Question

(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.

Answer 1

Answer: 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

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Answer 2

Answer:

k = 12Mg/h

Explanation:

For we to be able to solve this,

Making the spring is compressed by a distance to be x, before the will lift stop, then it will be

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

Kx - Mg = M ( 5g ) as equ(2)

Let's make x to be the subject of formula in equation (2)

We are going to have

Moving Mg to right hand side of the equation

Kx = 5Mg + Mg

Adding the like terms we have

Kx = 6Mg

Dividing both sides by k we have

x = 6Mg/k as equation (3)

Substituting equation 3 into 1 we have

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

Expanding the bracket

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

Subtracting the like terms

h =18Mg/k - 6Mg/h

We have k as

k = 12Mg/h

Which is our answer.