Power & Velocity Homework Solution | 450 kW

[elitewarr]

Homework Statement

The rocket sled has a mass of 4 Mg and travels along the smooth horizontal track such that it maintains a constant power output of 450 kW. Neglect the loss of fuel mass and air resistance, and determine how far the sled must travel to reach a speed of v = 60 m>s starting from rest.

Homework Equations

P = Fv
$E_k = \frac{1}{2} mv^2$
W=Fs

The Attempt at a Solution

Since the rocket sled is moving along a flat surface, and that there is no loss of energy, we can conclude that the energy coming from the power output will be converted to kinetic energy.

$E_k = W = Fs = \frac{P}{v} s$

Thus simplifying, we get

$\frac{1}{2} mv^3 = Ps$

But apparently, there is something wrong with the equations since if we go about from another way,

$P = Fv = (ma)v = (mv\frac{dv}{ds})v$

$\int_0^s P \, ds = m\int_0^v v^2 \, dv$

$Ps = \frac{1}{3} mv^3$

Is there something wrong with my concept or that I am not seeing?

Thank you.

 

[haruspex]

W=Fs

Force will not be constant here, so that needs to be an integral, not a simple product.

 

[elitewarr]

Force will not be constant here, so that needs to be an integral, not a simple product.

Thank you for the fast reply.

Okay. I was suspecting much. But how do I do the integration for the $\frac{1}{v}$ with respect to ds? Any clues as to where to start?

 

[NascentOxygen]

might be useful: Energy = power x time

 

[HallsofIvy]

Thank you for the fast reply.

Okay. I was suspecting much. But how do I do the integration for the $\frac{1}{v}$ with respect to ds? Any clues as to where to start?

Since v= ds/dt, [itex]\int \frac{P}{v}ds= P\int \d<br /> <blockquote data-attributes="member: 198689" data-quote="elitewarr" data-source="post: 5084413" cite="https://www.physicsforums.com/goto/post?id=5084413" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> elitewarr said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Thank you for the fast reply.<br /> <br /> Okay. I was suspecting much. But how do I do the integration for the $\frac{1}{v}$ with respect to ds? Any clues as to where to start? </div> </div> </blockquote> Since $v= \frac{ds}{dt}$,<br /> $$P\int \frac{1}{v}ds= P\int \frac{1}{\frac{ds}{dt}}ds= P\int \frac{dt}{ds}ds= P\int dt= P(t_1- t_0)$$[/itex]

 

[haruspex]

Thank you for the fast reply.

Okay. I was suspecting much. But how do I do the integration for the $\frac{1}{v}$ with respect to ds? Any clues as to where to start?

Relate F to dv/dt.
Edit: or better still, to vdv/ds.