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

   W₃ = 3310.49 J ,  W3 = 3310.49 J

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

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.

+15-35=-20

going in opposite directions