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

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

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Thái Dương 2 months 2021-09-02T15:39:51+00:00 1 Answers 3 views 0

Answers ( )

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    2021-09-02T15:41:21+00:00

    Answer:

       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

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