Lets say that a car engine can output a maximum power of $P$. Now initially, the car starts from rest would have zero velocity.

Now the tires start slipping and a force of $f_k = \mu mg$ will act on the car accelerating it forward where $\mu$ is the friction coefficient between the ground and the tires.

Now the car starts accelerating with $a = \frac{f_k}{m}$ and the power output of the car engine slowly rises until it reaches $P$. $f_k \cdot v = P \implies v = \frac{P}{f_k}$

Now I couldn't understand what exactly would happen after the power output of the engine reaches $P$ but these were my thoughts :

  1. Since there is a external force $f_k$ acting, it feels like it would still accelerate but that would increase its velocity, but that would increase its power beyond $P$ which is not possible.

  2. The velocity($v$) would increase and the external force($F$) would be less than the maximum kinetic friction($F < f_k$) such that $F \cdot v = P$. But I couldn't think about how that would work because the tire can only either slip or roll on the surface which would leave room for only 2 values of $F$ which are $f_k$ and $0$.

So please help me understand how exactly this situation proceeds after the engine has achieved power output $P$

  • 1
    $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Jan 11 at 20:02
  • $\begingroup$ I believe your thought #2 at the end is not correct. In the "roll on the surface case," your possible value of F is not just 0, but any value below the $\mu_s mg$ value. Does that help open doors for what could happen as the car approaches maximum power? $\endgroup$
    – Cort Ammon
    Commented Jan 11 at 20:12
  • $\begingroup$ @CortAmmon. That was my first thought. but it didnt make too much sense in my mind as when the tires are rolling, there is no friction force and when the tires are slipping, the force would be $f_k$. So I couldn't see how a force other than $f_k$ or $0$ acts on the car. Feels like my thinking could be wrong but came here for a better explanation to support this. $\endgroup$ Commented Jan 12 at 5:34
  • $\begingroup$ @CortAmmon Would like if you would create an answer explaining how the friction force could be anything in between $0$ and $\mu_s mg$. $\endgroup$ Commented Jan 12 at 5:48
  • $\begingroup$ Why do you think that the friction must be zero if the tires are rolling? $\endgroup$
    – Jakob KS
    Commented Jan 12 at 7:57

6 Answers 6


Regarding the car scenario, the size of the car's engine determines how powerful it is and hence how fast the car can accelerate when the engine is running at its maximum speed. It also establishes how fast the car will be going when it reaches its "terminal velocity" due to air drag.

A powerful engine will accelerate the car's mass faster, and achieve a higher maximum speed, than a less powerful engine.

You can buy an app with a plug-in acceleration sensor that runs on a laptop with GPS which solves for the acceleration of a car and its resulting speed in real time. If you insert the value of the car's mass in the app, it automatically solves for and graphs the power output of the engine.

  • $\begingroup$ This is an ideal situation with no air drag. the mass of the car $m$ is only the body and the wheels have negligible mass when compared to the body. Wanted to first understand how the stuff works before getting to air resistance and stuff. Thanks for the answer. $\endgroup$ Commented Jan 12 at 5:37
  • $\begingroup$ @DevNotTaken There is no way to understand this without including air resistance. $\endgroup$
    – my2cts
    Commented Jan 12 at 19:00
  • $\begingroup$ @my2cts, I got to know that. $\endgroup$ Commented Jan 12 at 19:03

For a car engine the torque and power as well as depending on the size of the engine depends on the speed (rpm) of the engine as shown below.
Gears are used to keep the engine running around the optimum speed (rpm) of the engine.

enter image description here

The torque determines the acceleration of a car.

The maximum power developed by the engine determines the car's maximum speed at which stage the maximum rate of working (power) of the engine is equal to the rate working against all the dissipative forces, eg air friction, friction in the bearings etc.

  • $\begingroup$ I have looked at this graph for a while and I do understand it properly right now. the friction force $f = \tau r$. That is why the friction force decreases after reaching the maximum torque while the velocity of the car keeps increasing. This sufficiently answered my question and I have deleted the 2 comments on your post that I had made without thinking hard enough. $\endgroup$ Commented Jan 12 at 7:46
  • $\begingroup$ @DevNotTaken No, seems you misinterpret chart. Engine torque decreases after maximum RPM, because parasitic reversal torque increases, which is ${\boldsymbol {\tau_{par} }}=\mathbf {r} \times \mathbf {F_{bearings}}$. And this is due to increasing friction in bearings and other parts in engine. $\endgroup$ Commented Jan 12 at 8:03
  • $\begingroup$ @AgniusVasiliauskas I seemed to have made a typo while writing the comment and couldn't edit it. Here, since we are ignoring friction everywhere except the ground, the $F = \frac{\tau}{r} = f$ and the friction force $f$ is the force accelerating the car. $\endgroup$ Commented Jan 12 at 8:15
  • $\begingroup$ @DevNotTaken If you are ignoring friction in bearings and other engine parts, then given charts in this post is irrelevant to you, because they exactly shows bearing friction effect. So seems like you apply "double standards" here. $\endgroup$ Commented Jan 12 at 10:36

Your question includes the key equation, $P=Fv$. Power is the force times the current velocity. Now in a real world setting, the power of an engine varies with speed (as many other answers have shown), but we'll use your idealized setting with an engine that outputs constant power at all speeds up to some maximum power limit, $P_{max}$. You've done the calculations for what happens up to that limit. Your question is what happens after that limit is achieved.

In this regime, we have a constant power, and an obviously increasing velocity. This implies the force must be decreasing. Indeed, the force applied by this idealized car on the road must be $F=\frac{P}{v}$ or else the engine is not achieving these ideals.

Your concern was that the car must either be spinning its tires, such that we use the dynamic friction equation $F=\mu_d mg$ (assuming a normal force of $mg$, as you did), or the car must have tires that are now sticking, such that static friction applies. In your question, you indicate that this means $F=0$, but this is not the case. Were that to be the case, static friction could never apply a force and nothing could be held in place by friction. In reality, the equation for static friction is $F\le \mu_s mg$. The force is whatever force is required to keep the objects stationary with respect to eachother, up to a max limit of $\mu_s mg$, at which point they start slipping. But any amount of force up to that point is valid.

So what will happen is that your car will accelerate off the line. At the start, velocity is zero, so force is infinite... that can't be right! Your idealized car is too idealized! So let's recognize that the spinning of tires at the start permits some "bleeding" of power into heat. So the car accelerates along some velocity curve. That curve will be based on the $f_k$ equations you did. What matters is that the velocity grows, and thus the corresponding force decreases.

At some velocity, the force is so low that the tires succeed in "sticking." At what force this occurs is a very complicated topic, especially with rubber tires when bend and flex and stick and do all sorts of complex physicsy things. But at some point this will occur, because your constant power engine will continue to apply less and less force, thus less and less acceleration.

Once the tires stick, we need to switch over to the static friction regime. In this regime, the force of friction is whatever is needed to satisfy the $F=ma$ equation for the car, so long as it doesn't exceed the maximum (which we've already seen because force only goes down in your scenario, never up). The fact that this force does not need to be 0 (as initially stated in the question) is how our car can keep going fast and faster with no limit (although its acceleration gets smaller and smaller)

  • $\begingroup$ This is the best answer I have received for this question. You have cleared all my queries. $\endgroup$ Commented Jan 12 at 18:47

Power is energy accumulation rate per time unit. In this case we are interested in car kinetic energy accumulation, so :

$$ \tag 1 K = Pt ,$$

substituting kinetic energy expression and solving for $t$ :

$$ \tag 2 t = \frac {mv^2}{2P},$$

(2) equation shows given car engine power how much time do you need in engine full throttle to reach speed $v$ starting from rest.

This is lovely metrics for sports-car lovers, they usually measure cars by how much seconds you need to drive car from a rest to $\approx 100~km/h$ speed.

For example Rimac Nevera can reach $97~km/h$ just in $1.74~s$. It can do so only because it's engine is like a little power station,- can generate up to $1.4~\text{MW}$ power.

  • $\begingroup$ So as per equation (2), as the time goes the velocity could keep increasing to eternity right. but $P = F \cdot v$ and since $v$ keeps increasing, $F$ has to decrease. $F$ here is nothing but the force due to friction. Now there are 2 cases. where the tires are pure rolling($F = 0$) and when the tires are slipping($F = \mu N$). Here I couldn't understand how the friction force could be anything except these 2 values. Would help if you could edit your answer explaining this. $\endgroup$ Commented Jan 12 at 5:46
  • $\begingroup$ No, velocity could not increase to infinity. In case velocities are high (substantial part of $c$), then you have to account for relativistic kinetic energy $K_{rel}$ in (1) equation instead of $K$, which would be : $$ K_{rel}=\left ({\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}} - 1 \right) mc^{2} $$. So I'm afraid it's not that simple. $\endgroup$ Commented Jan 12 at 7:13
  • $\begingroup$ And also your understanding about dissipation forces is incomplete. There's a rolling resistance force $F_{rr}=C_{rr}N$ and air drag force $F_{D}\,=\,{\tfrac {1}{2}}\,\rho \,v^{2}\,C_{D}\,A$ which car must compensate constantly. And air drag increases with increased car speed, so most power of car goes to conquer drag force. $\endgroup$ Commented Jan 12 at 7:35
  • $\begingroup$ Now that makes a lot of sense. the velocity wouldn't increase to infinity. $\endgroup$ Commented Jan 12 at 7:35
  • $\begingroup$ I do know about rolling friction and air drag. The scenario here is considered ideal where all those are ignored. $\endgroup$ Commented Jan 12 at 7:36

Power is defined as work over time $P=w/t$ or the rate at which work is done, work is defined as Force times displacement or $W=Fs$, Finally Force is defined as $F=m*a$, so another way to put that would be to define work as energy being transferred from one place to another. So power being work over time is basically how fast energy is transferred. So the maximum power would be the highest amount of work it could do in a given time frame. I didn't quite understand what was being asked in the scenario with the car but I hope this answers your question on what the variable power exactly is.


First, what is power? Power is the rate of change of energy.

For example $$\frac{d~}{dt} \frac{1}{2} mv^2={\vec f} \cdot {\vec v}$$ and $$\frac{dqV}{dt}=q{\vec \nabla}V \cdot {\vec v} ={\vec f} \cdot {\vec v}\,.$$

The power that the engine generates is the rate at which chemical energy is converted by fuel combustion. After internal losses a power P is delivered to the car. Initially this goes into tire friction and erosion, then into increasing the kinetic energy of the car until the car reaches maximum speed. At this point all power goes into air friction.

You can save a lot of energy by adapting your driving style ;-).


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