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