Calculating Force on Aircraft Carrier John F. Kennedy

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Homework Help Overview

The problem involves calculating the force of water resistance acting on the aircraft carrier John F. Kennedy, given its mass, engine power, and speed. The subject area includes dynamics and forces in motion.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting units of speed and power, and applying the power formula to find force. There is uncertainty about whether the calculated force represents the ship's thrust or the opposing water resistance. Some participants question the interpretation of the power distribution between thrust and resistance.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the application of power to find force, but there is no explicit consensus on the correct approach to determining the water resistance.

Contextual Notes

Participants mention that 30% of the power is lost in the drive train and propeller, which is a key assumption in the problem. There is also a focus on the implications of constant speed on the forces acting on the ship.

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Homework Statement


The aircraft carrier John F. Kennedy has mass 7.4*10^7kg. When its engines are developing their full power of 280000 hp, the John F. Kennedy travels at its top speed of 35 knots.

If 70% of the power output of the engines is applied to pushing the ship through the water, what is the magnitude of the force of water resistance that opposes the carrier's motion at this speed?


Homework Equations


P=F*v

The Attempt at a Solution



Not sure if I understand the question correctly. What I did was convert 35 knots into m/s which was 18.06m/s. I converted the power from hp to watts to get 208880000watts. I divided the watts by the velocity to get force. If this is correct, this would be the force of the boat and not the water correct? Would the resistance be the 30% that the force loses?
 
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Apparently 30% of the power is lost in the drive train and propeller.
Take 70% of the power and use it in the formula to find F.
 
Alright so couldn't I just find the total from what I did above, then dividing the power by the velocity to get force and then multiple that by .3 to get the water resistance? 70% is going forward while the other 30% is water resistance?

EDIT: Hmm that not correct.

I really don't know what I need to do. I found the power in watts, divided by the velocity in m/s and then that should be the power the ship is producing, in which 30% is lost to water resistance right?
 
Last edited:
Delphi51 said:
Apparently 30% of the power is lost in the drive train and propeller.
Take 70% of the power and use it in the formula to find F.
Would this just mean that the force of water resistance is equal to the force of the ship?
 
Would this just mean that the force of water resistance is equal to the force of the ship?
Yes. Motion at constant speed, so acceleration is zero and the total force on the ship is zero.
 

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