Question

A spring stretches 1.68cm vertically when a 2.50kg object is suspended from it.Find the distance (in cm) the spring stretches if you replace the 2.50kg object with a 4.40kg object. Assume that the spring obeys the Hooks Law

Answer 1

Answer:

2.96 cm

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extension(d)

 F = k × d

Force is the weight of the object, F = W = mg

So we get, mg = kd ⇒ m ∝ d

                                   2.5 ∝ 1.68  --------------(1)

                                   4.4 ∝ d'      --------------(2)

From (1) & (2),     4.4/2.5 = d'/1.68

                                   d' = 2.96 cm ⇒ the required extension.