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 in three stages, each requiring a vertical distance of 11.1 m: (1) the initially stationary spelunker is accelerated to a speed of 4.06 m/s; (2) he is then lifted at the constant speed of 4.06 m/s; (3) finally he is decelerated to zero speed. How much work is done on the 69.4 kg rescue by the force lifting him during each stage?(adsbygoogle = window.adsbygoogle || []).push({});

I have to find how much work is done by the force lifting him at each stage.

First I found the average velocity 4.06/2 = 2.03 (i think because of asuming constant acceleration). Then i divide the distance by the average velocity to get 11.1 / 2.03 = 5.4679 s . So this is how long it takes to make it to the end of the 1st stage. Then i found the acceleration which is .7425 m/s/s. but i plugged it into the work formula and i got 572 J... but it is wrong.

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# Homework Help: Work and acceleration of rescue team

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