A 0.16-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 140 N/m. The block is

Question

A 0.16-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 140 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = -1.2×10−2 m, find the acceleration of the block.

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Doris 5 years 2021-07-24T05:50:20+00:00 1 Answers 21 views 0

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  1. thienhuong
    0
    2021-07-24T05:51:35+00:00
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    Answer:

    The magnitude of the  acceleration  is  a = 10.5 m/s^2

    Explanation:

    From the question we are told that

       The mass of the block is m = 0.16 \ kg

       The spring constant is k = 140\ N/m

        At first displacement is  x = + 0.080 m

     At  second displacement is  x = -1.2 *10^{-2} \ m

    The acceleration at second displacement is mathematically represented as

                    a = \frac{k}{m} * x

                    a = - \frac{140}{0.16} * 1.2 *10^{-2}

                     a = - 10.5 m/s^2

    Therefore the magnitude of the acceleration is

                     a = 10.5 m/s^2

    And the direction is  in the negative x-axis

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )