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An engineer is designing a spring to be placed at the bottom of an elevator shaft.If the elevator cable breaks when the elevator is at a height h above the top of the spring, calculate the value that the spring constant k should have so that passengers undergo an acceleration of no more than 7.0 g when brought to rest. Let M be the total mass of the elevator and passengers.Express your answer in terms of the variables h, M, and appropriate constants.
k = ?

Respuesta :

Answer:

Explanation:

Acceleration will be in upward direction due to force by spring in upward direction .Let maximum compression be x .

maximum force  = k x -Mg

k x - Mg = 7Mg

kx = 8Mg

After falling by height h , compression in spring is x

1/2 k x² = Mg( h + x )

1/2 x 8Mg x = Mgh +Mgx

4Mgx = Mgh +Mgx

3 Mgx = Mgh

x = h / 3

kx = 8Mg

k h/3 = 8Mg

k = 24 Mg / h

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