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How far will the ship travel after the engine stops?

  1. Apr 20, 2017 #1
    1. The problem statement, all variables and given/known data
    A ship's engine supplies power of 85MW, which propels the ship of mass 5.3×106 kg across the sea at its top speed of 11##ms^{−1}##. The frictional force exerted on the ship by the sea is directly proportional to its speed. If it starts at top speed and then travels in a straight line, how far will it go after the engines stop?

    2. Relevant equations


    3. The attempt at a solution
    I thought the ship would start of with energy ##\frac{1}{2}mv^2##, and that the work done by friction has to equal this amount for the ship to stop. It says friction is proportional to speed and from dimensional analysis I need units of kgm##s^{-2}##, mass and time also have to be included in the equation for friction, and it has to be inversely proportional to time so that ##F = \frac{mv}{t}## and work done ##= \int F dx = \frac{mvx}{t}##. There should probably also be some kind of constant of proportionality in there.

    Equating the two energy equations gives me ##x = \frac{vt}{2}## but I don't know t, so that's no good. And I don't know what else to try. Thanks for any help!
     
  2. jcsd
  3. Apr 20, 2017 #2

    BvU

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    You have two expressions for the energy per second: one from the 85 MW and one from F ds.
    You know F = 0 if v = 0 and that F(v) is linear. So you can find F as a function of v
     
  4. Apr 20, 2017 #3

    haruspex

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    No, it just says that the constant of proportionality has dimension mass/time. This need not have anything to do with the mass of the ship nor the time since the engines stopped. F=kv for some constant k.
    How can you also relate F and v via acceleration?
     
  5. Apr 20, 2017 #4
    Oops. Ok so it's a case of F = kv maybe plus some constant c, except as BvU says for F=0 v=0 so then there can't be a ##c##. Then I think I can use P = Fv to find out the force of friction, so ##85MW = F \times 11##m##s^{-1}## meaning when the boat travels at max speed ##F = 7.7\times 10^6##, and from that ##k=7\times 10^5##. Then it's a differential equation. Thanks! :)
     
  6. Apr 20, 2017 #5

    BvU

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    Didn't want to give it away too much ..., but yes
     
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