Question

When a 2.40-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.92 cm.(a) What is the force constant of the spring

Answer 1

805.48N/m

Step-by-step explanation:

According to Hookes law

F = Ke

F is the force = mg

F = 2.4×9.8 = 23.52N

e is the extension = 2.92cm = 0.0292m

Force constant K = F/e

K = 23.52/0.0292

K = 805.48N/m

Hence the force constant of the spring is 805.48N/m