Will kinetic friction change when normal force and coefficient of friction remains the same?

Based on common knowledge, a car can do uniform motion in any velocity when the applied force(by engine) do not exceed the maximum kinetic force of friction which can be calculated by: $$\ F_f = \mu_k F_N$$ The object can move uniformly because of the force of friction can balance with the applied force as long as it does not exceed the maximum $\ F_f$.

My question is, isn't for a moving object, where its $\mu_k$ and $F_N$remains the same, the force of friction is the same as well (ignore the instability of friction). How can $\ F_f$ balance any magnitude of applied force if $\ F_f$ won't change?

• "the force of friction can balance" Are you sure? The friction force doesn't balance. The object can move uniformly, if the resultant force act on it is zero. Commented Apr 28, 2016 at 5:27

You are mixing up concepts.

The kinetic friction of a car has nothing to do with $F_N$.
The concept of friction as taught in school tells you, that friction is independent of velocity. This is clearly not realistic for a car! The friction - let's rather call it "air resistance", for this is the main part - has a very complicated dependence on the velocity - but it usually rises if you drive faster :)

You can keep at constant speed if all forces on you balance: the air resistance, that tries to stop you and the force that pushes you forward.
What is this force? Again friction. If there were no friction, the tires would move backwards on the road. So this is the force pushing you. For this friction the concept of static friction is suited in a good approximation.

So to you side-question: yes, for the concept of friction you learned in school, it will remain the same if you fix $F_N$ and $\mu$ - in particular independent of velociry. But this is not how reality works.

The resistance to motion in a car is not due to simple friction. At low speeds it is dominated by viscous losses in the engine and transmission while at high speeds it is mostly due to aerodynamic drag.

Viscous losses are approximately proportion to the velocity of the car:

$$F_v = Av$$

for some constant $A$. Aerodynamic drag is approximately proportional to the velocity squared:

$$F_d = Bv^2$$

for some constant $B$. So if the force the engine can produce is $F_e$ then the velocity of the car will be given by:

$$F_e = F_v + F_d = Av + Bv^2$$

This rearranges to a quadratic equation in $v$:

$$Bv^2 + Av - F_e = 0$$

And solving this will give you the velocity of the car. The constants $A$ and $B$ will vary from car to car because they depend on the design and the size of the car, and there is no simple way to guess what these constants are.