Question

A 30%-efficient car engine accelerates the 1800 kg car from rest to 11 m/s.

Part A
How much energy is transferred to the engine by burning gasoline in km?
Express your answer using two significant figures.

Answer 1

Concepts and reason

The concepts required to solve this problem are work done and efficiency of engine

Initially, calculate the the work done by the car engine using the work-energy theorem. Finally, calculate the energy transferred to the engine by using the efficiency of the engine.

Fundamentals

The work done by the engine in moving the car from initial velocity u to final velocity v is as follows:

$W = \frac{1}{2}m\left( {{v^2} - {u^2}} \right)$

Here, W is the work done, m is the mass of the car, v is the final velocity, and u is the initial velocity.

The efficiency of an engine is given by the following expression:

$\eta = \frac{W}{{\Delta Q}}$

Here, $\eta$ is the efficiency, W is the work done and $\Delta Q$ is the energy transferred.

The engine starts from rest hence, its initial velocity is 0 m/s.

The work done by the engine in moving the car from initial velocity to final velocity is as follows:

$W = \frac{1}{2}m\left( {{v^2} - {u^2}} \right)$

Substitute 1800 kg for m, 0 m/s for u, and 11 m/s for v in the above expression.

$\begin{array}{c}\\W = \frac{1}{2}\left( {1800{\rm{ kg}}} \right)\left( {{{\left( {11\;{\rm{m/s}}} \right)}^2} - {{\left( {0{\rm{ m/s}}} \right)}^2}} \right)\\\\ = 1.089 \times {10^5}{\rm{ J}}\\\end{array}$

The expression for the change in energy is,

$\Delta Q = \frac{W}{\eta }$

Substitute 30% for $\eta$ and $1.089 \times {10^5}{\rm{ J}}$ for W in the above expression.

$\begin{array}{c}\\\Delta Q = \frac{{1.089 \times {{10}^5}{\rm{ J}}}}{{30\% }}\\\\ = \frac{{1.089 \times {{10}^5}{\rm{ J}}}}{{\left( {\frac{{30}}{{100}}} \right)}}\\\\ = 3.6 \times {10^5}{\rm{ J}}\\\end{array}$

Ans: Part A

The energy transferred is $3.6 \times {10^5}{\rm{ J}}$ .