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9 October, 09:11

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

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Answers (2)
  1. 9 October, 09:40
    0
    The work to stretch a spring from its rest position is

    (1/2) (spring constant) (distance of the stretch) ²

    E = 1/2 k x².

    You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

    1700 joules = 1/2 k (3m) ²

    1 joule = 1 newton-meter

    1700 N-m = 1/2 k (3m) ²

    Multiply each side by 2: 3400 N-m = k · 9m²

    Divide each side by 9m² k = 3400 N-m / 9m²

    = (377 and 7/9) newton per meter
  2. 9 October, 09:41
    0
    W = (1/2) kx^2, where W=work required to stretch the spring from its equilibrium position

    k = spring constant

    x = displacement = 5 - 2 = 3m

    1800 J = (1/2) (k) (3) ^2

    by solving it we get

    k = 400 answer.
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