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Drag question

  1. Apr 11, 2005 #1
    A boat of mass 250kg is coasting, with its engine in neutral, through the water at speed 3.00m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0kg/hr. Assume that the boat is subject to a drag force due to water resistance. The drag is proportional to the square of the speed of the boat, in the form F=0.5v^2. What is the acceleration of the boat just after the rain starts?

    I've been approaching this question from a few angles and don't seem to be getting anywhere. I found the answer to the 1st part using conservation of momentum, but since momentum is no longer conserved I don't know what to do. I'd really appreciate some help with this.
     
  2. jcsd
  3. Apr 11, 2005 #2
    [tex]a = \frac{0.5v^2}{m}[/tex]

    But I don't see how you could use the other problem data.
     
  4. Apr 11, 2005 #3

    xanthym

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    SOLUTION HINTS:
    {Rain Accumulation Rate} = (10.0 kg/hr) = (2.7778e(-3) kg/sec)
    {Horizontal Boat Speed} = (3.00 m/s)

    From Conservation of Momentum between Boat and Rain Drops:
    {Change in Boat's Horizontal Momentum DUE TO RAIN ACCUMULATION} =
    = -{Change in Rain's Horizontal Momentum when Accumulating in Boat}

    ::: ⇒ -{FORCE on Boat from Rain Accumulation} = Frain =
    = -{RATE OF CHANGE in Boat's Horizontal Momentum DUE TO RAIN ACCUMULATION} =
    = {RATE OF CHANGE in Rain's Horizontal Momentum when Accumulating in Boat} =
    = {Rain Accumulation Rate}*{Horizontal Boat Speed} =
    = (2.7778e(-3) kg/sec)*(3.00 m/s) =
    = (8.3334e(-3) N)

    {Acceleration of Boat} = {Net Force on Boat}/{Mass of Boat} =
    = {-(Drag Force) - (Frain)}/{Mass of Boat}
    = {-0.5v^2 - (Frain)}/{Mass of Boat} =
    = { (-0.5)*(3.00 m/s)^2 - (Frain) }/{Mass of Boat} =
    Determine boat acceleration (negative because boat is slowing) from this last equation using values given and/or computed previously.


    ~~
     
    Last edited: Apr 11, 2005
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