial velocity is \(v_i = 0 \text{ m/s}\), the final velocity is \(v_f = 12 \text{ m/s}\), and the engine efficiency is \(e = 20\% = 0.20\). The energy transferred to the car manifests as an increase in its kinetic energy, denoted as \(\Delta K\). The efficiency of the engine, \(e\), is defined as the ratio of the useful output energy (\(\Delta K\)) to the total input energy (\(E_{in}\)) provided by the gasoline. Therefore, we can establish the relationship \(e = \frac{\Delta K}{E_{in}}\), which allows us to solve for the input energy as \(E_{in} = \frac{\Delta K}{e}\).
- Calculating the Change in Kinetic Energy
The change in kinetic energy \(\Delta K\) is calculated using the work-energy theorem, which states that the net work done on an object equals its change in kinetic energy:
\[\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\]
Substituting the known values \(m = 2000 \text{ kg}\), \(v_f = 12 \text{ m/s}\), and \(v_i = 0 \text{ m/s}\):
\[\Delta K = \frac{1}{2} \times 2000 \text{ kg} \times (12 \text{ m/s})^2 - \frac{1}{2} \times 2000 \text{ kg} \times (0 \text{ m/s})^2\]
\[\Delta K = 1000 \times 144 = 144,000 \text{ J}\]
The total change in kinetic energy of the car is \(144,000 \text{ Joules}\).
- Determining Total Energy Transferred to the Engine
Now that we have the change in kinetic energy, we apply the engine efficiency formula \(E_{in} = \frac{\Delta K}{e}\). With \(\Delta K = 144,000 \text{ J}\) and \(e = 0.20\), we calculate the total input energy:
\[E_{in} = \frac{144,000 \text{ J}}{0.20} = 720,000 \text{ J}\]
To convert this value to kilojoules (kJ), we divide by \(1000\):
\[E_{in} = \frac{720,000}{1000} = 720 \text{ kJ}\]
- Final Answer
The total energy transferred to the engine by burning gasoline is \(\boxed{720 \text{ kJ}}\).
Examples
Understanding energy efficiency is critical in real-world engineering, such as designing more fuel-efficient vehicles or optimizing electrical power systems. For instance, an engineer evaluating an electric motor must account for how much electrical energy is input versus how much mechanical work is output, which helps in calculating operational costs and environmental impact. By mastering these calculations, one can better analyze the performance and sustainability of various mechanical and electrical systems.