In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00 {\rm m/s} when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 {\rm m} as it does so. During the motion a safety clamp applies a constant 17000-{\rm N} frictional force to the elevator.
What is the speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?
2 answers
3.46 m/s
conservation of energy.
KE + PE --> Wf + PEspring
.5mv^2 + mgh = 17000*1 + .5kx^2
v = 3.65
KE + PE --> Wf + PEspring
.5mv^2 + mgh = 17000*1 + .5kx^2
v = 3.65
0 answers