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 5.90 m: (a) the initially stationary spelunker is accelerated to a speed of 2.30 m/s; (b) he is then lifted at the constant speed of 2.30 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 69.0 kg rescue by the force lifting him during each stage

Answer 1

W₃ = 3310.49 J , W3 = 3310.49 J

Step-by-step explanation:

We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections

We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics

v2 = v₀² + 2 a₁ y

as they rest part of the rest the ricial speed is zero

v² = 2 a₁ y

a₁ = v² / 2y

a₁ = 2.3² / (2 5.90)

a₁ = 0.448 m / s²

with this acceleration we can calculate the applied force, using Newton's second law

F -W = m a₁

F = m a₁ + m g

F = m (a₁ + g)

F = 69 (0.448 + 9.8)

F = 707.1 N

Work is defined by

W₁ = F.y = F and cos tea

As the force lifts the man, this and the displacement are parallel, therefore the angle is zero

W₁ = 707.1 5.9

W₁ = 4171.89 J W3 = 3310.49 J

Let's calculate for the second part

the speed is constant, therefore they relate it to zero

F - W = 0

F = W

F = m g

F = 60 9.8

F = 588 A

the job is

W² = 588 5.9

W2 = 3469.2 J

finally the third part

in this case the initial speed is 2.3 m / s and the final speed is zero

v² = v₀² + 2 a₂ y

0 = vo2₀² + 2 a₂ y

a₂ = -v₀² / 2 y

a₂ = - 2.3²/2 5.9

a2 = - 0.448 m / s²

we calculate the force

F - W = m a₂

F = m (g + a₂)

F = 60 (9.8 - 0.448)

F = 561.1 N

we calculate the work

W3 = F and

W3 = 561.1 5.9

W3 = 3310.49 J

total work

W_total = W1 + W2 + W3

W_total = 4171.89 +3469.2 + 3310.49

w_total = 10951.58 J