The max power an engine puts out is what determines the max speed a car travel at right, obviously it's more nuanced than that because there are losses in the drive train, and the shape of the car determines it's drag coefficient, but the power output of the engine is the still the top limit and the rest falls out of that.
So here's my question, if I had a manual transmission car and I put it into first gear and let's say the max power occurs at 6000 RPM and the gear ratio in first gear is 3:1 and the final drive ratio is 2:1.
Now let's say I cruise in first gear at the max power output, so my tries are moving at 1000 RPM and for arguments sake lets say that transmits to 40 MPH. And the torque at the tires is 6 times the value it would be at 6000 RPM, I understand that higher gears if I run the engine at max power I get to the top speed the car is capable of and this limit is imposed by the drag power increasing proportionally to velocity cubed, ignoring the losses due to transmission friction and rolling friction and what not.
But in first gear I am going to slow for the power of the drag to be enough to counter that of the engine. But where does the excess power go while in first gear?
There's a lot of torque being produced at the tire and that represent a large force against the pavement and if there's no slipping this is what causes the car to move forward right?
But once we're cruising with the tries at 1000 RPM the force at the tires is still large the engine just can't turn any faster, so what is the offsetting force that causes the car to stop accelerating if it isn't drag?
Is it the engine itself braking the while simultaneously providing the force to keep the car going?
Sorry for rambling, I've been thinking a lot about this question the past couple of days and I have not been able to figure out exactly how these forces are being balanced.