Given:
efficiency of engine, [tex]\eta = 25% = 0.25[/tex]
Average Force, F = 1000 N
Energy content of gasoline, E = 40000000 J/L
Solution:
Part of energy that is utilized is given by:
[tex]E_{1} = \eta \times E[/tex]
[tex]E_{1} = 0.25\times 40000000[/tex]
[tex]E_{1} = 10000000 = 1\times 10^{7} J[/tex]
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:
[tex]E_{1} = F\times s[/tex]
⇒[tex]s = \frac{E_{1}}{F} = \frac{10^{7}}{1000} = 10^{4}m = 10 km[/tex]
Therefore, the distance moved by the car per liter in km is 10 km
