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Sagot :
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.
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