Homework Help: physics

Posted by alexandra on Wednesday, April 21, 2010 at 2:49pm.

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?

physics - Damon, Wednesday, April 21, 2010 at 3:02pm
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

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