I'm not sure that the rate of energy loss is proportional to drag but it is a reasonable assumption. All the other factors should be constants. Substitute 1 for the constants then the ratio of 120²/100² = 1.44.volvolugnut wrote: ↑13 Dec 2019, 08:27 Then the drag increase is square of 120 MPH and 100 MPH. 14,400 divided by 10,000 is 1.44 or 44% increase in drag.
The energy lost due to drag = Fdrag⋅displacement. To check your assumption, you would need to calculate the drag for each speed, know the displacement, calculate the energy lost each speed, then get the ratio.
I do think you are on the correct path. Another equation states: The power needed to push an object through a fluid increases as the cube of the velocity.
What surprised me is Drag and Kinetic energy increased at the same rate. Kinetic energy would increase 44% if there was no drag. Then again it is expected if all the other variables can be substituted with 1.
Kinetic energy = 1/2 * Mass * Velocity² .
Ke 100 = 1/2 * M * 100²
Ke 120 = 1/2 * M * 120²
(0.5 * M * 100²) / (0.5 * M * 120²) =
(0.5/0.5) * (M/M) * (100²/120²) =
1 * 1 * 100²/120² =
100²/120²
Same ratio you got in the drag equation.
