7
$\begingroup$

I am calculating power to accelerate an object of mass M in time T. My intuition tells me to calculate kinetic energy of the object and divide by T. Say a mass of 40 kg is accelerated from 0 m/s to 10 m/s in 4 s. That's a kinetic energy of 0.5 * 40 * 10^2 = 2000 J so power required is 2000J/4s = 500 W. Acceleration of 2.5 m/s^2. But I am seeing another equation, P = maΔv. Using that equation gives 40 * 2.5 * 10 = 1000 W. What am I missing?

Yes I am a student, an 81 year old student of many things. Have been out of the classroom for 60 years so may have forgotten a thing or two.

$\endgroup$

2 Answers 2

4
$\begingroup$

Power required for acceleration increases with velocity. Your first intuition is correct for average power, while the second expression is a (slightly problematic) expression for instantaneous power.

Mechanical Power is the rate of change of kinetic energy; that is, at any moment, it equals the corresponding slope of the kinetic energy curve.

Given constant acceleration $a$ and initial velocity $v=0$ at time $t=0$, then velocity $v=at$ and kinetic energy $T = .5 mv^2 = .5 ma^2t^2$ is a parabola, with a parabola's characteristic increasing slope.

The instantaneous power is the derivative. For this special case of $T$:

$$P(t) = \frac{d}{dt}T = ma^2t = m(at)a = mva$$

Applying the chain rule for derivatives to the general expression of kinetic energy $T = 0.5m(\frac{d}{dt}x)^2$, we can prove that this is generalizable to all functions with constant mass, no matter how they accelerate or how fast they started out (note instantaneous velocity $v$, not change in velocity $\Delta v$):

$$P(t) = \frac{d}{dt}T = .5m\frac{d}{dt}(\frac{d}{dt} x)^2 = m\frac{dx}{dt} \frac{d^2x}{dt^2} = mva$$

If you average this function over the relevant time period, you will find that it comes out to the same average power given by $P_{avg} = \frac{\Delta T}{\Delta t}$.

$\endgroup$
1
  • $\begingroup$ Thank you. Seems my question has been closed because it looks like a homework question. About as far from the truth as possible. $\endgroup$
    – Bill
    Commented May 30, 2023 at 18:13
2
$\begingroup$

This problem runs into the double duty that forces play. It appears not only in Newton's $2^\text{nd}$ Law, but also in the Work-Energy Theorem.

Namely, let me first define the momentum $\vec p=m\vec v$, and then the proper statement of N2L is $$\text{N2L}:\quad\sum\vec F=\frac{\mathrm d\vec p}{\mathrm dt}$$ This is the $\vec F=m\vec a$ that you are so much more familiar with.

The other expression is for the energy. Power is the rate of change of energy, i.e. $$P=\frac{\mathrm dE}{\mathrm dt}=\frac{\mathrm dE}{\mathrm d\vec x}\cdot\frac{\mathrm d\vec x}{\mathrm dt}=\vec F\cdot\vec v$$ That is, we have the double duty that $$\frac{\mathrm d\vec p}{\mathrm dt}=\vec F=\frac{\mathrm dE}{\mathrm d\vec x}$$


Since we are now interested in the power, we should directly use the power expression. $$P=\vec F\cdot\vec v=\frac{\mathrm d\vec p}{\mathrm dt}\cdot\frac{\vec p}m=\frac1{2m}\frac{\mathrm d\ }{\mathrm dt}\left(\vec p^2\right)$$ Nota Bene the kinetic energy is KE $=\frac12mv^2=\frac{p^2}{2m}$ so this is just nice.

Anyway, from the fact that it is actually going to be the time derivative of a squared quantity, it is clear that, even if we assume that the momentum, or velocity, of the particle is going to increase linearly with time (corresponding to constant acceleration), the energy of the particle will increase quadratically with time, leading to a power expenditure that is increasing linearly with time.

So, really, what you have done is that your first expression, the 500W, is average (mean) power, whereas the latter expression, the 1000W, is the instantaneous power at the very end of the acceleration, when it is the most difficult to accelerate. Both expressions are correct in their own usage.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.