Learning Goal: To practice Problem-Solving Strategy: Simple Harmonic Motion II: Energy. A child’s toy consists of a spherical object of mass

Question

Learning Goal: To practice Problem-Solving Strategy: Simple Harmonic Motion II: Energy. A child’s toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby’s crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 4.1 cm from its equilibrium position

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niczorrrr 5 years 2021-08-23T10:34:53+00:00 1 Answers 138 views 0

Answers ( )

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    2021-08-23T10:35:58+00:00

    Answer:

     K = 0.1357 J

    Explanation:

    This is a problem of the simple harmonium movement, where the system this former for ball and spring. This is represented by relate

              x = A cos (wt + Ф)

    They indicate the maximum speed of the system, let’s find the speed

              v = dx / dt

              v = – A w sin (wt + Ф)

    for maximum speed the sine is ±1

              v = A w

              w = v / A

    let’s reduce to the SI system

              W = 6cm (1m / 100cm) = 0.06 m

              w = 3.2 / 0.06

               w = 53.33 rad / s

    let’s substitute the values ​​in the equation

               x = 0.06 cos (53.34 t)

    phase angle is zero since the system is released from maxillary elongation

    let’s find the time for ax = 4.1 cm = 0.041 m

               cos (53.34t) = x / 0.06

               t = 1 / 53.34 cos -1 (x / 0.06)

               t = 1 / 53.35 cos-1 (0.04 / 0.06)

               t = 0.0187 0.818

               t = 0,0153

              v = Aw sin wt

               v = 0.06 53.34 sin (53.34 0.0153)

                v = 3.2 0.728

                v = 2.33 m / s

    now we can search for kinetic energy

               K = ½ m v²

                K = ½ 0.050 2.33²

                K = 0.1357 J

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