Question

An elevator cable breaks when a 970 kg elevator is 23.0 m above a huge spring (k = 2.10×10⁵ N/m ) at the bottom of the shaft. Calculate the work done by gravity on the elevator before it hits the spring. Calculate the speed of the elevator just before striking the spring. Calculate the magnitude of the amount the spring compresses (note that work is done by both the spring and gravity in this part.

Answer 1

a) Work done by weight on elevator prior to hitting spring = mgh
mgh = (970)(9.81)(23) =249311.34 J ANS (a)

b) V = √2gh = √[(2)(9.81)(23)] = √514 = 21.24 m/s ANS (b)

c) Work done by weight is stored as spring P.E. and further work by gravity:
249311.34 = 1/2kx² +mgx {where k = 2.10e5, x = compression distance}
0 = -249311.34 + (0.5)(2.10e5)x² + (970)(9.81)x
0 = - 249311.34 + 9515.7 x + 2.20e10 x²

Use formula

x = -b² () sqrt((b²-4ac) /(2a) we get
solve for x, positive root = 9.05*10⁷ m ANS (c)