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 6.00 m: (a) the initially stationary spelunker is accelerated to a speed of 4.50 m/s; (b) he is then lifted at the constant speed of 4.50 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 55.0 kg rescue by the force lifting him during each stage? (a) Number Units (b) Number Units (c) Number Units

Explanation / Answer

Applying work - energy theorem,

Work done by force + work doneby gravity = change in KE

(A) W + (-55 x 9.8 x 6) = 55 (4.50^2 - 0^2) /2

W - 3234 = 556.9

W = 3790.9 J

(B) W - 3234 = 55 ( 4.50^2 - 4.50^2)/2

W = 3234 J

(c) W - 3234 = 55(0^2 - 4.50^2) /2

W = 3234 - 556.9

W = 2677.1 J

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