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Power and velocity

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations

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

    3. 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.
     
  2. jcsd
  3. Apr 22, 2015 #2

    haruspex

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    Force will not be constant here, so that needs to be an integral, not a simple product.
     
  4. Apr 22, 2015 #3
    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?
     
    Last edited by a moderator: Apr 22, 2015
  5. Apr 22, 2015 #4

    NascentOxygen

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    might be useful: Energy = power x time
     
  6. Apr 22, 2015 #5

    HallsofIvy

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    Since v= ds/dt, [itex]\int \frac{P}{v}ds= P\int \d
    Since [itex]v= \frac{ds}{dt}[/itex],
    [tex]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)[/tex]
     
  7. Apr 22, 2015 #6

    haruspex

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    Relate F to dv/dt.
    Edit: or better still, to vdv/ds.
     
    Last edited: Apr 22, 2015
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