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A ball of mass 500 g hangs from a spring whose stiffness is 125 newtons per meter. A string is attached to the ball and you are pulling the string to the right, so that the ball hangs motionless 8cm from it's original position. In this situation the spring is stretched, and its length is 15 cm.


What would be the relaxed length of the spring, if it were detached from the ball and laid on a table?

Answer :

Hamzaahmedsgo

This question involves the concept of Hooke's Law.

The relaxed length of the spring would be "3 cm".

First, we will find out the stretched length of spring due to the weight of the ball.

Stretched Length = Final Length - Stretch from the original position due to the pull force

Stretched Length = 15 cm - 8 cm = 7 cm

Now, we will use Hooke's Law, to find out the relaxed length:

[tex]F = k\Delta x[/tex]

where,

F = Weight of the ball = mg = (0.5 kg)(9.81 m/s²) = 4.91 N

k = spring constant = 125 N/m

Δx = change in length due to weight of ball

Δx = stretched length - relaxed length = 7 cm - relaxed length

Therefore,

[tex]4.91\ N = (125\ N/m)(7\ cm -\ relaxed\ length)\\\\relaxed\ length = 7\ cm - \frac{4.91\ N}{125\ N/m}\\relaxed\ length = 7\ cm - 0.04\ m = 7\ cm - 4\ cm[/tex]

relaxed length = 3 cm

Learn more about Hooke's Law here:

https://brainly.com/question/13348278?referrer=searchResults

The attached picture illustrates Hooke's Law.

View image Hamzaahmeds

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