Question

n a "worst-case" design scenario, a 2000-kg elevator with broken cables is falling at 4.00 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 m as it does so. During the motion a safety clamp applies a constant 17000-N frictional force to the elevator.

How would I go about finding the velocity and acceleration of the elevator after it has compressed the spring 1 meter?

Answer 1

initial kinetic energy of elevator wen touching the cushion is 1/2mv^2 that is converted to potential energy of spring=k*2+frictioanl force*2
find k then using same concept
1/2mv^2=k(1)+frictioanl force*1
v=?
2as=v^2-u^2
v=0 as elevator comes to rest
this is all based on moment retriceumptions

Answer 2

and we are comfortably ignoring gravity........

Answer 3

to make problem simple

Answer 4

oh and the spring constant is 1.06 x 10 ^4 N/m.

so im confused as to which formula i should use. should i use the 1/2mv^2?

Answer 5

are u able to follow:
1/2*mv^2=kx+frict force*x
v^2-u^2=2ax
tahts all
see wat energy is converted to what thats all
equate forces if equilibrium conserve energy tahts all wat nature and physics is about

Answer 6

thank you salini

Answer 7

:)