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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?
Zywrx
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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∙ 12y ago
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