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## Homework Statement

An elevator is traveling at 30 m/s just as it touches a cushioning spring. The safety clamps engage at that moment and provide 20,000N of frictional force. The mass of the elevator is 3000kg and the spring constant is 15000 N/m. How fast will the elevator be traveling after the spring compresses 0.8m ?

## Homework Equations

K

_{1}+W

_{f}= U

_{g2}+U

_{s2}+K

_{2}

Allow the point that the elevator touches the spring to be y=0

## The Attempt at a Solution

1/2*m(v

_{1})

^{2}+F

_{f}d=mgy

_{2}+1/2*kx

^{2}+1/2*mv

_{2}

^{2}

Manipulating this around and plugging in numbers gives an answer of 30.03 m/s for me. Am I missing something here?

Edit: d is just the distance along which friction occurs, -.8 m?

x is the compression of the spring, also .8m

y

_{2}is the distance traveled by the elevator, so -.8m

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