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 5.30 m/s. (2) Then he is then lifted at the constant speed of 5.30 m/s. (3) Finally he is decelerated to zero speed. How much work is done on the 84.0 kg rescuee by the force lifting him during each stage? Work done on rescuee during stage (1)? __ kJ Work done on rescuee during stage (2)? ___ kJ Work done on rescuee during stage (3)? ___kJ

Explanation / Answer

Each stage is 10.0m.

Stage 1: Work is required to lift him 10m AND give him a velocity of 5.30 m/s
W1 = m*g*s + 1/2*m*v^2 = m*(g*s + 1/2*v^2)
= 84*(9.8*10 + 1/2*5.3^2) = 9.41 KJ

Stage 2: Work is required to move him 10m
W2 = m*g*s = 84*9.8*10 = 8232 J =8.32KJ

Stage 3: His kinetic energy is converted to potential, so he only needs the remainder from the winch.

W3 = m*g*s - 1/2*m*v**2 = m*(g*s - 1/2*v**2)
= 84*(9.8*10 - 1/2*5.3**2) = 7.05 KJ

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