Find force on a particle at time t

[Istiak]

Homework Statement

A particle of mass ##m## is driven by a machine that delivers a constant power of k watt. If the particle starts from rest, the force on the particle at time t is...................

Relevant Equations

##\vec F=m\frac{d^2 \vec s}{dt^2}##
##P=\frac{W}{t}##
##W=\vec F\cdot \vec s##

My attempt :

##\frac{\vec Ft^2}{2}=m\vec s##
##s=\frac{Ft^2}{2m}##
##P=\frac{W}{t}##
##k=\frac{\vec F\cdot \vec s}{t}##
##k=\frac{F^2t^2}{2mt}##
##k=\frac{F^2t}{2m}##
##F=\sqrt{\frac{2mk}{t}}##

But there was an option which was ##2\sqrt{\frac{mk}{t}}##. And my assumption was that it was correct. They approximately equal but not completely. Do you get the same answer? Or do you get ##2## outside the square root?

 

[PeroK]

Have you assumed acceleration is constantfor your first equation?

 

[Istiak]

Have you assumed acceleration is constant for your first equation?

Yes

 

[BvU]

But it isn't !
You want to start from ##W = P t\ ##
And use ##F = {dp\over dt} ##

 

[PeroK]

Yes

The power is constant; not the force. Hence the acceleration is time dependent.

 

[Istiak]

But it isn't !
You want to start from ##W = P t\ ##
And use ##F = {dp\over dt} ##

##W=Pt##
##Fs=Pt##
##\frac{kt}{s}=\frac{dp}{dt}##

 

[BvU]

Excellent. What is ##W## in terms of ##p ## and ##m## ?
Remember, you don't want ##s##. You want ##{dp\over dt}##

 

[Istiak]

Excellent. What is ##W## in terms of ##p ## and ##m## ?
Remember, you don't want ##s##. You want ##{dp\over dt}##

If I take ##p## then I don't get ##m## cause I must use ##p=mv## to find out ##W## in terms of ##m##. I can write that ##W=\frac{dp}{dt}vdt##, but I don't want to use velocity also.

 

[BvU]

You sure don't. But ##W = kt## and you can write ##W## in terms of ##p## and ##m##, I should hope

[edit] ##W=\frac{dp}{dt}vdt## says ## W={\rm d}p\,v \ ##, that can't be right !

##\ ##

 

[Istiak]

You sure don't. But ##W = kt## and you can write ##W## in terms of ##p## and ##m##, I should hope

[edit] ##W=\frac{dp}{dt}vdt## says ## W={\rm d}p}\,v ## can't be right !

##\ ##

Oops I remember, ##W=\frac{p^2}{2m}##..

 

[BvU]

Good. So now you want ##{\rm d}p\over dt## out of ##W = kt## and bingo !

##\ ##

 

[Istiak]

##W=kt##
##\frac{p^2}{2m}=kt##
##p^2=2mkt##
##\frac{d(p^2)}{dt}=2mk##
##2p\dot p=2mk##
##F=\frac{mk}{p}##
blah :|

Instead of doing that I found ##p=\sqrt{2mkt}##
##F=\frac{1}{2}(2mkt)^{-\frac{1}{2}}2mk##
##=\sqrt{\frac{mk}{2t}}##

Done! Thanks. but sad :)

 

[BvU]

Done! Thanks. but sad :)

Nonsense. Well done! Do you think others just do this kind of thing 'on the fly'? Not me and not many others either.

Sorry @PeroK for somewhat intruding.

##\ ##

 

[PeroK]

You could also have used ##Fv = P = k## and ##\frac 1 2 m v^2 = Pt = kt##.

In general we have:
$$E = \frac 1 2 m \vec v \cdot \vec v$$$$\Rightarrow \ P \equiv \frac{dE}{dt} = m \vec a \cdot \vec v = \vec F \cdot \vec v$$