Elevator free fall

  • #1
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.
 

Answers and Replies

  • #2
960
0
it looks very good (and a well posted question and attempt at solution!) up until part c:

The energy stored in a spring is 1/2kX^2

Equate this to the energy lost by "friction",
 
  • #3
Thank you Denverdoc.


Is this what you mean?


MGH= 50,000N* -1 = -50,000N
SPE = 1/2Kx2 = 1/2 K (1)

-50,000 + 1/2 K (1) = 60,000 N (??)

1/2 K = 110,000

K= 220,000
 
  • #4
960
0
not quite: mgh = 1/2 kx^2. The kneticenergy gets converted to the potemtial energy of the spring.
 
  • #5
Is this what you mean?

50,000 = 1/2 K X ^2

100,000 = K
 

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