Mechanical power as $P = F v$

Question

I understand that, in the particular case of a constant force $F$ applied to an object, the speed increases linearly, both the instantaneous power $P$ and the kinetic energy also increase linearly.

Often, we deal with the situation in which power $P$ is set to be constant and we hear that if the force $F$ is large the speed $v$ is low and vice versa. Even if this makes sense mathematically, I have to admit that it is still does not make full sense conceptually.

Constant power $P$ means that we supply the object with energy at a constant rate (ex: say 2 Joule every second). Why would a large force imply a small velocity and vice versa?

I think that the underlying assumptions are:

a) the net power is zero: a constant positive input power that injects energy in the system per unit time is matched by a constant negative power (due to resistive forces) that removes an equal amount of energy from the system

b) The speed is constant since the net power is zero and there is not kinetic energy change

I just cannot wrap my head around the idea that when the object moves at small constant speed $v$ the force $F$ on it is large and vice versa. I am stuck with thinking that a large force has to correlate with a large speed.

Answer 1

$\rm power = force \times velocity$ so if the power is constant a large force implies a small velocity.

The power equation comes from $\rm work \,\,done = force \times displacement$ and dividing both sides of the equation by $\rm time$.
So if a body is travelling fast it undergoes a large displacement per second and so to keep the power at some constant value the force must be small.

Answer 2

Constant power 𝑃 means that we supply the object with energy at a constant rate (ex: say 2 Joule every second). Why would a large force imply a small velocity and vice versa?

The engine can't cause velocity. All it can do is supply a force. This relation (assuming it holds) tells us that a constant power engine can only provide small forces at high velocity. We can't pick both the power and the force.

Remember we don't need a large force to travel quickly. A small force over time can do that if drag is small. Forces cause changes in velocity, not velocity itself.

So, say we start from rest with our car in low gear (1s gear) pressing the gas pedal all the way. The car speeds up (i.e. accelerates, changes velocity). What happens if we don't shift and stay in the 1st low gear? Does the car reach a constant speed for the 1st gear because the force on the ground is matched by the net resistive force? I don't think the car speeds up any further...Why?

You can keep accelerating for a while, but normally you don't want to because of damage to the engine. But if your engine can't blow up, you'll accelerate until your RPMs get so high that the power output of the engine drops. Above a certain point, the engine can no longer deliver much force. When this happens, it's balanced against the sources of drag and the acceleration stops.

For a simpler scenario, imagine a merry-go-round. While it's stopped, you can push pretty hard on it. Your force is able to accelerate it quite rapidly. As it goes faster, you can no longer push as hard. The same thing happens in any moving system. As the velocity increases, the force you can develop to push it decreases.

Answer 3

Just a short answer but maybe this helps:
Your question is equivalent to asking why kinetic energy $E=\frac{1}{2}mv^2$ increases with $v^2$ instead of $v$ because $dE/dt=ma\cdot v=P$. So the amount of energy needed to increase the velocity of an object $dv$ at high velocities is greater than it is for small velocities because $dE(v)=mv\cdot dv$. Vice versa this means that the power for a given force (that tells you how much the object changes its velocity per time unit) is greater at high velocities because $dE/dt=P=Fv$.
This is kind of a circular argument but usually one does not question the fact that kinetic energy increases with $v^2$ (which also simply follows from $F=ma$ and $dE:=F\cdot ds$) and maybe you are more used to this.

Answer 4

I think the confusion is treating everything constant between power, force, and velocity, but this isn't possible.

If power is constant, then $P=Fv\to F=P/v$, meaning $F$ is a monotonically decreasing function of $v$. So when $v$ is "small", $F$ is "large". This causes an acceleration, which increases $v$ and decreases $F$. Note that this has to happen for constant power. There are no other forces present. You have said the power is constant for this process (whatever it is), and so the force has to be a function of $v$ as $F=P/v$.

If it helps, you can work out the velocity as a function of time in the case of a constant power:

$$F=\frac Pv=m\frac{\text dv}{\text dt}$$ $$\int_{v_0}^{v}v'\,\text dv'=\int_0^t\frac Pm\,\text dt'$$ $$\frac12\left(v^2-v_0^2\right)=\frac Pmt$$ $$v(t)=\sqrt{\frac{2P}{m}t+v_0^2}$$

And we can determine the force as well

$$F(t)=m\frac{\text dv}{\text dt}=\frac{P}{\sqrt{(2P/m)t+v_0^2}}$$

which is just $P/v$ as we come full circle.

So as you can see, over time the velocity increases and the force decreases in such a way that the power is constant. No other forces or $0$ net power is needed to explain this. Note that all of the above assumes a positive power.