(III) 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 5.0 g when brought to rest. Let M be the total mass of the elevator and passengers.

12Mg/h

Explanation:

Let the spring is compressed by a distance x,before the lift stops,then

Mg(h+x)= 1/2 kx^2 1

Kx - Mg = M ( 5g ) 2

Make x the subject in equation 2

Kx = 5Mg + Mg

Kx = 6Mg

x = 6Mg/k 3

Put equation 3 into 1

Mg ( h + x ) = 1/2 kx^2

Mgh + Mgx = 1/2kx^2

Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2

Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2

h =18Mg/k - 6Mg/h

K = 12Mg/h

the body's situation won't change the forces are equal opposit and on the same body

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