When a 2.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.88 cm. (a) What is the force constant of the spring?
+36916 F= -kx where k = spring constant x = stretch distance
F=ma = 2 kg x 9.8 m/sec^2 = 19.6 N
19.6 = - k (.0288 m) k = - 680 N/m
