Posted by alexandra on .
When a 2.60kg 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 2.60kg object is removed, how far will the spring stretch if a 1.30kg 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,
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