Question

5. An elevator filled with passengers has a mass of 1570 kg a.) The elevator accelerates upward from rest at a rate of 11.2 m/s² for 1.95, Calculate the tension in the cable Supporting the elevator. b.) The elevator continues upward at constant velocity for 5.ls. What is the tension in the cable during this time? TE_ Units c.) The elevator decelerates at a rate of 0.8m/s² for 3.2s. What is the tension in the cable during deceleration ? T= units. d.) How high has the elevator moved above its original starting point and what is its final velocity ? O - units - units

Answer 1

- @ T-w=ma T-15709. 1570*102 T= 1570(9.8+102) T=1727ON Aus - T-ng-mo) T=157089,8 =15386 N Aus & © mg - Tama 5 Tem(g-a) = 1570x9=14130N

SE St S. 9 U=0 - J a=1.2 t=1.95€ azo t=s.tsee a=0.8m/2 t=3.2 sec. = urat V = 0+1.2X109 = 2.28 mis V2 = %= 2.28 m/s Vz = Vztat = 2.28 - 3.2X0.8 = -0.28 m/s Se ut tyg at? = 0 + x1.2 *(1.93= 216 mg Są= 2.28% 5.7 = 13m $z= 2,28*3-2 + { (-0-8)( 3.2} = 3.2m Stotal = 2. 16+13+ 3.2 = 17:36m ] dus