How far will the ship travel after the engine stops?

1. Apr 20, 2017

Kara386

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. Apr 20, 2017

BvU

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

3. Apr 20, 2017

haruspex

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?

4. Apr 20, 2017

Kara386

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! :)

5. Apr 20, 2017

BvU

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