A ship going upstream and experiences a frictional force

AI Thread Summary
The discussion centers on a problem involving a ship moving upstream against a frictional force that depends on its speed relative to the water. The key equation presented is P = F_r x U, where F_r is the frictional force and U is the water's speed relative to the shore. Participants clarify that the power required to maintain position and to move upstream should not be differentiated, as they are effectively the same. There is confusion regarding the use of relative speeds, with emphasis on the need to correctly interpret the relationship between the ship's speed and the water's current. The conversation highlights the importance of accurately applying physics principles to solve the problem efficiently.
fara0815
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Hello there!

This problem is giving me a hard time and I thought maybe one of you could give me a hint:

"A ship is going upstream with a constant power P. Its speed V is relative to the water and the water's speed U is relative to the shore. The ship needs tp overcome a frictional force of F_r=cv^2 caused by the water which depends on its relativ speed. How big does v have to be so that the ship goes from A to B with the lowest energy consumption?"

Where I would start is that the ship needs at least the power which is neccesary to stay in the same place. Which would be
P=F_r x U = cU^2U=cU^3 and than in addition some extra power to go upstream. That would be P=F_r x (v-u)=c(V-U)^2 x (V-U). But from here, if it's really correct, I do not know how to continue :(
 
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I assume the "x" in your formula is a multiplication sign. Don't do that, it's confusing. Just P= F_rU would be sufficient.
I don't believe you can or should distinguish between the power necessary to "stand still" and the power necessary to move forward against the current. They are the same. Note also that U is speed of the boat relative to the water so it makes no sense to use V-U (which would be negative: if the boat is moving forward against the current, U> V).
 
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