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In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 m as it does so. Spring coefficient is 10.6 kN/m. During the motion a safety clamp applies a constant 17000-N frictional force to the elevator. When the elevator is 1.00 m below point where it first contacts a spring, what is its acceleration?

Respuesta :

Answer:

4 m/s²

Explanation:

 When the elevator is 1 m below point of contact , compression will be 1 m.

Restoring force in the spring will be 10600 N. Friction force of 17000N  will also act in upward direction . The weight of 2000 x 9.8 N will act downwards

Force in down ward direction = 2000 x 9.8

= 19600 N

Force in upward direction

= 10600 + 17000

= 27600 N

Net force in upward direction

= 27600 - 19600

= 8000 N

Acceleration in upward direction

= 8000 / 2000

= 4 m/s²

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