Motion + Spring problem

In summary, the jet fighter lands on the aircraft carrier at 52 m/s and its tail hook snags the cable, which stretches to stop the jet.
  • #1
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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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  • #2
jjd101 said:

The Attempt at a Solution


too hard to put on computer

No effort, no help!
 
  • #3
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.
 
  • #4
Show us what you tried, so we can do a better job helping you.
 
  • #5
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
 
  • #6
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?
 
  • #7
KE=1/2mv2 and Us=1/2k(deltaX)^2 correct? so do i set these equal to each other?
 
  • #8
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?
 
  • #9
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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