(adsbygoogle = window.adsbygoogle || []).push({}); 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 V^{2}- Friction

1/2 5000 V^{2}=1,350,000-60,000

V^{2}=564

V=24

Total energy: 1290000 ?????

(C)Use work energy principal to determine spring constant K.

MGH= 50,000N* -1 = -50,000N

SPE = 1/2Kx^{2}= 1/2 K

-50,000+1/2 K = 1,290,000

K= 2,680,000

This doesn't seem correct.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Elevator free fall

**Physics Forums | Science Articles, Homework Help, Discussion**