Question

A 2000-kg elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of the shaft. The spring is intended to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17,000-N friction force to the elevator. What is the necessary force constant k for the spring? (1.06 x 104 N/m)

Name: a 2000 kg elevator with broken cables in a test rig is falling at 400 ms when it contacts a cushioning spring at the bottom of the shaft the spring is intended to stop the elevator compressing 200 m a
Uploaded: 2021-11-16T19:05:55-08:00
Duration: 2 min 29 s
Channel: Nishant Kumar
Description: a 2000 kg elevator with broken cables in a test rig is falling at 400 ms when it contacts a cushioning spring at the bottom of the shaft the spring is intended to stop the elevator compressing 200 m a

          A 2000-kg elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts 
a cushioning spring at the bottom of the shaft. The spring is intended to stop the elevator, 
compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 
17,000-N friction force to the elevator. What is the necessary force constant k for the 
spring?  (1.06 x 104 N/m)

Added by Victoria W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Nishant Kumar

11/16/2021

Step 1

The formula for kinetic energy is 1/2mv^2, where m is the mass and v is the velocity. So, the kinetic energy of the elevator is 1/2 * 2000 kg * (4.00 m/s)^2 = 16,000 J. Show more…

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A 2000-kg elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of the shaft. The spring is intended to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17,000-N friction force to the elevator. What is the necessary force constant k for the spring? (1.06 x 104 N/m)