Question

An elevator cable breaks and the 1550 kg elevator drops 30 m onto a large spring at the bottom of the shaft. The stiffness of the spring is k = 90, 000 N/m. Calculate the kinetic energy and velocity of the elevator after falling the 30 m distance. Find the distance of maximum compression of the spring. Determine the maximum acceleration of the elevator. What is the fastest velocity reached by the elevator during the fall?

Answer 1

(A) Applying energy conservation,

PE + KE = constant

m g h + 0 = 0 + KE

KE = 1550 x 9.8 x 30 = 455700 J

KE = m v^2 /2

455700 = 1550 v^2 /2

v = 24.25 m/s

(B) applying energy conservation,

KE + spring PE +gravitational PE = constant

455700 + 0 + 0 = 0 + k d^2 /2 + m g (-d)

45000 d^2 - 15190 d - 455700 = 0

d = 3.35 m

(D) KE = 455700 - 45000d^2 + 15190d

d(KE)/dd = - 2(45000)d + 15190 = 0

d = 0.17 m

KE will be max at this distance.

KE = 455700 - 45000(0.17^2) + 15190(0.17) = 456981.8

v = 24.3 m/s