Calculating Force on Aircraft Carrier John F. Kennedy

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SUMMARY

The discussion focuses on calculating the force of water resistance acting on the aircraft carrier John F. Kennedy, which has a mass of 7.4 x 107 kg and operates at a top speed of 35 knots (18.06 m/s) with an engine power output of 280,000 hp (208,880,000 watts). It is established that 70% of the engine power is utilized for propulsion, while 30% is lost to water resistance and other factors. The force of water resistance is equal to the force produced by the ship at constant speed, leading to the conclusion that the water resistance force is equivalent to the effective force generated by the ship's engines.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of power conversion from horsepower to watts
  • Familiarity with basic physics equations, specifically P=F*v
  • Ability to convert units, such as knots to meters per second
NEXT STEPS
  • Learn about the principles of fluid dynamics as they apply to naval engineering
  • Study the effects of resistance on ship performance and efficiency
  • Explore advanced calculations involving propulsion systems and energy losses
  • Investigate the design considerations for optimizing ship hulls to reduce water resistance
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Students in physics or engineering, naval architects, and anyone interested in the mechanics of large vessels and their operational efficiency.

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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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