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Confused why both answers are different. P = F.v

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data
    A fire hose for use in urban areas must be able to shoot a stream of water to a maximum height of 34m . The water leaves the hose at ground level in a circular stream 3.0cm in diameter.
    What minimum power is required to create such a stream of water? Every cubic meter of water has a mass of 1.00×103kg.


    2. Relevant equations

    Ek = 0.5*m*v^2
    Ep = m*g*h
    Volume cylinder = A*h
    P=F*v (?)


    3. The attempt at a solution

    I attempted twice, and one of them was wrong (almost double the value), so I was wondering what went wrong in the second attempt.

    1st attempt:

    First I went to find the speed of the water needed to leave the nozzle.
    I used:
    Ek(leaving nozzle) = Ep(at 34m)
    0.5*m*v^2 = m*g*h

    and got
    velocity = 25.827 ms^-1

    Then I went to find the volume, then the mass of water leaving the nozzle in 1s.
    Mass = A*h*1000
    pi*0.015*0.015*25.827*1000
    = 18.2566kg

    With that, I went to find the energy per second required (hence power) to give this mass of water speed V.
    Ek = 0.5*18.2566*25.827*25.827 = 6089J

    So in Watts, that's the correct answer.

    ----------------------------------------

    But the stupid me went and tried another way of solving it (i dont know why i'd ever do that)

    So with v = 25.827 ms^-1
    I used equation P = F.v for the power needed
    P = F.v = (v.dm/dt)v
    = v*v*(dm/dt)
    So in one second,
    Energy to water -> 25.827*25.827*(18.2566) = 12177

    Which is the wrong answer, and coincidentally (or not), is twice as large.

    ----------------------------------------

    So... I can't figure out what went wrong with the second attempt.
     
  2. jcsd
  3. Sep 7, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    That is the mass of water leaving the nozzle in 1 second.

    That is the energy /second ΔEk/Δt = 6089J/s =6089 W.

    an excellent solution!

    ----------------------------------------

    I see you applied the impulse law F=Δ(mv)/Δt. Be careful: mass is not created: mass of water is accelerated up to the given speed. P=Fv can be applied when a certain force acts on a certain object with constant mass.


    ehild
     
  4. Sep 7, 2014 #3
    Hi! Thanks for the quick reply!

    Do you mean that, only if the force is acting on a constant mass, i.e. F=m*a, can we then use P=F*v?

    So if mass is moved/changed and requires dm/dt in the force equation, P=F*v cannot be considered"? (I'm still not too sure about the idea behind P=F*v, so...)
     
  5. Sep 7, 2014 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    P=Fv comes from the definition of work. If the force F acts on an object along a path, the work is equal to the integral ##W=\int{\vec F \cdot\vec{dr}}##. ##\vec {dr}=\vec v dt##,
    so
    ##W=\int{\vec F \cdot\vec{v}dt}##.
    The instantaneous power is ##dW/dt = \vec F \cdot\vec{v}##
    To apply the formula, you need to know the force. And the force is F=ma. In Classical Physics, the mass is constant and it can not be created from nothing. F=d(mv)/dt has to be used with caution.
    In the problem, a certain amount of water has to be accelerated to a certain speed in 1 s. dm/dt is not change of mass, it is the mass flown across the cross-section of the hose in unit time.

    ehild
     
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