Given:
efficiency of engine,
![\eta = 25% = 0.25](https://img.qammunity.org/2020/formulas/physics/college/k2xihffjgb29kpj25ytmp00tryobc9kfld.png)
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 * E](https://img.qammunity.org/2020/formulas/physics/college/tgsqvz164uuprgr7b44onzaoekmgm1zxt4.png)
![E_(1) = 0.25* 40000000](https://img.qammunity.org/2020/formulas/physics/college/j841nargsxwbzhprp84k3drp2042ysgbn0.png)
![E_(1) = 10000000 = 1* 10^(7) J](https://img.qammunity.org/2020/formulas/physics/college/jretz2j2sihg02u6nhdg0vv15ybj0i9ndk.png)
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* s](https://img.qammunity.org/2020/formulas/physics/college/dx9i2mbmi87r36xzkdr09vyyaemhtf9zny.png)
⇒
![s = (E_(1))/(F) = (10^(7))/(1000) = 10^(4)m = 10 km](https://img.qammunity.org/2020/formulas/physics/college/apyxkwwik7790eyqgwwuzxn6sg22jqa9ch.png)
Therefore, the distance moved by the car per liter in km is 10 km