An elevator mass of 1000 kg falls from a height of 12 m after a sudden failure i

Question

An elevator mass of 1000 kg falls from a height of 12 m after a sudden failure in the hoisting cable. The mass is stopped by a spring at the bottom of the shaft having a spring constant of 80 kN/m.

A) Determine the spring's compression distance in m that brings the falling elevator momentarily to rest.

B) Determine the acceleration in m/s/s that the elevator undergoes when brought to rest by the spring.

C) Determine the spring constant in kN/m that will bring the elevator and occupants to rest at an acceleration of 5 g.

The ansers are a)1.843 b)6.51 c)65400

Please show me all steps.

Explanation / Answer

A)

x = compression of spring

k = spring constant = 80,000 N/m

m = mass = 1000 kg

h = height = 12 m

using conservation of energy

spring potential energy = Potential energy of elevator mass

(0.5) k x2 = mgh

(0.5) (80000) x2 = 1000 x 9.81 x 12

x = 1.71 m

b)

Vi = velocity of elevator mass just before hitting the spring

using conservation of energy

kinetic energy = potential energy

(0.5) m Vi2 = mgh

Vi= sqrt(2gh) = sqrt(2 x 9.8 x 12) = 15.34 m/s

Vf = final velocity = 0 m/s

x = displacement = 1.71 m

a = acceleration

using the equation

Vf2 = Vi2 + 2 a x

02 = 15.342 + 2 a (1.71)

a = - 68.81 m/s2

c)

a = - 5 x 9.8 = - 49 m/s2

Vi= sqrt(2gh) = sqrt(2 x 9.8 x 12) = 15.34 m/s

Vf = final velocity = 0 m/s

x = displacement

using the equation

Vf2 = Vi2 + 2 a x

02 = 15.342 + 2 (-49) (x)

x = 2.40 m

using conservation of energy

spring potential energy = Potential energy of elevator mass

(0.5) k x2 = mgh

(0.5) k (2.4)2 = 1000 x 9.8 x 12

K = 40833.3 N/m

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