Задача id9825

A lift cage weighing 5000 kg serves a shaft 900 m deep. When the cage is at the bottom of the shaft, a traction force of 60 kN begins to act vertically upward on it. 150 m after the start of the ascent, the traction force changes so that over the next 600 m the movement of the cage becomes uniform. Finally, the traction force changes again so that the cell stops when it reaches the top of the shaft. The friction force is considered constant and equal to 5 kN. Observe the movement of the cell in these areas and determine the time of rise.

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