Question

A spring is attached to the ceiling, and when a brick is suspended from the spring at rest, it stretches by 50 cm. If instead two identical copies of the same spring are placed side-by-side and both attached to the brick and the ceiling, how much will each of the springs stretch by

Answer 1

This question involves the concepts of the parallel combination of springs, Hooke's Law and equivalent spring constant.

Each spring will stretch by "25 cm".

For static condition of spring:

Weight of Brick = Extension Force

using Hooke's Law:

$mg = kx$

Consider the following data for a single spring:

K₁ = spring constant of single spring = K

x₁ = stretch in single spring = 50 cm

m = mass of brick

g = acceleration due to gravity

Therefore,

$mg = k_1x_1\\mg = K(50\ cm) ---- eqn(1)$

Now, for the second scenario, where two identical copies of the single spring are placed side by side in a parallel combination:

K₂ = Equivalent spring constant of the parallel combination of springs

K₂ = K + K = 2K

x₂ = stretch in each spring in the parallel combination of springs = ?

m = mass of brick

g = acceleration due to gravity

Therefore,

$mg = k_1x_1\\mg = 2K(x_2) ---- eqn(2)$

dividing eqn (1) and eqn(2), we get:

$1 = \frac{50\ cm}{2x_2}$

x₂ = 25 cm

Learn more about Hooke's Law here:

brainly.com/question/13348278?referrer=searchResults

The attached picture illustrates Hooke's Law.