Calculating Spring Stretch in Aircraft Carrier Landing

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
jjd101
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0

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
 

What is the formula for calculating the period of a spring?

The formula for calculating the period of a spring is T = 2π√(m/k), where T is the period in seconds, m is the mass attached to the spring in kilograms, and k is the spring constant in newtons per meter.

How does the mass of an object affect the period of a spring?

The mass of an object affects the period of a spring by increasing or decreasing its inertia. A larger mass will have a longer period, while a smaller mass will have a shorter period. This is because a larger mass requires more force to accelerate, resulting in a longer time to complete one oscillation.

What is the relationship between the spring constant and the period of a spring?

The spring constant and the period of a spring are inversely proportional. This means that as the spring constant increases, the period decreases, and vice versa. This is because a higher spring constant indicates a stiffer spring, which requires more force to stretch or compress, resulting in a shorter period.

How does gravity affect the motion of a spring?

Gravity affects the motion of a spring by providing a force that opposes the spring's oscillations. This force, known as weight, increases with the mass of the object attached to the spring and can slow down the spring's motion and affect its period. However, for most practical spring problems, the effect of gravity is negligible.

What factors can affect the accuracy of calculations for a motion + spring problem?

Some factors that can affect the accuracy of calculations for a motion + spring problem include air resistance, friction, and the elasticity of the spring. These factors can introduce additional forces that can alter the motion of the spring and affect the accuracy of the calculated period.

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