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 4.60 m/s. (2) Then he is then lifted at the constant speed of 4.60 m/s. (3) Finally he is decelerated to zero speed. How much work is done on the 87.0 kg rescuee by the force lifting him during each stage?

Work done on rescuee during stage (1)?

Work done on rescuee during stage (2)?  = 8.526

Work done on rescuee during stage (3)?

Explanation / Answer

1st stage:ht = 10 m,v = 4.6 a= v^2/2h=1.058
W=m(acc)h=87*(1.058+9.8)*10=9533.46 J=9.53346 kJ
2nd stage = mgh=8526 J=8.526 kJ
3rd stage a=-1.058
W= m(9.8-1.058)h=7605.54 J =7.60554 kJ

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