The winding cages in mine shafts are used to move workers in and out of the mines. These cages move much faster than any of the other commercial elevators. In on South African mine, speeds of up to 65.0 km/hr area attained. The mine has a depth of 2072 m. Suppose two cages start their downward journey at the same moment. The first cage immediately attains the maximum speed (an unrealistic situation), the proceeds to descend uniformly at that speed all the way to the bottom. The second cage starts at rest and ten increases its speed with a constant acceleration of magnitude 4.00 x 102 m/s2. How long will the trip take for each cage? Which cage will reach the bottom of the mine shaft first?

The unit system allows us to exchange magnitudes with precision and without antiques, for which we will use the intentional system of measurements (SI), let's reduce

v = 65 km / h ( [tex]\frac{1000m}{1 km}[/tex] ) ( [tex]\frac{1h}{ 3600 s}[/tex] ) = 18.06 m / s

Kinematics allows finding the relationships between position, velocity and acceleration.

In this exercise we have two cages, work each one separately

Cage 1

They indicate that it is going at a constant speed, for which we can use the relations of uniform motion

[tex]v = \frac{d}{t} \\t = \frac{d}{v} \\t_1 = \frac{2072}{18.06}[/tex]

t₁ = 114.76 s

Cage 2

Stop from rest so your initial velocity is zero

y = v₀ t + ½ a t²

y = 0 + ½ a t²

t = [tex]\sqrt{\frac{2y}{a} }[/tex]

t₂ = [tex]\sqrt\frac{2 \ 2072}{4 \ 10^2}[/tex]

t₂ = 0.32 s

In conclusion using the kinematics relations we find the results are:

a) The time of cage 1 is t1 = 114.76 s and the time of cage 2 e t2 = 0.32 s

b) Cage 2 reaches the bottom first

learn more about kinematics here:

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