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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**

F

Also possibly the Equation for kinetic energy

K=1/2mv

And also posibly

v

where v is velocity and a is acceleration

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

F

_{sp}=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=1/2mv

^{2}And also posibly

v

_{f}^{2}=v_{i}^{2}+2a[tex]\Delta[/tex]swhere 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 (v

_{f}

^{2}=v

_{i}

^{2}+2a[tex]\Delta[/tex]s) because v final should be zero, but I got the wrong answer. My incorrect answer was 82m/s.