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electriceel
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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.
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.