When a 2.50-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.13 cm.

Question

a)What is the force constant of the spring?

b) If the 2.50-kg object is removed, how far will the spring stretch if a 1.25-kg block is hung on it?

Answer 1

k= force/x= mg/.0313
b) x= force/k=mg/k it better be half as far.