Question

g suppose a spring with spring constant of 50 N/m is hanging from the ceiling. You hang 2.0 kg mass from the spring. How far is the spring stretched

Answer 1

The extension of the spring is 0.392 m.

Step-by-step explanation:

Given;

spring constant, k = 50 N/m

mass attached to the spring, m = 2.0 kg

let the extension of the spring = x

The extension of the spring is calculated by applying Hook's law;

F = kx

mg = kx

`x = (mg)/(k) \\\\x = (2 * 9.8)/(50) \\\\x = 0.392 \ m`

Therefore, the extension of the spring is 0.392 m.