-S e The Cable of The 2400 Kg E WileyPLUS https://edugenxvileplus.com, ed To see
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-S e The Cable of The 2400 Kg E WileyPLUS https://edugenxvileplus.com, ed To see favorites here, select then , and drag to the Favorites Bar folder. Or import from another browser. Import favorites Assignment> Open Assignment uni FULL SCREEN PRINTER VERSION BACK NEXT ASSIGNMENT The cable of the 2400 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d - 2.7 m above a spring of spring constant k- 0.15 MN/m. A safety device clamps the cab against quide rails so that a constant frictional force of 2.8 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.) 006 022 049 077 (a) N (b) NumberT1.0134 (c) Number[2.923 (d) NumberT23.36 6.83 Un Review Results by Activate Windows All Rights Reserved. A Division of Version 4.24.5.1 O Type here to search ^P41) 6:54 PM 18-Mar-18 4Explanation / Answer
d)
Let D be the total distance that the cab will move before coming to rest.
From part (c), you know that the elastic potential energy stored in the spring at its maximum compression (after the first fall) is:
EPE = k×X²/2
EPE = (150000 N/m)×(1.0134 m)²/2
EPE = 77023.47 J
All that energy must be 'consumed by friction'.
So
(2800 N)×D = (77023.47 J)
D = 27.51 m
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