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

Answer:

The extension of the spring is 0.392 m.

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

$x = \frac{mg}{k} \\\\x = \frac{2 \times 9.8}{50} \\\\x = 0.392 \ m$

Therefore, the extension of the spring is 0.392 m.