A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an

Question

A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back toward equilibrium. What is the amplitude A, of the motion? Enter your answer in units of cm but do not include units

in progress 0
Kim Cúc 3 months 2021-08-02T15:03:31+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-08-02T15:05:11+00:00

    Answer:

    7.1 cm.

    Explanation:

    Given,

    mass of the block, m = 5 Kg

    Spring constant, k = 2000 N/m

    moved position, x = 5 cm

    initial speed,v = 1 m/s

    Amplitude,A = ?

    We know,

    \omega = \sqrt{\dfrac{k}{m}}

    \omega = \sqrt{\dfrac{2000}{5}}

    \omega = 20\ rad/s

    Relation between velocity, Amplitude is given by

    v = \omega \sqrt{A^2-x^2}

    1 = 20\times \sqrt{A^2-0.05^2}

    0.05^2+0.05^2= A^2

    A =7.1\ cm

    Amplitude of the motion is equal to 7.1 cm.

Leave an answer

Browse

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