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Hooke's Law tells use that a spring will deform linearly as

y = m x + b

were y is the length of the spring under loading

m is the spring constant (slope of a line)

b is the initial length of the spring

first find the slope m:

m = (y₂ - y₁) / (x₂ - x₁)

m = (40 - 30) / (50 - 25) = 10 / 25 = 2/5

now use the point slope for the line equation and work to the slope intercept form

y - y₁ = m(x - x₁)

y - 30 = 2/5 (x - 25)

y - 30 = 2/5x-10

y = 2/5x + 20

the spring is initially 20 cm long.

~I got this from another internet source, so i have no idea if this is right. but I'm pretty sure it is

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Q: A load of 25 kg stretches a spring to a total length of 30 cm and a load of 50 kg stretches it to a total length of 40 cm so what is the natural length of the spring and how do you find it?
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