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12 February, 18:01

A 10-kg mass, hung by an ideal spring, causes the spring to stretch 2.0 cm. what is the spring constant (force constant) for this spring?

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  1. 12 February, 19:53
    0
    From Hooke's law the spring constant k is given by F = ke; then k = F/e. But F = Ma = Mg = 10 * 9.8 = 98N where g = 9.8ms^ (-2) Extension e = 2cm = 0.02m. So k = 98/0.02 = 4900 N/m.
  2. 12 February, 21:17
    0
    The computation for this problem is calculated by:F = kx

    F = (10) (9.8)

    x =.02 meters after converting from cm to m

    98 = (k) (.02) So, k=4900 N m.

    Then back to cm by dividing by 100 which is now 49 N cm would be the spring constant for this problem.
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