The stiffness of a particular spring is 730 N/m. One end of the spring is attached to a wall. When you pull on the other end of the spring a

Question

The stiffness of a particular spring is 730 N/m. One end of the spring is attached to a wall. When you pull on the other end of the spring and hold it stretched with a steady force of 133 N, the spring elongates to a total length of 79 cm. What was the relaxed length of the spring?

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Ben Gia 5 months 2021-09-05T11:52:20+00:00 1 Answers 3 views 0

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    2021-09-05T11:53:33+00:00

    Answer:

    The relaxed length of spring was 61 cm

    Explanation:

    To solve this problem, we will use Hooke’s Law. Hooke’s law is stated as:

    F = k Δx

    F = k(x – x₀)

    where,

    F = Force applied to the spring = 133 N

    k = Spring Constant = 730 N/m

    x = Total Length after elongation = 79 cm = 0.79 m

    x₀ = Initial or Relaxed Length of Spring = ?

    Therefore,

    133 N = (730 N/m)(0.79 m – x₀)

    (0.79 m – x₀) = (133 N)/(730 N/m)

    (0.79 m – x₀) = 0.18 m

    x₀ = 0.79 m – 0.18 m

    x₀ = 0.61 m = 61 cm

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