The cable of an elevator of mass M = 3560 kg snaps when the elevator is a rest at one of the floors of a skyscraper. At this point the elevator is a distance d = 13.4 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 = 6542 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.

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jsimpson11000 jsimpson11000 · Answer 1 · 2024-03-23T17:01:53+01:00

Answer:

Explanation:

Upward friction force = 6542 N

Downward force due to gravity = m g = 3560 kg * 9.81 m/s^2 = 34923.6 N

Net downward force = 32923.6 - 6542 = 28,381.6 N

Use F = ma to find acceleration = 7.97236 m/s^2

Velocity at the bottom of the shaft

d = 1/2 at^2

shows t = 1.8334 seconds

vf = a t = 7.97236 x 1.8334 = 14.617 m/s

Kinetiic energy is then 1/2 m v^2 = 1/2 3560 x 14.617^2 380313.48 Joules

The spring must do this much work to stop the elevator ( 380318.48 J)

1/2 k x^2 = 380313.48

1/2 (23100) x^2 = 380313.48 shows x = 5.74 meters

Alternative :

Net downward force = 28381.6 N for an equivalent mass of 2893.129

Potential energy is converted to Kinetic Energy and is what the spring work must be to stop the elevator

PE = mgh = 2893.129 * 9.81 * 13.4 = 380313.48 ( as before)

then work of spring = 1/2 kx^2 again shows x = 5.74 m