Is it possible to lift an object from rest with constant power?

Question

This is inspired by the following question.

Consider some object which I want to lift from rest with a constant power throughout the whole process; the power I apply when lifting the object from rest is the same power I apply to keep lifting it. The force $F$ and the speed $v$ may change, but power may not.

The power is $P=Fv$. If we want constant power, then $dP/dt=0$.

First, differentiate wrt time, $$\dfrac{dP}{dt} = F \dfrac{dv}{dt}+v \dfrac{dF}{dt}.$$

Set equal to zero, this guarantees constant power, which implies:

$$F \dfrac{dv}{dt} = -v\dfrac{dF}{dt}.$$

From there, I use $F=\dfrac{d(mv)}{dt}\implies F = m\dfrac{dv}{dt}$ and get $$\left( \dfrac{dv}{dt} \right)^2 + v\dfrac{dv}{dt} = 0.$$

How may I solve this non-linear ODE?

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EDIT: Initially, I got to the wrong conclusion that $F(v-v_0) = -v(F-F_0)$ is a solution. That's what the answer by Vilvanesh addresses.

Answer 1

Your discussion is valid till $\displaystyle{F\frac{dv}{dt}=-v\frac{dF}{dt}}$.

I think the only way to reach $F(v-v_o)=v(F-F_o)$, is by assuming $F$ is constant while integrating on the left side and by assuming $v$ as constant while integrating on the right side. This is inconsistent because on the left side $F$ is being treated as a constant and on the right side it is treated as a variable. Same is applicable for $v$.

Answer 2

$$\left( \dfrac{dv}{dt} \right)^2 + v\dfrac{dv}{dt} = 0$$ $$\frac{dv}{dt}\times\Big[\frac{dv}{dt}+v\Big]=0$$

When $\displaystyle{\frac{dv}{dt}=0}$, $v$ has to be a constant. But that isn't possible because initially it was at rest but while being lifted, it is not. $$\frac{dv}{dt}+v=0$$ $$\frac{dv}{v}=-dt$$ $$\int\limits_{v_o}^v\frac{dv}{v}=-\int\limits_0^tdt$$ $$\ln{v}-\ln{v_o}=-t$$ But since $v_o=0$, $\ln{v_o}$ won't be defined.

Hence a solution doesn't exist for the above differential equation.