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 10.0 m. (1) First, the initially stationary spelunker is accelerated to a speed of 4.80 m/s. (2) Then he is then lifted at the constant speed of 4.80 m/s. (3) Finally he is decelerated to zero speed. How much work is done on the 80.0 kg rescuee by the force lifting him during each stage? Work done on rescuee during stage (1) Work done on rescuee during stage (2) Work done on rescuee during stage (3)

Explanation / Answer

a) In stage 1 a=0.288 so work done is m*a*x=256.32J b) In stage 2 a=0 so work done is 0 c) In stage 3 a=-0.288 so work done is -256.32 J

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