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 7.70 m: (a) the initially stationary spelunker is accelerated to a speed of 3.80 m/s; (b) he is then lifted at the constant speed of 3.80 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 62.0 kg rescue by the force lifting him during each stage?

Answer 1

Force on cable = m (g+a)
work done = force * distance
phase 1
average speed up v = (0+3.8)/2 =1.9 /s
t = 7.7/1.9 = 4.05 seconds
a = change in speed/change in time = 3.8/4.05 = .938 m/s^2
F = 62 (9.8+ .938) = 667 N
work = 667*7.7 = 5126 Joules
phase 2
F = m g = 62*9.8 = 608 N
F * 7.7 = 4678 Joules
phase 3
F = 62 (9.8 -.938) = 549 N
work = 549*7.7 = 4230 Joules