Energy Problem Involving Spring Force

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SUMMARY

The discussion centers on calculating the landing speed of a jet plane using spring force principles. A jet plane weighing 1.7×104 kg lands on an aircraft carrier, where its tail hook engages a cable attached to a spring with a spring constant of 6.0×104 N/m, stretching 31 m to halt the plane. The correct approach involves applying the conservation of energy principle, equating kinetic energy to potential energy stored in the spring, leading to the formula v2 = kx2/m. The accurate landing speed is calculated to be approximately 60 m/s.

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  • Understanding of spring force and Hooke's Law
  • Familiarity with kinetic energy and potential energy equations
  • Knowledge of basic kinematics and motion equations
  • Ability to apply conservation of energy principles
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1. As a 1.7×104 jet plane lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable is attached to a spring with spring constant 6.0×104 . If the spring stretches 31 to stop the plane, what was the plane's landing speed? (Answer in m/s)



2. Homework Equations
Fsp=k[tex]\Delta[/tex]s where k is the spring constant and [tex]\Delta[/tex]s is the change in position.
Also possibly the Equation for kinetic energy
K=1/2mv2
And also posibly
vf2=v i2+2a[tex]\Delta[/tex]s
where v is velocity and a is acceleration

Also the conservation of energy and momentum equations are possible along with f=ma




The Attempt at a Solution



I found the spring force to be 1860000 N and then I lost where to go exactly. I tried to use F=ma to find acceleration, and then use kinematics (vf2=v i2+2a[tex]\Delta[/tex]s) because v final should be zero, but I got the wrong answer. My incorrect answer was 82m/s.
 
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Maybe consider using the KE and PE of spring?

1/2*m*v2 = 1/2*k*x2

v2 = k*x2/m = 60,000*312/17,000
 

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