Question

9. How many kilometers per liter will a car obtain if its engine is 25 percent efficient and it encounters an average force of 1000 Newtons? Assume the energy content of gasoline is 40,000,000 joules per liter (J/L).

Answer 1

Given:

efficiency of engine, $\eta = 25% = 0.25$

Average Force, F = 1000 N

Energy content of gasoline, E = 40000000 J/L

Solution:

Part of energy that is utilized is given by:

$E_{1} = \eta \times E$

$E_{1} = 0.25\times 40000000$

$E_{1} = 10000000 = 1\times 10^{7} J$

Nw,

To calculate the distance 's', we know that the amount of work done is the energy utilized and it can be given as the product of Force and the distance:

$E_{1} = F\times s$

⇒$s = \frac{E_{1}}{F} = \frac{10^{7}}{1000} = 10^{4}m = 10 km$

Therefore, the distance moved by the car per liter in km is 10 km