A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a mov

Question

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by 0.065 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 11.3 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.048 m relative to its unstrained length

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Khoii Minh 3 years 2021-09-01T12:33:07+00:00 1 Answers 21 views 0

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  1. Cherry
    0
    2021-09-01T12:34:25+00:00
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    Answer:

    Explanation:

    Given that:

    angular frequency = 11.3 rad/s

    Spring constant (k) = = \omega^2 \times m

    k = (11.3)² m

    k = 127.7 m

    where;

    x_1 = 0.065 m

    x_2  = 0.048 m

    According to the conservation of energies;

    E_1=E_2

    ∴

    \Big(\dfrac{1}{2} \Big) kx_1^2 =\Big(\dfrac{1}{2} \Big) mv_2^2 + \Big(\dfrac{1}{2} \Big) kx_2^2

    kx_1^2 = mv_2^2 + kx_2^2

    (127.7 \ m) \times 0.065^2 = v_2^2 + (127.7 \ m) \times 0.048^2

    0.5395325 = v_2^2 +0.2942208 \\ \\ 0.5395325 - 0.2942208 = v_2^2 \\ \\ v_2^2 = 0.2453117 \\ \\ v_2 = \sqrt{0.2453117} \\ \\ \mathbf{ v_2 \simeq0.50 \ m/s}

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