The cable of a 4400 lb elevator in the figure below snaps when the elevator is a

Question

The cable of a 4400 lb elevator in the figure below snaps when the elevator is at rest at the first floor so that the bottom is a distance d - 12.0 ft above a cushioning spring whose force constant is k 10 removing 1000 ft-lb of mechanical energy for each 1.00 ft that the elevator moves. ,000 lb/ft. A safety device clamps the guide rails Figure 13-21 (a) Find the speed of the elevator just before it hits the spring 24.4ft/s (b) Find the distance that the spring is compressed. (3.217 )× ft (c) Find the distance that the elevator will bounce back up the shaft. (d) Calculate approximately the total distance that the elevator will move before coming to rest. 52.8 Additional Materials Section 13.1

Explanation / Answer

b] Let the distance be x.

The final Kinetic energy is also zero.

total work done by all forces is zero.

mg(d+x) - frictional heat -0.5kx^2 = 0

4400*(12+x) - (12+x)*1000 - 0.5*10000*x^2 = 0

x = 3.22 ft answer

c] Let the distance be y

Work done by all forces is zero as final KE is zero,

-4400y -1000y +0.5*10000*3.21673^2 = 0

5400 y = 51736

y = 51736/5400 = 9.58 ft answer

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