Question

Suppose a miracle car has a 100% efficient engine and burns fuel that has a 40-megajoules-per-liter energy content. If the air drag and overall frictional forces on this car traveling at highway speeds total 1000 N, what is the overall limit in distance per liter it could be driven on the highway

Answer 1

Answer:

The car can travel at a limit of 40 km/liter on the highway.

Explanation:

We can solve this by equating the energy per liter of fuel to the product of distance and force of resistance. This means the following:

Energy = Force * Distance

(40000000 Joules) = (1000 N) * Distance

To travel the maximum amount of distance, our car needs to maintain a constant speed, where the only energy needed is to push back against air resistance and friction.

Solving the above equation, we get:

Distance = 40000000 / 1000

Distance = 40000 meters

Distance = 40 km

Thus, the car can travel 40 km/liter on the highway.

Answer 2

Answer:

The overall limit in distance per liter it could be driven on the highway is 40,000 m/L.

Explanation:

It was assumed the efficiency of the miracle car engine is 100%, therefore, energy input equals energy output.

Energy input = 40 MJ/L = 40×10^6 = 4×10^7 J/L

Energy output = 4×10^7 J/L = 4×10^7 Nm/L

Frictional force = 1000 N

Overall limit in distance per liter = energy output ÷ frictional force = 4×10^7 Nm/L ÷ 1000 N = 40,000 m/L