An elevator cable snaps when the elevator is at rest at a height of d=6.00 m abo

Question

An elevator cable snaps when the elevator is at rest at a height of d=6.00 m above an ideal spring. The mass of the elevator is 1700 kg. The force constant of the spring is 146,000 N/m. Due to its safety mechanism, the elevator encounters a constant friction of 4450 N as it moves.

(a) What is the speed of the elevator just before it makes contact with the spring?

(b) By how much is the spring compressed when the elevator (momentarily) comes to rest?

(c) The elevator bounces back up the shaft. What’s the maximum height it reaches above its position in part b?

Explanation / Answer

(A) APplying work - energy conservation,

Work done by gravity + work done by friction = change in KE

(1700 x 9.81 x 6) - (4450 x 6) = 1700(v^2 - 0) / 2

v = 9.29 m/s

(B) Applying work - energy theorem again.

(1700 x 9.81 x d) - (4450 d) - (146000 d^2 / 2 ) = 0 - 1700 x 9.29^2 /2

16677 d - 4450 d - 73000 d^2 = - 73362

73000 d^2 - 12227 d - 73362 = 0

d = 0.868 m

(C) Applying work- energy theorem again,

- (1700 x 9.81 x 0.868) - (4450 x 0.868) + (146000 x 0.868^2 /2) = 1700 v^2 /2 - 0

v = 6.57 m/s

Work - energy theorem to find the height,

- (1700 x 9.81 x h) - 4450h = 1700 (6.57^2 /2 - 0 )

h = 1.74 m

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