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I need some closure on this problem:

An engineer is designing a spring to be palced at the bottom of an elevator shaft. If the elevator cable should happen to break when the elevator is at a height h above the top of the spring, calculate the value of the spring constant k so that the passengers undergo an acceleration of no more than 5.0 g when brought to a rest. Let M be the total mass of the elevator and passengers.

Let y be the distance of the elevator from the spring. There are three locations of interest here:

Loc. 1 - When the elevator is at a height y = h.

Loc. 2 - When the elevator is at a height y = 0.

Loc. 3 - When the elevator is at a height of y = -x, i.e. when the elevator is at rest a distance x below the uncompressed height of the spring.

The total mechanical energy at

Loc. 1 is Mgh

Loc. 2 is 1/2Mv^2

Loc. 3 is 1/2kx^2 - Mgx

Now assuming the acceleration of the elevator is constant when on the spring, then

0 = v^2 - 2ax => v^2 = 2ax

Using conservation of energy, then

Mgh = 1/2Mv^2 = Max => x = gh/a

1/2Mv^2 = Max = 1/2kx^2 - Mgx => 1/2kx^2 = M(ax + gx) so

kx = 2M(a + g) => kgh/a = 2M(a + g) => k = 2Ma(a + g)/ (gh)

Since a <= 5g, then k <= 10Mg (6g) / (gh) = 60Mg/h. The book where I got this problem from claims the anwer to be 12Mg/h. I don't think I did anything wrong here though. What gives?

e(ho0n3

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# Homework Help: Need some closure

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