A solid metal block with a mass of 1.30 kg is attached to a spring and is able to oscillate horizontally with negligible friction. The block

Question

A solid metal block with a mass of 1.30 kg is attached to a spring and is able to oscillate horizontally with negligible friction. The block is pulled to a distance of 0.200 m from its equilibrium position, held in place with a force of 19.0 N, and then released from rest. It then oscillates in simple harmonic motion. (The block oscillates along the x-axis, where x = 0 (e) What is the maximum acceleration of the block? (Enter the magnitude in m/s2.) m/s2 (f) At what position(s) (in m) on the x-axis does the maximum acceleration occur? (g) What is the total mechanical energy of the oscillating spring-block system (inJ)? h) What is the speed of the block (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium? m/s () What is the magnitude of the acceleration of the block (in m/s) when its position is equal to one-third of the maximum displacement from equilibrium? m/s2 is the equilibrium position.)

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Ladonna 3 years 2021-07-22T04:55:52+00:00 1 Answers 300 views 0

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  1. RobertKer
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    2021-07-22T04:57:16+00:00
    Reply

    Answer:

    e)   a = 14.62 m / s², f)  the maximum acceleration occurs at the points x = ± A
    , g)    Em = 1.9 J
    , h)   v = 1.61 m / s, i) a = -4.82 m / s²

    Explanation:

    A system of a spring with a mass is described by the equation

                 x = A cos (wt + Ф)

    Where

               w = √ k / m

               

    e) ask to find the acceleration, for this we use the definition of acceleration

                v = dx / dt

                a = dv / dt

                v = -A w cos (wt + Ф)

                a = -A w² cos (wt + Ф)

    The acceleration is maximum when the cosine is ± 1

                a = A w²

             

    In the exercise, indicate that the amplitude is A = 0.200 m

    To find the constant k, let’s use Newton’s equilibrium equation

              F – Fe = 0

              F = k x

              k = F / x

              k = 19 / 0.2

              k = 95 N / m

    Now we can calculate the angular velocity

              w = √ k / m

             w =√ (95 / 1.3)

             w = 8.55 rad / s

    Substitutes in the maximum acceleration equation

              a = 0.2 8.55²

              a = 14.62 m / s²

    f) the acceleration depends on the cosine and the cosine is maximum in the express of the displacement, therefore the maximum acceleration occurs at the points x = ± A

    g) mechanical energy is given by the equation

             Em = ½ k A²

             Em = ½ 95 0.2²

             Em = 1.9 J

    h) what is the speed when the position is x = 1/3 A

            Em = K + U

            Em = ½ m v² + ½ k x²

            v² = (Em – ½ k x²) 2 / m

            v² = (1.9 – ½ 95 (0.2 /3)²)  2 / 1.30

            v² = 2,598

            v = 1.61 m / s

    i) block acceleration when x = A / 3

    For this we must find fi and the time to reach this position

    Let’s look fi

    As the system starts from rest the speed starts is zero

             V = A w sin (0+ Ф)

              0 = sin Ф

             Ф = 0

    Now we look for the time to reach the displacement x = A / 3

               A / 3 = A cos 8.5 t

               1/3 = cos 8.5t

               8.5t = cos-1 1/3

               t = 1 / 8.5 cos-1 1/3

               t = 0.1448 s

    Now we substitute in the acceleration equation

              a = – A w² cos wt

              a = – 0.2 8.5² cos (8.5 0.1448)

              a = -4.82 m / s²

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