Answer:
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
Answer:
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
