an elevator car weighing 50,000 N is 27m from the ground. the cable snaps and the elevator goes into free fall. Shock absorbers have been provided, the coil length is 1 m. Assume elevator is able to lock onto the shaft at the point of contact w/ the absorber so that it does not bounce back. When elevator comes into contact w/ absorbers there is friction force of 60,000N acting on it as it descends that 1 m distance. (A) Total energy of the elevator 27m above ground? MGH= 50,000N*27 MGH = 1,350,000 (B)At the bottom of the shaft,when elevator comes to rest what is the total energy (take into account work done by friction) MGH= 0 KE = 1/2 M V2- Friction 1/2 5000 V2=1,350,000-60,000 V2=564 V=24 Total energy: 1290000 ????? (C)Use work energy principal to determine spring constant K. MGH= 50,000N* -1 = -50,000N SPE = 1/2Kx2 = 1/2 K -50,000+1/2 K = 1,290,000 K= 2,680,000 This doesn't seem correct.