When a 3.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.07 cm.

Question

(a) What is the force constant of the spring?
N/m

(b) If the 3.00-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?
cm

(c) How much work must an external agent do to stretch the same spring 9.10 cm from its unstretched position?
J

Answer 1

k=force/stretch= 3g/.0307 in N/m

b. x=force/k=1.5/k I bet it is half of 3.07

c. work= 1/2 k x^2 watch units, x has to be in m.