When is power constant in this scenario?

[rakesh]

Homework Statement

The question says to find the relation between displacement and time when power remains constant.

Relevant Equations

Power P = F x v,where F is force and v is velocity.

Power P = F x v,where F is force and v is velocity,

if power remains constant then i think force can not remain constant as it will change the velocity v,

but the solution I found is,
F = ma, v = at,
so, P = F x v = ma x at = ma2 t,

after that calculus comes to show that displacement is directly proportional to t to the power 3/2,

my question is if a is constant, how can power remain constant,

thanks.

 

[erobz]

Homework Statement:: The question says to find the relation between displacement and time when power remains constant.
Relevant Equations:: Power P = F x v,where F is force and v is velocity.

Power P = F x v,where F is force and v is velocity,

if power remains constant then i think force can not remain constant as it will change the velocity v,

but the solution I found is,
F = ma, v = at,
so, P = F x v = ma x at = ma2 t,

after that calculus comes to show that displacement is directly proportional to t to the power 3/2,

my question is if a is constant, how can power remain constant,

thanks.

Under constant power if the the mass is accelerating then its velocity is changing, if the velocity is changing, the force ( either propelling it or retarding it) is changing.

An example /specific scenarios where this happens in reality currently escapes me, but the math works.

 

[DaveE]

$v(t) = a(t)⋅t$ is only true if $a(t)$ is constant. Have you studied calculus yet?

 

[rakesh]

$v(t) = a(t)⋅t$ is only true if $a(t)$ is constant. Have you studied calculus yet?

yes, i know calculus basics but will request if u can expalin it in the simplest possible way, thanks.

 

[erobz]

$v(t) = a(t)⋅t$ is only true if $a(t)$ is constant. Have you studied calculus yet?

Yeah, that is the crux of the issue. The apparent contradiction is from working under the assumption of constant acceleration ( which implies constant force). A force cannot both change and be constant.

However, if we allow changing mass ( leaky cart perhaps) we can get instances of constant acceleration under constant power?

 

[rakesh]

Under constant power if the the mass is accelerating then its velocity is changing, if the velocity is changing, the force ( either propelling it or retarding it) is changing.

An example /specific scenarios where this happens in reality currently escapes me, but the math works.

so, the solution given in this link is wrong as he has probably taken a constant ?

he discusses at 2.00 in the video,

 

[erobz]

so, the solution given in this link is wrong as he has probably taken a constant ?

he discusses at 2.00 in the video,

Yeah, that’s not correct IMOP. It’s a contradiction unless the mass is changing.

The only thing I can imagine having constant accelerating under constant power is a leaky cart.

 

[rakesh]

Yeah, that’s not correct IMOP. It’s a contradiction unless the mass is changing.

I think the discussion in this video is bit different but I dnt understand the difference, he solves it in two ways, in the second method at 2.00, he discusses through calckus,

 

[erobz]

I think the discussion in this video is bit different but I dnt understand the difference, he solves it in two ways, in the second method at 2.00, he discusses through calckus,

In this video constant acceleration is not assumed. It is declining with increasing velocity.

 

[rakesh]

In this video constant acceleration is not assumed.

how to know that its not assumed ?

 

[rakesh]

moreover if a and v are in opposie directions, power will be negative and if both in the same direction its positive, how can a case arise only that power can remain constant.

 

[erobz]

how to know that its not assumed ?

Because they state:

$$ a = \frac{P}{m v}$$

It is a function of velocity.

Don’t ask about what happens at $v=0$ , they will shut the thread down…I know from experience!

 

[DaveE]

yes, i know calculus basics but will request if u can expalin it in the simplest possible way, thanks.

OK, rewrite the equations you had before, like F=ma, but this time indicate any variable as time varying except those that you know are constant from the problem description. So, for example use f(t), not F. Then add in the calculus based relationships between displacement, velocity and acceleration. That will give you a set of differential equations for the variable d(t) that should be pretty straightforward to solve.

 

[rakesh]

seems the case arises only in the retarded motion, smaller a higher v and vice versa, in the accelerated motion whatever be a, v will go on increasing.

 

[erobz]

seems the case arises only in the retarded motion, smaller a higher v and vice versa, in the accelerated motion whatever be a, v will go on increasing.

Which case arises in retarded motion?

 

[rakesh]

Which case arises in in retarded motion?

like when u press the breaks of ur car less, its velocity will be high and when pressed hard it will become low, force and displcement will be in opposite directions in these scenarios and powe will always be negative.

 

[erobz]

like when u press the breaks of ur car less, its velocity will be high and when pressed hard it will become low, force and displcement will be in opposite directions in these scenarios and powe will always be negative.

Negative, but not constant with constant acceleration. If you are holding power constant ( positive or negative) , and the acceleration constant, you must be losing or gaining mass.

 

[rakesh]

Negative, but not constant with constant acceleration. If you are holding power constant ( positive or negative) , and the acceleration constant, you must be losing or gaining mass.

acceleration is not constant, u press more when velocity is less, u press infinite when velocity is zero.

 

[erobz]

acceleration is not constant, u press more when velocity is less, u press infinite when velocity is zero.

Ok, I was getting mixed signals.

 

[haruspex]

Yeah, that’s not correct IMOP.

Quite.. it is utter rubbish.
So how did he get the right answer to the question? Because to answer the question posed only needs dimensional analysis, so as long as he committed no dimensional violations he was going to get there.
[P]=ML²T^-3, [t]=T, [m]=M, [d]=L.
The only dimensionless combo is $\frac{Pt^3}{md^2}$.

He compounds his error by later using d=1/2 at².

The easy way is to consider energy. Since power is constant, $\frac 12 mv^2=Pt$, $v=\sqrt{\frac{2Pt}m}$. Integrate wrt t.

 

[jack action]

I'll try something:
$$P = F \frac{(\Delta x)_{avg}}{\Delta t}= ma \frac{(\Delta x)_1 + (\Delta x)_2}{2 \Delta t}$$
$$P = m \frac{\frac{(\Delta x)_2}{\Delta t} - \frac{(\Delta x)_1}{\Delta t}}{\Delta t}\frac{(\Delta x)_1 + (\Delta x)_2}{2 \Delta t} = m \frac{[(\Delta x)_2 - (\Delta x)_1] [(\Delta x)_1 + (\Delta x)_2]}{2 (\Delta t)^3}$$
$$P = m \frac{\left((\Delta x)_2\right)^2 \left[1 - \left(\frac{(\Delta x)_1}{(\Delta x)_2}\right)^2\right]}{2 (\Delta t)^3}$$
And:
$$\frac{(\Delta x)_1}{(\Delta x)_2} = \frac{v_1 \Delta t}{v_2 \Delta t} = \frac{v_1}{v_2}$$
$$(\Delta x)_2 = \sqrt{\frac{2P}{m\left[1 - \left(\frac{v_1}{v_2}\right)^2\right]}}(\Delta t)^{3/2}$$
Which assumes velocity must change but not that acceleration is constant. So we can still say:
$$\Delta x \propto (\Delta t)^{3/2}$$

 

[rakesh]

Quite.. it is utter rubbish.
So how did he get the right answer to the question? Because to answer the question posed only needs dimensional analysis, so as long as he committed no dimensional violations he was going to get there.
[P]=ML²T^-3, [t]=T, [m]=M, [d]=L.
The only dimensionless combo is $\frac{Pt^3}{md^2}$.

He compounds his error by later using d=1/2 at².

The easy way is to consider energy. Since power is constant, $\frac 12 mv^2=Pt$, $v=\sqrt{\frac{2Pt}m}$. Integrate wrt t.

is my assumption correct ?

"seems the case arises only in the retarded motion, smaller a higher v and vice versa, in the accelerated motion whatever be a, v will go on increasing."

 

[rakesh]

I'll try something:
$$P = F \frac{(\Delta x)_{avg}}{\Delta t}= ma \frac{(\Delta x)_1 + (\Delta x)_2}{2 \Delta t}$$
$$P = m \frac{\frac{(\Delta x)_2}{\Delta t} - \frac{(\Delta x)_1}{\Delta t}}{\Delta t}\frac{(\Delta x)_1 + (\Delta x)_2}{2 \Delta t} = m \frac{[(\Delta x)_2 - (\Delta x)_1] [(\Delta x)_1 + (\Delta x)_2]}{2 (\Delta t)^3}$$
$$P = m \frac{\left((\Delta x)_2\right)^2 \left[1 - \left(\frac{(\Delta x)_1}{(\Delta x)_2}\right)^2\right]}{2 (\Delta t)^3}$$
And:
$$\frac{(\Delta x)_1}{(\Delta x)_2} = \frac{v_1 \Delta t}{v_2 \Delta t} = \frac{v_1}{v_2}$$
$$(\Delta x)_2 = \sqrt{\frac{2P}{m\left[1 - \left(\frac{v_1}{v_2}\right)^2\right]}}(\Delta t)^{3/2}$$
Which assumes velocity must change but not that acceleration is constant. So we can still say:
$$\Delta x \propto (\Delta t)^{3/2}$$

did not understand much, just one question when we say v = ds/dt, dt is very small tending to zero, is ds also very small, if so why dnt they cancel out and give v = 1.

 

[haruspex]

did not understand much, just one question when we say v = ds/dt, dt is very small tending to zero, is ds also very small, if so why dnt they cancel out and give v = 1.

Just because they are both very small doesn’t mean they have to be equal, or anything like equal. 0.001 and 0.000000000001 are both very small numbers, but what is their ratio?
If an incline has a smooth steady slope of 1/100, going forwards 1m takes you up 1cm; going forward 1cm takes you up 0.1mm; going forward 01.mm takes you up 1μm, and so on. dy and dx can be made arbitrarily small but dy/dx remains 1/100.

 

[haruspex]

is my assumption correct ?

"seems the case arises only in the retarded motion, smaller a higher v and vice versa, in the accelerated motion whatever be a, v will go on increasing."

What exactly do you mean by "this case"? Do you mean constant power, or the combination of constant power and constant acceleration?

Since power = velocity x force, and the mass is constant, constant power and constant acceleration would imply constant velocity, so the force would be zero.

 

[jack action]

did not understand much, just one question when we say v = ds/dt, dt is very small tending to zero, is ds also very small, if so why dnt they cancel out and give v = 1.

$\Delta t$'s cancel each other out because they have the exact same value, not because they are close to zero. $\Delta x$'s do not cancel each other because - although small - they are not necessarily of the same value.

What I did is expand the acceleration $a$ definition between 2 points, namely point $1$ and $2$:
$$a = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{\Delta t} = \frac{\frac{(\Delta x)_2}{\Delta t} - \frac{(\Delta x)_1}{\Delta t}}{\Delta t}$$
The average velocity $v_{avg}$ somewhere between those 2 points is:
$$v_{avg} = \frac{v_1 + v_2}{2} = \frac{\frac{(\Delta x)_1}{\Delta t} + \frac{(\Delta x)_2}{\Delta t}}{2} = \frac{(\Delta x)_1 + (\Delta x)_2}{2 \Delta t}$$
And then you put those definitions into your original equation:
$$P = F v_{avg} = m a v_{avg}$$