A spring with k = 83 N????m hangs vertically next to a ruler. The end of the spring is next to the 15-cm mark on the ruler. If a 2.5-kg mass

Question

A spring with k = 83 N????m hangs vertically next to a ruler. The end of the spring is next to the 15-cm mark on the ruler. If a 2.5-kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with the ruler marks when the mass is at its lowest position?

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Doris 1 year 2021-08-29T20:45:39+00:00 1 Answers 34 views 0

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    2021-08-29T20:47:13+00:00

    Answer:

    [tex]0.4452m[/tex]

    Explanation:

    Spring constant is given as:

    [tex]k=F/x[/tex]

    But F is mass times velocity(gravitational acceleration[tex]([/tex]=[tex]9.8m/s^2)[/tex]

    83N/m=[tex]2.5kg([/tex][tex]9.8m/s^2)/x[/tex]

    [tex]x=0.2952m[/tex]

    Therefore the ruler reading will be [tex]x+0.15m[/tex]. Hence the ruler reading is 0.4451m

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