A body weighing 100 g (rope length l = 25 cm) was suspended to a rubber cord, and the rigidity of the cord

A body weighing 100 g (rope length l = 25 cm) was suspended to a rubber cord, and the rigidity of the cord was k = 25N / M. Determine the initial length of a non-deformed body.

m = 100 g = 0.1 kg.

k = 25 N / m.

l = 25 cm = 0.25 m.

g = 10 m / s2.

l0 -?

The condition for finding the load at rest, according to Newton’s 1 law, is the compensation of all the forces that act on it. The load is acted upon by the force of gravity m * g and and the force of elasticity of the rubber band Fcont, which are oppositely directed.

m * g – Fcont = 0 – the condition of the balance of the load on the rubber band.

The force of elasticity Fel, which arises at small deformations, can be expressed by Hooke’s law: Fel = k * x, where k is the rigidity of the rope, x is the change in the length of the rope.

The change in the length of the bundle is expressed by the formula: x = l – l0, where l0 is the length of the non-deformed bundle, l is the length of the stretched bundle.

k * (l – l0) = m * g.

l0 = l – m * g / k

l0 = 0.25 m – 0.1 kg * 10 m / s2 / 25 N / m = 0.21 m.

Answer: a non-deformed rubber band had a length of l0 = 0.21 m.