A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mech

Question

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 0.12 J. What is the greatest extension of the spring from its equilibrium length? What is the maximum speed of the block?

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RobertKer 5 years 2021-08-02T11:30:55+00:00 1 Answers 787 views 1

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  1. Sigrid
    1
    2021-08-02T11:32:38+00:00
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    Answer:

    The greatest extension of the spring is \bf{0.055~m} and the maximum speed of the block is \bf{0.695~m/s}.

    Explanation:

    Given:

    The mass of the block is, m = 0.50~kg

    The spring constant of the spring is, k = 80~N/m

    The mechanical energy of the block is, E = 0.12~J

    When a particle is oscillating in a simple harmonic way, its total energy is given by

    E = \dfrac{1}{2}m\omega^{2}a^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

    where \omega is the angular velocity of the mass and a is the amplitude of its motion.

    The relation between angular velocity and spring constant is given by

    \omega = \sqrt{\dfrac{k}{m}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

    Substituting equation (2) in equation (1), we have

    ~~~~~~&& E = \dfrac{1}{2}ka^{2}\\&or,& a = \sqrt{\dfrac{2E}{k}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

    Substituting 0.12~J for E and 80~N/m for k in equation (3), we can write

    a &=& \sqrt{\dfrac{2(0.12~J)}{80~N/m}}\\~~~&=& 0.055~m

    The relation between the maximum velocity and the amplitude is given by

    v_{m} &=& \omega a\\~~~~&=& \sqrt{\dfrac{k}{m}}a~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)

    Substituting 80~N/m for k , 0.50~kg for m and 0.055~m for a in equation (4), we have

    v_{m} &=& \sqrt{\dfrac{80~N/m}{0.50~kg}}(0.055~m)\\~~~&=& 0.695~m/s

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