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 15.0 m: (a) the initially stationary spelunker is accelerated to a speed of 2.40 m/s; (b) he is then lifted at the constant speed of 2.40 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 85.0 kg rescue by the force lifting him during each stage

Answer 1

A) 12752.55 J

B) 12507.75 J

C) 12262.95 J

Step-by-step explanation:

We are given;

Mass;m = 85 kg

Vertical distance; d = 15 m

From change in kinetic energy, the work done by the applied force to pull the spelunker is given by;

Change in kinetic energy = Wa + Wg

Where Wg = -mgd

A) In the first stage;the the work done is given by;

Wa = mgd + ½m(v_f)² - ½m(v_i)²

Since initially stationary, v_i = 0

So, we have;

Wa = mgd + ½m(v_f)²

v_f = 2.4 m/s

So,

Wa = (85 × 9.81 × 15) + ((1/2) × 85 × 2.4²)

Wa = 12752.55 J

B) for the second stage, there is a constant speed of 2.4 m/s

So, v_f = v_i

Thus; Wa = mgd

Wa = (85 × 9.81 × 15)

Wa = 12507.75 J

C) he finally decelerated to zero.

So v_f = 0 while v_i is 2.4 m/s

Thus;

Wa = mgd - ½m(v_i)²

Wa = (85 × 9.81 × 15) - (½ × 85 × 2.4²)

Wa = 12507.75 - 244.8

Wa = 12262.95 J