If a mass of 0.55kg attached to a vertical spring stretches the spring 32.0 cm from its original equilibrium, what is the spring constant?

Question

If a mass of 0.55kg attached to a vertical spring stretches the spring 32.0 cm from its original equilibrium, what is the spring constant?

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Euphemia 4 months 2021-09-05T13:49:33+00:00 1 Answers 6 views 0

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    2021-09-05T13:51:22+00:00

    G weight of mass m=0.55kg.

    G=m×g=k×dx, whre dx=32cm is the spring stretch, computed as the difference between the actual length and original length of the spring.

    dx=32cm=0.32m

    g=9.81m/s²

    Thus we have:

    m×g=k×dx

    k=m×g/dx=0.55×9.81/0.32=16.86N/m

    Answer: k=16.86N/

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