A cave rescue team lifts an injured spelunker directly upward and out of a sinkh

Question

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 11.2 m. (1) First, the initially stationary spelunker is accelerated to a speed of 5.20 m/s. (2) He is then lifted at the constant speed of 5.20 m/s. (3) Finally he is decelerated to zero speed. How much work is done on the 86.0 kg rescuee by the force lifting him during each stage?

Explanation / Answer

1) W1=F.d=ma.d=86*(5.22/2*11.2)*11.2=43*5.2*5.2 J

2)W2=0 because F=0

3)W3=-W1

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