How Do You Calculate Elevator Motion and Energy After a Cable Snap?

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The discussion revolves around calculating the motion and energy of a 1800 kg elevator after its cable snaps. Key calculations include determining the elevator's speed just before hitting a spring, the maximum compression of the spring, and the subsequent bounce-back distance. Participants are using conservation of energy principles but are struggling with signs and the application of friction in their equations. There is confusion over whether to consider compression as negative and how to properly account for friction during the elevator's upward motion. The conversation highlights the need for clarity in applying physics concepts to solve the problem accurately.
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elevator problem - urgent help!

The cable of the 1800 kg elevator snaps when the elevator is at rest at the first floor, where the cab bottom is a distance d = 3.7 m above a cushioning spring whose spring constant is k = 0.15 MN/m. A safety device clamps the elevator against guide rails so that a constant frictional force of 4.4 kN opposes the motion of the elevator.

(image attached - i don't know how to make it embedded, sorry)

(a) Find the speed of the elevator just before it hits the spring._____ m/s
(b) Find the maximum distance x that the spring is compressed.______ m
(c) Find the distance that the elevator will bounce back up the shaft._____m
(d) Using conservation of energy, find the approximate total distance that the elevator will move before coming to rest._____m


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alrite let's see...

a) I've gotten the first answer by dKinetic + dPotGrav (or Ug) + E thermal (from friction) = 0 from conservation of energy... solve for V final from the kinetic part... 7.37ms-1

b) i believe it should be -1/2mv^2 - mgS + 1/2kS^2 + F(of friction)*S = 0 but I'm not sure if I'm getting the signs wrong... should my answer be negative because its compressing or not?... not getting the right answer... help please!

c) now i need my result from b) as x and is it 1/2kx^2 = mg(d up) + Friction(s +d up) solve for (d up)... Note: not sure if should just use answer from b) times the friction force or is it friction * (d up)... instead of what my equation says...?

d) to be honest I'm sorry but i have no clue to this one... and without b) and c)... I'm having trouble understanding it...

i hope some1 can help me out?! thanks in advance!
 

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Hmm, I am stuck on c) and d) too. but i can help you on b) you have it right but use the velocity you got in a) and if you have a TI-86 or higher just use solver to find distance from the quadratic equation you get.

I used: .5kS^2=.5mv^2-(force of friction)S+mgS
 
question?

i solved the quadratic, but is it the negative s because its in compression or not?...

i got either 0.90 or -0.72... I'm thinking its the positive... but i only have 1 try left... which one did u use? just want to make sure!
 
It is .9

but your equations for C and D don't work (I did the same thing as you did in C and the answer was wrong) so try it yourself and if it is still wrong were going to have to brainstorm.
 
actually, ur right... friction only acts during d up... my original equation had friction acting at (s + d up)

it should be:

1/2kx^2 = mg(d up) + Friction(d up)
 
other ?

hey randomphyre, i was wondering... have u gotten around to ch8 #66 part (c)... any clues? its the crates falling on the conveyor belt...

energy supplied by motor?
 
haha can get all the other parts except that. Its got me stumped too.
 
did u get d) for the elevator?

Ugrav (as a function of S + dup) = Uspring (as a function of S) + F(friction) L

solve for L... i thought it made sense... but i didn't get the right answer!... wut do u think??
 
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