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. (1) First, the initially stationary spelunker is accelerated to a speed of 4.10 m/s. (2) Then he is then lifted at the constant speed of 4.10 m/s. (3) Finally he is decelerated to zero speed. How much work is done on the 70.0 kg rescuee by the force lifting him during each stage

Answer 1

The workdone for the stage 1,2,3 are;

1) Wa = 7455.35 J

2) Wb = 6867 J

3) Wc = 6278.65 J

Step-by-step explanation:

Shown in the attachment....

Answer 2

(1)7448.35J

(2)6680 J

(3) 6091.65 J

Step-by-step explanation:

Given:

distance 'd'= 10.0 m

mass 'm' = 70kg

Workdone by the applied force to pull spelunker can be determined by this equation, i.e

Wa + Wg = Kf - Ki

Where Wg=-mgd

(1) In the first stage, workdone by the applied force can be determined by

Wa= mgd + 1/2 mVf² - 1/2mVi²

Wa= (70 x 9.8 x 10) + (1/2 x 70 x 4.10²)

Wa= 6680+588.35

Wa= 7448.35J

(2) Then he is then lifted at the constant speed, so Kf=Ki

Workdone by the applied force will be

Wa= mgd + 1/2 mVf² - 1/2mVi²

Wa= mgd

Wa=70 x 9.8 x 10

Wa=6680 J

(3) In third part, he is decelerated to zero speed. So, Kf=0

Workdone by the applied force will be

Wa= mgd - 1/2mVi²

Wa= (70 x 9.8 x 10) - (1/2 x 70 x 4.10²)

Wa= 6680-588.35

Wa= 6091.65 J