Definition of Power

Question

I wanted to clear my doubts regarding the true definition of power. Imagine a mass falling from a height and reaching the ground thanks to gravity. The power of this event would be the work done by gravity on the mass divided by the amount of time it took to reach the ground.

However another definition of power would be the force applied times the velocity, however since this mass is accelerating, the power would constantly increase until the mass reached the ground.

How do I connect these two definitions of power? Is power force times velocity even when the velocity is constantly changing?

Answer 1

$$P_\text{avg} = \dfrac{\Delta P}{\Delta t}$$ is the average power, whereas $$P=F\cdot v$$ is the instantaneous power derived from the fact that $P\equiv \dfrac{\mathrm d W}{\mathrm dt}$ where $W$ is the work done

Answer 2

So we know the potential energy converted to kinetic energy is:

$$ U=mgz $$

where $z$ is the vertical coordinate. The power is the rate at which this conversion takes place:

$$ P\equiv \frac{dU}{dt}=mg\frac{dz}{dt}=F_gv_z$$

It's that straight forward. At higher speed, more height is traversed per unit time.

From the kinetic energy side, we have:

$$T=\frac 1 2 mv_z^2 $$

with

$$ P\equiv \frac{dT}{dt}=\frac 1 2 m\frac{d(v_z^2)}{dt}=mv_z\dot v_z$$

and:

$$ \dot v_z = \frac{F_g} m $$

so that:

$$p= F_gv_z $$

Answer 3

In your example, assuming the right approximations, force is simply given by -

$F = ma$ with the usual terminology. So the force is constant.

The power exerted on the falling object is given by the rate of change of energy.

The definition of energy in terms of force is:

$E = \int F . ds$

again, with the usual notations.

Therefore, power (the rate of change of energy) will be -

$P = \frac{dE}{dt} = \int F. \frac{ds}{dt} = F . v$ which is true because $s = s(t)$ and $F$ is constant.

As this resource says,

In the straightforward cases where a constant force moves an object at constant velocity, the power is just P = Fv

So the answer depends on the question you are asking.

If you would like to calculate the instantaneous power, the product of force and the velocity at that time would indeed be correct.

If you are going to calculate power by taking the difference of the initial and final energies and dividing that by the time difference, that is the average power and here, you could calculate it as the force times the average velocity measured over the duration of the experiment.

Answer 4

Work done by a mass falling

integral f . dr

however in falling dr is v(t) dt as dr/dt = r'(t)

So integral f. v(t) dt

Power in this case is the rate at which work is being done by gravity on the mass

so d/dt (work) = d/dt (integral f.v(t) dt)

is obviously f.v(t)