When a 0.30 kg mass is suspended from a massless spring, the spring stretches a distance of 2.0 cm. Let 2.0 cm be the rest position for the

Question

When a 0.30 kg mass is suspended from a massless spring, the spring stretches a distance of 2.0 cm. Let 2.0 cm be the rest position for the mass-spring system. The mass is then pulled down an additional distance of 1.5 cm and released.Calculate the total mechanical energy of the system in SI Units.
Spring constant can be found using Hooke’s Law.

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RI SƠ 6 months 2021-07-21T01:28:30+00:00 1 Answers 25 views 0

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    2021-07-21T01:29:37+00:00

    Answer:

    E = 17 mJ

    Explanation:

    kx = mg

    k = 0.30(9.8)/0.02 = 147 N/m

    E = ½kA²

    E = ½(147)0.015²

    E = 0.0165375 J

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