An engineer is designing a spring to be placed 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 that the spring constant k should have so that passangers undergo an acceleration of no more than 5.0G when brought to rest. Let M be the total mass of the elevator and passengers.

Value that the spring constant k = 12Mg / h

Explanation:

According to 2nd law of Newton:

upward force of the spring= F

The weight of the elevator W = mg

            F = Mg = M(5g)

==> F =6Mg.

As the spring is compressed to its maximum distance ie s,the maximum upward acceleration comes just , Hence

  F =ks = 6Mg

==>   s = 6Mg/k

We have gravitational potential energy turning into elastic potential of the spring as the elevator starts at the top some distance h from the spring, and undergoes a total change in height equal to h + s, so:

            Mg(h+s) = 1/2ks2

And plugging in our expression for s:

         Mg(h+6Mg/k)= 1/2k(6Mg / k)2

          gh + 6M2g2/k = 1/2k(36M2g2 /k2)

        Mgh +6M2g2/k = 1/2k(36M2g2 /k2)

             gh + 6Mg2/k = 18Mg2 / k

                             gh = 12Mg2 / k

                               h = 12Mg / k

                               k = 12Mg / h

k = 18.0Mg/h

Explanation:

Considering the forces acting on the elevator, the weight of the elevator and the restoring force of the spring

From Newtown's second law of motion,

Kx – Mg = Ma

Given that a = 5.0g

Kx – Mg = M(5.0g)

Kx = 5.0Mg + Mg

Kx = 6.0Mg(1)

From the energy consideration

The potential energy change in falling through the height of h is converted in to the elastic potential energy of the spring

Elastic potential energy = Gravitational potential energy

1/2Kx² = Mgh

Kx² =2Mgh

x² = 2Mgh/K

x = √(2Mgh/K)

So substituting x in the equation (1) above

K× √(2Mgh/K) = 6.0Mg

Squaring both sides

K² × 2Mgh/K = 36.0M²g²

K× 2Mgh = 36.0M²g²

Dividing through by 2Mgh

We have that

K = 18.0Mg/h