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: (a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s; (b) he is then lifted at the constant speed of 4.70 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage

(a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s - 6106 J

(b) he is then lifted at the constant speed of 4.70 m/s - 5488 J

(c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage - 4869 J

Explanation:

knowing

d = 10 m

m = 56 kg

The work done by the applied force to pull the spelunker is given by

Wa + Wg = Kf - Ki

Wg = -mgd

First

Wa = mgd + 0.5 - 0.5

Wa = (56*9.8*10) + (0.5*56*)

Wa = 6106 J

Second

Kf = Ki

Wa = mgd + 0.5 - 0.5

Wa = 56*9.8*10

Wa = 5488 J

Third

Kf = 0

Wa = mgd + 0.5 - 0.5

Wa = (56*9.8*10) - (0.5*56*)

Wa = 4869 J

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!