Posted by alexandra on .
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?
2.30 cm = .0230 meters
2.60 kg * 9.8 m/s^2 = 25.5 Newtons
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