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A descending elevator of mass 420 kg is uniformly decelerated to rest over a distance of 4 m by a cable in which

the tension is 5519 N. The acceleration due to gravity is 9.8 m/s 2

a) Calculate the speed vi of the elevator at the beginning of the 4 m descent. Answer in units of m/s

b) Calculate the acceleration of the elevator during the fourth 5 s interval (from 15 s to 20 s). Answer in units of m/s 2 . 

c)What is the velocity of the elevator at the end of the fourth 5 s interval (at 20 s)? Answer in units of m/s.

Top Answer

a) The initial velocity is -5.2 m/s (downward). b) The acceleration... View the full answer

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