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 13.0 m: (a) the initially stationary spelunker is accelerated to a speed of 3.40 m/s; (b) he is then lifted at the constant speed of 3.40 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 78.0 kg rescue by the force lifting him during each stage?

Total work done = = 29811.60 J

Explanation:

Since the person is being moved upward, the person’s potential and kinetic energy are increasing.

To determine the increase in potential energy, the following equation is used;

∆ PE = m * g * ∆ h

m = 78.0 Kg, g = 9.8 m/s, ∆ h = 13.0 m

∆ PE = 78.0 * 9.8 * 13.0 = 9937.20 J

Increase in kinetic energy is given by the following equation;

∆ KE = ½ * m * (vf² – vi²)

vf = 3.4, vi = 0

∆ KE = ½ * 78.0 * 3.4² = 450.84 J

Total work = 9937.20 + 450.84 = 10388.04 J

(b) He is then lifted at the constant speed of 3.40 m/s

Velocity is constant, therefore, there is no increase in kinetic energy. The only work done is the increase of potential energy

∆ PE = 78.0 * 9.8 * 13.0

∆ PE = 9937.20 J

(c) He is then decelerated to zero speed.

Since his velocity decreased from 3.40 m/s to 0 m/s, his kinetic energy decreased.

∆ KE = ½ * 78.0 * (0² - 3.4²)

∆ KE = - 450.84 J

∆ PE = 78.0 * 9.8 * 13.0 = 9937.20 J

Total work = 9937.20 - 450.84 = 9486.36 J

Sum of works = 10388.04 + 9937.20 + 9486.36 = 29811.60 J

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