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: (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

Answer 1

Answer:

(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 $mv^{2}_{f}$ - 0.5 $mv^{2}_{i}$

Wa = (56*9.8*10) + (0.5*56*$4.7^{2}$)

Wa = 6106 J

Second

Kf = Ki

Wa = mgd + 0.5 $mv^{2}_{f}$ - 0.5 $mv^{2}_{i}$

Wa = 56*9.8*10

Wa = 5488 J

Third

Kf = 0

Wa = mgd + 0.5 $mv^{2}_{f}$ - 0.5 $mv^{2}_{i}$

Wa = (56*9.8*10) - (0.5*56*$4.7^{2}$)

Wa = 4869 J