Spring and elevator

  • Thread starter NAkid
  • Start date
  • #1
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Homework Statement


The cable of an elevator of mass M = 1920 kg snaps when the elevator is at rest at one of the floors of a skyscraper. At this point the elevator is a distance d = 15.6 m above a cushioning spring whose spring constant is k = 23100 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of f = 13217 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.


Homework Equations


K1 + U(grav,1) + U(elastic,1) + W(other) = K2 + U(grav, 2) + U(elastic, 2)


The Attempt at a Solution


I tried using the above formula for work-energy. I set my origin at the point at which the elevator initially hits the spring. So,
0 (initially at rest so K1=0) + (1920)(9.8)(15.6) + 0 (spring not yet compressed) -(13217)y2 = 0 (v2=0 so K2=0) + (1920)(9.8)y2 + .5(23100)(y2)^2

Basically I then used quadratics to solve for negative value of y2. What am I doing wrong?
 

Answers and Replies

  • #2
Doc Al
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0 (initially at rest so K1=0) + (1920)(9.8)(15.6) + 0 (spring not yet compressed) -(13217)y2 = 0 (v2=0 so K2=0) + (1920)(9.8)y2 + .5(23100)(y2)^2
Recalculate the work done by the friction force. Over what distance does it act?
 
  • #3
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oh, should it be y1 - y2, because y2 is negative?
 
  • #4
Doc Al
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oh, should it be y1 - y2, because y2 is negative?
Sounds right.
 
  • #5
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ok now i think my algebra is just screwy because i keep getting an incorrect answer. i have

.5k(y2)^2 + (mg - f)y2 - mgy1 + fy1 = 0

solve for negative value of y2..
 
  • #6
Doc Al
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Looks good to me. Just plug in the numbers and solve.
 

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