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

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?

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.

• If power is constant then this must mean that the quantity Fv is constant. What does this tell you about the derivatives above? Mar 19, 2021 at 5:41
• if F increased by a factor of x, then v is multiplied by a factor of 1/x. However, this becomes tricky when the initial speed is 0, which is why I ask the question Mar 19, 2021 at 5:43
• So the F and v are changing, so how can dP/dt=0? Mar 19, 2021 at 5:47
• Why did you assume the force is not a function of the velocity? Mar 19, 2021 at 5:59

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 $$v$$.

• Updated my question. I accepted your post since it did answer my initial question, but any advice for my updated post is much appreciated Mar 19, 2021 at 6:12

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

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.

• so it is therefore impossible to lift an object with constant power. thank you. Mar 19, 2021 at 6:48