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A 0.100 kg ball hangs from a spring of negligible mass. When the ball is hung on the spring and is at rest, the spring is stretched by 0.200 m from its equilibrium to a new equilibrium, in which the tension in the spring balances the weight of the ball. What is the spring constant of this spring

Respuesta :

Answer:

4.9 N/m

Explanation:

m = 0.1 kg

x = 0.2 m

Let k be the spring constant.

As teh weight is balanced by the tension in the spring

Weight = force = mass x acceleration due to gravity

F = m x g = 0.1 x 9.8 = 0.98 N

This force si balanced by spring force

F = k x

Where, k be the spring constant and x be the displacement in spring

0.98 = k (0.2)

k = 4.9 N/m