Calculating Spring Stretch in Aircraft Carrier Landing

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Homework Help Overview

The problem involves calculating the stretch of a spring when a jet fighter lands on an aircraft carrier and engages a cable attached to the spring. The scenario includes a mass of 16,000 kg and a landing speed of 52 m/s, with a spring constant of 60,000 N/m.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods to relate velocity, distance, and acceleration, with some expressing difficulty in applying standard kinematic equations due to changing forces. There is mention of using conservation of energy principles to equate kinetic and potential energy.

Discussion Status

The discussion includes attempts to derive equations and clarify concepts related to the problem. Some participants have provided guidance on using energy conservation, while others are exploring different interpretations of the problem setup. There is no explicit consensus on a final approach, but productive dialogue is ongoing.

Contextual Notes

Participants note the absence of time and distance information in their attempts, which complicates the application of traditional motion equations. There is also a recognition of the need to consider the variable nature of forces involved with the spring.

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Homework Statement



A 16,000 kg F-18 (jet fighter) lands at 52 m/s on an aircraft carrier, its tail hook snags the cable to slow it down. The cable is attached to a spring with a spring constant of 60,000 N/m. How far does the spring stretch to stop the jet?

Homework Equations



f=-ks
motion equations


The Attempt at a Solution


too hard to put on computer
 
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jjd101 said:

The Attempt at a Solution


too hard to put on computer

No effort, no help!
 
i tried a lot of different things, all the methods i tried didn't seem to work because it doesn't give neither time nor distance.
 
Show us what you tried, so we can do a better job helping you.
 
i tried taking the velocity V=m/s and setting V=deltaX/deltaT. Then i solved delta T and got deltaT=deltaX/V. I then tried substituting this into the first motion equation vf=vi+adeltaT and solving for a, then used a in another motion equation V^2 = Vi^2 +2adeltaX after all this i think i got a =36.75 and t = 1.415 which i don't think is right
 
jjd101 said:
i tried taking the velocity V=m/s and setting V=deltaX/deltaT. Then i solved delta T and got deltaT=deltaX/V. I then tried substituting this into the first motion equation vf=vi+adeltaT and solving for a, then used a in another motion equation V^2 = Vi^2 +2adeltaX after all this i think i got a =36.75 and t = 1.415 which i don't think is right

You're right, it's not right :smile:

As a spring stretches, the force it supplies increases. If the force is changing with distance, then so is the acceleration. If they are not constant, then you can't apply the 'usual' kinematic formulas that assume constant force or acceleration.

Problems involving springs are often best approached from a conservation of energy point of view. Kinetic and potential energy get exchanged via the spring. Do you have the formulas for kinetic energy and spring potential energy?
 
KE=1/2mv2 and Us=1/2k(deltaX)^2 correct? so do i set these equal to each other?
 
jjd101 said:
KE=1/2mv2 and Us=1/2k(deltaX)^2 correct? so do i set these equal to each other?

Yes, you'll be setting them equal to each other. Do you know why?
 
yeah i understand now, i just have trouble figuring out what equations to use. I just did it and i got distance is 26.85m. Thanks a lot i really appreciate it
 

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