Questions: A 24% efficient engine accelerates a 1,808 kg car from rest to 9.5 m / s. How much energy is transferred to the engine by burning gasoline?

Solution Steps

First, we need to calculate the kinetic energy (\( KE \)) of the car when it reaches a speed of \( 9.5 \, \text{m/s} \). The formula for kinetic energy is:

\[ KE = \frac{1}{2} m v^2 \]

where \( m = 1808 \, \text{kg} \) is the mass of the car and \( v = 9.5 \, \text{m/s} \) is the velocity.

\[ KE = \frac{1}{2} \times 1808 \, \text{kg} \times (9.5 \, \text{m/s})^2 \]

\[ KE = \frac{1}{2} \times 1808 \times 90.25 \]

\[ KE = 81444.2 \, \text{J} \]

The engine is 24% efficient, meaning only 24% of the energy from burning gasoline is converted into kinetic energy. To find the total energy transferred to the engine, we use the efficiency formula:

\[ \text{Efficiency} = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \]

Rearranging for the total energy input:

\[ \text{Total Energy Input} = \frac{\text{Useful Energy Output}}{\text{Efficiency}} \]

Substituting the known values:

\[ \text{Total Energy Input} = \frac{81444.2 \, \text{J}}{0.24} \]

\[ \text{Total Energy Input} = 339350.8333 \, \text{J} \]

Final Answer