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20 September, 10:56

How much force is required to pull a spring 3.0 cm from its equilibrium position if the spring constant is 3.7 x 103 n/m?

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Answers (2)
  1. 20 September, 11:02
    0
    From Hooke's Law

    We have F = ke where k is the spring constant and e is the extension.

    So e = 3cm = 0.03m and k is 3.7 * 10^3

    So it follows that F = 3.7 * 10^3 * 0.03 = 3.7 * 10^3 * 3 * 10^-2

    So F = 11.1 * 10^ (3-2) = 11.1 * 10 = 111N
  2. 20 September, 11:02
    0
    Hooke's law states that for a helical spring the extension is directly proportional to the force applied provided the elastic limit is not exceeded.

    Therefore; F = ke, where k is the spring constant, F is the force and e is the extension.

    k = 2700 N/m and e = 3 cm or 0.03 M

    therefore, F = 2700 * 0.03

    = 81 N

    Thus, the force required will be 81 N
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