physics

When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.30 cm.
(a) What is the force constant of the spring?
(b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it?
(c) How much work must an external agent do to stretch the same spring 8.40 cm from its unstretched position?

asked by alexandra
  1. 2.30 cm = .0230 meters
    2.60 kg * 9.8 m/s^2 = 25.5 Newtons
    so
    k = 25.5/.0230 = 1108 N/m

    1.30*9.8 = 1108 x
    x = .0115 m (half as far of course)

    Energy stored = work done = (1/2) k x^2
    = .5*1108*(.0840)^2

    posted by Damon

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